给你一个整数数组 nums ，数组中的元素 互不相同 。返回该数组所有可能的子集（幂集）。

解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。

示例 1：

输入：nums = [1,2,3]
输出：[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
示例 2：

输入：nums = [0]
输出：[[],[0]]

提示：

1 <= nums.length <= 10
-10 <= nums[i] <= 10
nums 中的所有元素 互不相同

给定一个不含重复数字的数组 nums ，返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。

示例 1：

输入：nums = [1,2,3]
输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
示例 2：

输入：nums = [0,1]
输出：[[0,1],[1,0]]
示例 3：

输入：nums = [1]
输出：[[1]]

数字 n 代表生成括号的对数，请你设计一个函数，用于能够生成所有可能的并且 有效的 括号组合。

示例 1：

输入：n = 3
输出：[“((()))”,“(()())”,“(())()”,“()(())”,“()()()”]
示例 2：

输入：n = 1
输出：[“()”]

提示：

1 <= n <= 8

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例 1：

输入：digits = “23”
输出：[“ad”,“ae”,“af”,“bd”,“be”,“bf”,“cd”,“ce”,“cf”]
示例 2：

输入：digits = “”
输出：[]
示例 3：

输入：digits = “2”
输出：[“a”,“b”,“c”]

提示：

0 <= digits.length <= 4
digits[i] 是范围 [‘2’, ‘9’] 的一个数字。

峰值元素是指其值严格大于左右相邻值的元素。

给你一个整数数组 nums，找到峰值元素并返回其索引。数组可能包含多个峰值，在这种情况下，返回 任何一个峰值 所在位置即可。

你可以假设 nums[-1] = nums[n] = -∞ 。

你必须实现时间复杂度为 O(log n) 的算法来解决此问题。

示例 1：

输入：nums = [1,2,3,1]
输出：2
解释：3 是峰值元素，你的函数应该返回其索引 2。
示例 2：

输入：nums = [1,2,1,3,5,6,4]
输出：1 或 5
解释：你的函数可以返回索引 1，其峰值元素为 2；
  或者返回索引 5， 其峰值元素为 6。

提示：

1 <= nums.length <= 1000
-231 <= nums[i] <= 231 - 1
对于所有有效的 i 都有 nums[i] != nums[i + 1]

整数数组 nums 按升序排列，数组中的值 互不相同 。

在传递给函数之前，nums 在预先未知的某个下标 k（0 <= k < nums.length）上进行了 旋转，使数组变为 [nums[k], nums[k+1], …, nums[n-1], nums[0], nums[1], …, nums[k-1]]（下标 从 0 开始 计数）。例如， [0,1,2,4,5,6,7] 在下标 3 处经旋转后可能变为 [4,5,6,7,0,1,2] 。

给你 旋转后 的数组 nums 和一个整数 target ，如果 nums 中存在这个目标值 target ，则返回它的下标，否则返回 -1 。

你必须设计一个时间复杂度为 O(log n) 的算法解决此问题。

示例 1：

输入：nums = [4,5,6,7,0,1,2], target = 0
输出：4
示例 2：

输入：nums = [4,5,6,7,0,1,2], target = 3
输出：-1
示例 3：

输入：nums = [1], target = 0
输出：-1

提示：

1 <= nums.length <= 5000
-104 <= nums[i] <= 104
nums 中的每个值都 独一无二
题目数据保证 nums 在预先未知的某个下标上进行了旋转
-104 <= target <= 104

给你一个按照非递减顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。

如果数组中不存在目标值 target，返回 [-1, -1]。

你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。

示例 1：

输入：nums = [5,7,7,8,8,10], target = 8
输出：[3,4]
示例 2：

输入：nums = [5,7,7,8,8,10], target = 6
输出：[-1,-1]
示例 3：

输入：nums = [], target = 0
输出：[-1,-1]

提示：

0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums 是一个非递减数组
-109 <= target <= 109

给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。

请必须使用时间复杂度为 O(log n) 的算法。

示例 1:

输入: nums = [1,3,5,6], target = 5
输出: 2
示例 2:

输入: nums = [1,3,5,6], target = 2
输出: 1
示例 3:

输入: nums = [1,3,5,6], target = 7
输出: 4

给你一个非负整数 x ，计算并返回 x 的 算术平方根 。

由于返回类型是整数，结果只保留 整数部分 ，小数部分将被 舍去 。

注意：不允许使用任何内置指数函数和算符，例如 pow(x, 0.5) 或者 x ** 0.5 。

示例 1：

输入：x = 4
输出：2
示例 2：

输入：x = 8
输出：2
解释：8 的算术平方根是 2.82842…, 由于返回类型是整数，小数部分将被舍去。

提示：

0 <= x <= 231 - 1

Time Complexity: O(logn) | due to binary search using while loop.
Space Complexity: O(1) | as only 4 variables are initialized at the beginning. Which is constant irrespective of given input.

long long s=0, e=x, ans, mid;   //long long due to some of test cases overflows integer limit.
        while(s<=e){             
            mid=(s+e)/2;
            if(mid*mid==x) return mid;     //if the 'mid' value ever gives the result, we simply return it.
            else if(mid*mid<x){             
                s=mid+1;        //if 'mid' value encounterted gives lower result, we simply discard all the values lower than mid.
                ans=mid;        //an extra pointer 'ans' is maintained to keep track of only lowest 'mid' value. 
            }
            else e=mid-1;       //if 'mid' value encountered gives greater result, we simply discard all the values greater than mid. 
        }
        return ans;   

给定一个包含 n + 1 个整数的数组 nums ，其数字都在 [1, n] 范围内（包括 1 和 n），可知至少存在一个重复的整数。

假设 nums 只有 一个重复的整数 ，返回 这个重复的数 。

你设计的解决方案必须 不修改 数组 nums 且只用常量级 O(1) 的额外空间。

示例 1：

输入：nums = [1,3,4,2,2]
输出：2
示例 2：

输入：nums = [3,1,3,4,2]
输出：3

提示：

1 <= n <= 105
nums.length == n + 1
1 <= nums[i] <= n
nums 中 只有一个整数 出现 两次或多次 ，其余整数均只出现 一次

进阶：

如何证明 nums 中至少存在一个重复的数字?
你可以设计一个线性级时间复杂度 O(n) 的解决方案吗？

Overview
Finding the Duplicate Number is a classic problem, and as such there are many different ways to approach it; a total of 7 approaches are presented here. The first 4 approaches involve rearranging or modifying elements of the array, and hence do not meet the constraints specified in the problem statement. However, they are included here since they are more feasible to come up with as the first approach in an interview setting. Since each approach is independent of the other approaches, they can be read in any order.

Proof
Proving that at least one duplicate must exist in numsnums is an application of the pigeonhole principle. Here, each number in numsnums is a "pigeon" and each distinct number that can appear in numsnums is a "pigeonhole." Because there are n+1n+1 numbers and nn distinct possible numbers, the pigeonhole principle implies that if you were to put each of the n + 1n+1 pigeons into nn pigeonholes, at least one of the pigeonholes would have 2 or more pigeons.

Approach 1: Sort
Note: This approach modifies individual elements and does not use constant space, and hence does not meet the problem constraints. However, it utilizes a fundamental concept that can help solve similar problems.

Intuition

In an unsorted array, duplicate elements may be scattered across the array. However, in a sorted array, duplicate numbers will be next to each other.

Algorithm

Sort the input array (numsnums).

Iterate through the array, comparing the current number to the previous number (i.e. compare nums[i]nums[i] to nums[i - 1]nums[i−1] where i > 0i>0).

Return the first number that is equal to its predecessor.
Complexity Analysis

Time Complexity: O(n \log n)O(nlogn)

Sorting takes O(n \log n)O(nlogn) time. This is followed by a linear scan, resulting in a total of O(n \log n)O(nlogn) + O(n)O(n) = O(n \log n)O(nlogn) time.

Space Complexity: O(\log n)O(logn) or O(n)O(n)

The space complexity of the sorting algorithm depends on the implementation of each programming language:

In Java, Arrays.sort() for primitives is implemented using a variant of the Quick Sort algorithm, which has a space complexity of O(\log n)O(logn)
In C++, the sort() function provided by STL uses a hybrid of Quick Sort, Heap Sort and Insertion Sort, with a worst case space complexity of O(\log n)O(logn)
In Python, the sort() function is implemented using the Timsort algorithm, which has a worst-case space complexity of O(n)O(n)
Approach 2: Set
Note: This approach does not use constant space, and hence does not meet the problem constraints. However, it utilizes a fundamental concept that can help solve similar problems.

Intuition

As we traverse the array, we need a way to "remember" values that we've seen. If we come across a number that we've seen before, we've found the duplicate. An efficient way to record the seen values is by adding each number to a set as we iterate over the numsnums array.

Algorithm

In order to achieve linear time complexity, we need to be able to insert elements into a data structure and look them up in constant time. A HashSet/unordered_set is well suited for this purpose. Initialize an empty hashset, seenseen.

Iterate over the array and first check if the current element exists in the hashset (seenseen).

If it does exist in the hashset, that number is the duplicate and can be returned right away.
Otherwise, insert the current element into seenseen, move to the next element in the array and repeat step 2.

Complexity Analysis

Time Complexity: O(n)O(n)

HashSet insertions and lookups have amortized constant time complexities. Hence, this algorithm requires linear time, since it consists of a single for loop that iterates over each element, looking up the element and inserting it into the set at most once.

Space Complexity: O(n)O(n)

We use a set that may need to store at most nn elements, leading to a linear space complexity of O(n)O(n).

给你一个整数数组 nums ，请你找出一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。

子数组 是数组中的一个连续部分。

示例 1：

输入：nums = [-2,1,-3,4,-1,2,1,-5,4]
输出：6
解释：连续子数组 [4,-1,2,1] 的和最大，为 6 。
示例 2：

输入：nums = [1]
输出：1
示例 3：

输入：nums = [5,4,-1,7,8]
输出：23

提示：

1 <= nums.length <= 105
-104 <= nums[i] <= 104

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为 “Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为 “Finish” ）。

问总共有多少条不同的路径？

示例 1：

输入：m = 3, n = 7
输出：28
示例 2：

输入：m = 3, n = 2
输出：3
解释：
从左上角开始，总共有 3 条路径可以到达右下角。

1. 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右
3. 向下 -> 向右 -> 向下
   示例 3：

输入：m = 7, n = 3
输出：28
示例 4：

输入：m = 3, n = 3
输出：6

提示：

1 <= m, n <= 100
题目数据保证答案小于等于 2 * 109

Since the robot can only move right and down, when it arrives at a point, it either arrives from left or above. If we use dp[i][j] for the number of unique paths to arrive at the point (i, j), then the state equation is dp[i][j] = dp[i][j - 1] + dp[i - 1][j]. Moreover, we have the base cases dp[0][j] = dp[i][0] = 1 for all valid i and j.

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> dp(m, vector<int>(n, 1));
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
};
The above solution runs in O(m * n) time and costs O(m * n) space. However, you may have noticed that each time when we update dp[i][j], we only need dp[i - 1][j] (at the previous row) and dp[i][j - 1] (at the current row). So we can reduce the memory usage to just two rows (O(n)).

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<int> pre(n, 1), cur(n, 1);
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                cur[j] = pre[j] + cur[j - 1];
            }
            swap(pre, cur);
        }
        return pre[n - 1];
    }
};
Further inspecting the above code, pre[j] is just the cur[j] before the update. So we can further reduce the memory usage to one row.

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<int> cur(n, 1);
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                cur[j] += cur[j - 1];
            }
        }
        return cur[n - 1];
    }
};
Now, you may wonder whether we can further reduce the memory usage to just O(1) space since the above code seems to use only two variables (cur[j] and cur[j - 1]). However, since the whole row cur needs to be updated for m - 1 times (the outer loop) based on old values, all of its values need to be saved and thus O(1)-space is impossible. However, if you are having a DP problem without the outer loop and just the inner one, then it will be possible.

给定一个包含非负整数的 m x n 网格 grid ，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。

说明：每次只能向下或者向右移动一步。

示例 1：

输入：grid = [[1,3,1],[1,5,1],[4,2,1]]
输出：7
解释：因为路径 1→3→1→1→1 的总和最小。
示例 2：

输入：grid = [[1,2,3],[4,5,6]]
输出：12

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 200
0 <= grid[i][j] <= 100

his is a typical DP problem. Suppose the minimum path sum of arriving at point (i, j) is S[i][j], then the state equation is S[i][j] = min(S[i - 1][j], S[i][j - 1]) + grid[i][j].

Well, some boundary conditions need to be handled. The boundary conditions happen on the topmost row (S[i - 1][j] does not exist) and the leftmost column (S[i][j - 1] does not exist). Suppose grid is like [1, 1, 1, 1], then the minimum sum to arrive at each point is simply an accumulation of previous points and the result is [1, 2, 3, 4].

Now we can write down the following (unoptimized) code.
class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size(); 
        vector<vector<int> > sum(m, vector<int>(n, grid[0][0]));
        for (int i = 1; i < m; i++)
            sum[i][0] = sum[i - 1][0] + grid[i][0];
        for (int j = 1; j < n; j++)
            sum[0][j] = sum[0][j - 1] + grid[0][j];
        for (int i = 1; i < m; i++)
            for (int j = 1; j < n; j++)
                sum[i][j]  = min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];
        return sum[m - 1][n - 1];
    }
};
As can be seen, each time when we update sum[i][j], we only need sum[i - 1][j] (at the current column) and sum[i][j - 1] (at the left column). So we need not maintain the full m*n matrix. Maintaining two columns is enough and now we have the following code.

假设你正在爬楼梯。需要 n 阶你才能到达楼顶。

每次你可以爬 1 或 2 个台阶。你有多少种不同的方法可以爬到楼顶呢？

示例 1：

输入：n = 2
输出：2
解释：有两种方法可以爬到楼顶。

1. 1 阶 + 1 阶
2. 2 阶
   示例 2：

输入：n = 3
输出：3
解释：有三种方法可以爬到楼顶。

1. 1 阶 + 1 阶 + 1 阶
2. 1 阶 + 2 阶
3. 2 阶 + 1 阶

提示：

1 <= n <= 45

The problem seems to be a dynamic programming one. Hint: the tag also suggests that!
Here are the steps to get the solution incrementally.

Base cases:
if n <= 0, then the number of ways should be zero.
if n == 1, then there is only way to climb the stair.
if n == 2, then there are two ways to climb the stairs. One solution is one step by another; the other one is two steps at one time.

The key intuition to solve the problem is that given a number of stairs n, if we know the number ways to get to the points [n-1] and [n-2] respectively, denoted as n1 and n2 , then the total ways to get to the point [n] is n1 + n2. Because from the [n-1] point, we can take one single step to reach [n]. And from the [n-2] point, we could take two steps to get there.

The solutions calculated by the above approach are complete and non-redundant. The two solution sets (n1 and n2) cover all the possible cases on how the final step is taken. And there would be NO overlapping among the final solutions constructed from these two solution sets, because they differ in the final step.

Now given the above intuition, one can construct an array where each node stores the solution for each number n. Or if we look at it closer, it is clear that this is basically a fibonacci number, with the starting numbers as 1 and 2, instead of 1 and 1.

一条包含字母 A-Z 的消息通过以下映射进行了 编码 ：

‘A’ -> “1”
‘B’ -> “2”
…
‘Z’ -> “26”
要 解码 已编码的消息，所有数字必须基于上述映射的方法，反向映射回字母（可能有多种方法）。例如，“11106” 可以映射为：

“AAJF” ，将消息分组为 (1 1 10 6)
“KJF” ，将消息分组为 (11 10 6)
注意，消息不能分组为  (1 11 06) ，因为 “06” 不能映射为 “F” ，这是由于 “6” 和 “06” 在映射中并不等价。

给你一个只含数字的 非空 字符串 s ，请计算并返回 解码 方法的 总数 。

题目数据保证答案肯定是一个 32 位 的整数。

示例 1：

输入：s = “12”
输出：2
解释：它可以解码为 “AB”（1 2）或者 “L”（12）。
示例 2：

输入：s = “226”
输出：3
解释：它可以解码为 “BZ” (2 26), “VF” (22 6), 或者 “BBF” (2 2 6) 。
示例 3：

输入：s = “0”
输出：0
解释：没有字符映射到以 0 开头的数字。
含有 0 的有效映射是 ‘J’ -> “10” 和 ‘T’-> “20” 。
由于没有字符，因此没有有效的方法对此进行解码，因为所有数字都需要映射。

提示：

1 <= s.length <= 100
s 只包含数字，并且可能包含前导零。
通过次数235,094提交次数721,617

给你一个整数 n ，求恰由 n 个节点组成且节点值从 1 到 n 互不相同的 二叉搜索树 有多少种？返回满足题意的二叉搜索树的种数。

示例 1：

输入：n = 3
输出：5
示例 2：

输入：n = 1
输出：1

提示：

1 <= n <= 19

给定一个数组 prices ，它的第 i 个元素 prices[i] 表示一支给定股票第 i 天的价格。

你只能选择 某一天 买入这只股票，并选择在 未来的某一个不同的日子 卖出该股票。设计一个算法来计算你所能获取的最大利润。

返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润，返回 0 。

示例 1：

输入：[7,1,5,3,6,4]
输出：5
解释：在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。
注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格；同时，你不能在买入前卖出股票。
示例 2：

输入：prices = [7,6,4,3,1]
输出：0
解释：在这种情况下, 没有交易完成, 所以最大利润为 0。

提示：

1 <= prices.length <= 105
0 <= prices[i] <= 104

I hope now question, approach is absolute clear.

code each line explained : Similar for C++, Java

{
        int lsf = Integer.MAX_VALUE; // least so far
        int op = 0; // overall profit
        int pist = 0; // profit if sold today
        
        for(int i = 0; i < prices.length; i++){
            if(prices[i] < lsf){ // if we found new buy value which is more smaller then previous one
                lsf = prices[i]; // update our least so far
            }
            pist = prices[i] - lsf; // calculating profit if sold today by, Buy - sell
            if(op < pist){ // if pist is more then our previous overall profit
                op = pist; // update overall profit
            }
        }
        return op; // return op 

给你一个整数数组 prices ，其中 prices[i] 表示某支股票第 i 天的价格。

在每一天，你可以决定是否购买和/或出售股票。你在任何时候 最多 只能持有 一股 股票。你也可以先购买，然后在 同一天 出售。

返回 你能获得的 最大 利润 。

示例 1：

输入：prices = [7,1,5,3,6,4]
输出：7
解释：在第 2 天（股票价格 = 1）的时候买入，在第 3 天（股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5 - 1 = 4 。
  随后，在第 4 天（股票价格 = 3）的时候买入，在第 5 天（股票价格 = 6）的时候卖出, 这笔交易所能获得利润 = 6 - 3 = 3 。
总利润为 4 + 3 = 7 。
示例 2：

输入：prices = [1,2,3,4,5]
输出：4
解释：在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5 - 1 = 4 。
  总利润为 4 。
示例 3：

输入：prices = [7,6,4,3,1]
输出：0
解释：在这种情况下, 交易无法获得正利润，所以不参与交易可以获得最大利润，最大利润为 0 。

提示：

1 <= prices.length <= 3 * 104
0 <= prices[i] <= 104

给你一个字符串 s 和一个字符串列表 wordDict 作为字典。请你判断是否可以利用字典中出现的单词拼接出 s 。

注意：不要求字典中出现的单词全部都使用，并且字典中的单词可以重复使用。

示例 1：

输入: s = “leetcode”, wordDict = [“leet”, “code”]
输出: true
解释: 返回 true 因为 “leetcode” 可以由 “leet” 和 “code” 拼接成。
示例 2：

输入: s = “applepenapple”, wordDict = [“apple”, “pen”]
输出: true
解释: 返回 true 因为 “applepenapple” 可以由 “apple” “pen” “apple” 拼接成。
  注意，你可以重复使用字典中的单词。
示例 3：

输入: s = “catsandog”, wordDict = [“cats”, “dog”, “sand”, “and”, “cat”]
输出: false

提示：

1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s 和 wordDict[i] 仅有小写英文字母组成
wordDict 中的所有字符串 互不相同

We use a boolean vector dp[]. dp[i] is set to true if a valid word (word sequence) ends there. The optimization is to look from current position i back and only substring and do dictionary look up in case the preceding position j with dp[j] == true is found.

给你一个整数数组 nums ，请你找出数组中乘积最大的非空连续子数组（该子数组中至少包含一个数字），并返回该子数组所对应的乘积。

测试用例的答案是一个 32-位 整数。

子数组 是数组的连续子序列。

示例 1:

输入: nums = [2,3,-2,4]
输出: 6
解释: 子数组 [2,3] 有最大乘积 6。
示例 2:

输入: nums = [-2,0,-1]
输出: 0
解释: 结果不能为 2, 因为 [-2,-1] 不是子数组。

提示:

1 <= nums.length <= 2 * 104
-10 <= nums[i] <= 10
nums 的任何前缀或后缀的乘积都 保证 是一个 32-位 整数

There can be Mutliple ways to frame the solution once we get the intuition right !! So, what should the intuition be ? Let's discuss that out !

Let's consider array to have no 0s (for the moment)......
So, on what factor does the answer depends now ?? It surely depends on the count of negative numbers in the array !!

There are 2 possibilities - either the count of -ve numbers is even or odd.... --->>>

If the count is even, then obviously we would want to include all of them(in fact the whole array) to maximise the product. As multiplying an even number of -ve numbers would make the result +ve.

If the count is odd, then we would want to exclude one -ve number from our product, so that the product gets maximised. So, now the question is, which -ve number to exclude? Example ---> arr={-2,-3,-1,-4,-5} which number should be excluded ? On observing it , we should get one fact clear, that the number which is going to get ignored is either going to be the first one or the last one.

Note that, we cannot exclude a -ve number that is not the first or the last, because, if we do so, we will need to exclude all(because you are breaking the product at this point) other -ve nums following that -ve number and then that needn't result in the maximum product.
Having said all that, now the question is whether to exclude the first -ve num or the last -ve num in the array. We can only know the answer by trying both.
So, firstly we will take the product from the beginning of the array and we will include the first -ve number and will leave out the last one !!
And will do the vice-versa for checking the other scenario !!
So , in that example we would leave the first -ve number... (-2 and then total_product will be product of rest of the numbers in array) or we would leave the last number...(-5) ... And maximum of those 2 cases will be the answer !!
Now, what if array has zeroes? Well, it changes nothing much to be honest, we can consider the part on both the side of 0 as the subarrays and the maximum product that way will be the max(subarray1_ans, subarray2_ans) .... And how to mark the division point ? How do we seperate the subarrays????...
Thats pretty simple and we have done it in kadane's algo, just make the curr_ongoing_prod=1 !! And maintain one maxm_prod variable seperately ....

Example -->>> arr={-2,1,4,5,0,-3,4,6,1,-2} .... so we can consider subarray1={-2,1,4,5} and subarray2={-3,4,6,-2} and then the max_ans(subarray1,subarray2) will be our answer !!

Let's have a look on our code now ....

你是一个专业的小偷，计划偷窃沿街的房屋。每间房内都藏有一定的现金，影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统，如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。

给定一个代表每个房屋存放金额的非负整数数组，计算你 不触动警报装置的情况下 ，一夜之内能够偷窃到的最高金额。

示例 1：

输入：[1,2,3,1]
输出：4
解释：偷窃 1 号房屋 (金额 = 1) ，然后偷窃 3 号房屋 (金额 = 3)。
  偷窃到的最高金额 = 1 + 3 = 4 。
示例 2：

输入：[2,7,9,3,1]
输出：12
解释：偷窃 1 号房屋 (金额 = 2), 偷窃 3 号房屋 (金额 = 9)，接着偷窃 5 号房屋 (金额 = 1)。
  偷窃到的最高金额 = 2 + 9 + 1 = 12 。

提示：

1 <= nums.length <= 100
0 <= nums[i] <= 400

This is a classic 1D-DP problem where at every step we have a choice to make ...
So the first and foremost thing in any DP problem is to find the reccurence relation !!
At every ith house robber has 2 options: a) rob current house i. b) don't rob current house.

In case he is robbing the (i)th house, the money he can get till the i-th house == money robbed till (i-2)th house + money robbed at (i)th house....let's say total money robbed in this case equals to X.
In case he is not robbing, money robbed till i-th house==money robbed till (i-1)th house...lets say total money robbed in this case equals to Y.
So , the maxm money he gets till i-th house is the max(X,Y).
Example of case (a) --> nums={2,3,2} ... Here, the robber will rob the house at index-2 as nums[index-2] + nums[index-0] > nums[index-1]
Example of case (b)--> nums={2,7,3} ... here maximum money robbed till index-2 will not be equal to nums[index-2] + nums[index-0]... as nums[index-1] is greater than the sum of money at both those houses ...

We can achieve the desired solution to this problem via mutliple ways, let's start with the simpler ones and then will look forward to optimize the Time and Space Complexities

Simple Recursion
Time Complexcity : O ( 2^n ) Gives us TLE
Space Complexcity : O( 1 )

Approach 1: Brute Force
The simplest approach consists of trying to find out every possible square of 1’s that can be formed from within the matrix. The question now is – how to go for it?

We use a variable to contain the size of the largest square found so far and another variable to store the size of the current, both initialized to 0. Starting from the left uppermost point in the matrix, we search for a 1. No operation needs to be done for a 0. Whenever a 1 is found, we try to find out the largest square that can be formed including that 1. For this, we move diagonally (right and downwards), i.e. we increment the row index and column index temporarily and then check whether all the elements of that row and column are 1 or not. If all the elements happen to be 1, we move diagonally further as previously. If even one element turns out to be 0, we stop this diagonal movement and update the size of the largest square. Now we, continue the traversal of the matrix from the element next to the initial 1 found, till all the elements of the matrix have been traversed.


Complexity Analysis

Time complexity : O((mn)^2）
In worst case, we need to traverse the complete matrix for every 1.
Space complexity : O(1)O(1). No extra space is used.

给你一个整数 n ，返回 和为 n 的完全平方数的最少数量 。

完全平方数 是一个整数，其值等于另一个整数的平方；换句话说，其值等于一个整数自乘的积。例如，1、4、9 和 16 都是完全平方数，而 3 和 11 不是。

示例 1：

输入：n = 12
输出：3
解释：12 = 4 + 4 + 4
示例 2：

输入：n = 13
输出：2
解释：13 = 4 + 9

Idea:
This question is very similar to coin change https://leetcode.com/problems/coin-change/discuss/1104203/C%2B%2B-Super-Simple-and-Short-Dynamic-Programming-Solution.
The only difference is that in coin change we get a vector of coins and here we know that the coins are all the perfect squares.
So our first step will be to construct a "coin" vector.
Then, we do it the same way as coin change.

给你一个整数数组 nums ，找到其中最长严格递增子序列的长度。

子序列 是由数组派生而来的序列，删除（或不删除）数组中的元素而不改变其余元素的顺序。例如，[3,6,2,7] 是数组 [0,3,1,6,2,2,7] 的子序列。

示例 1：

输入：nums = [10,9,2,5,3,7,101,18]
输出：4
解释：最长递增子序列是 [2,3,7,101]，因此长度为 4 。
示例 2：

输入：nums = [0,1,0,3,2,3]
输出：4
示例 3：

输入：nums = [7,7,7,7,7,7,7]
输出：1

提示：

1 <= nums.length <= 2500
-104 <= nums[i] <= 104

进阶：

你能将算法的时间复杂度降低到 O(n log(n)) 吗?

✔️ Solution 1: Dynamic Programming

This is a classic Dynamic Programming problem.
Let dp[i] is the longest increase subsequence of nums[0..i] which has nums[i] as the end element of the subsequence.

Complexity

Time: O(N^2), where N <= 2500 is the number of elements in array nums.
Space: O(N)
✔️ Solution 2: Greedy with Binary Search

Let's construct the idea from following example.
Consider the example nums = [2, 6, 8, 3, 4, 5, 1], let's try to build the increasing subsequences starting with an empty one: sub1 = [].
Let pick the first element, sub1 = [2].
6 is greater than previous number, sub1 = [2, 6]
8 is greater than previous number, sub1 = [2, 6, 8]
3 is less than previous number, we can't extend the subsequence sub1, but we must keep 3 because in the future there may have the longest subsequence start with [2, 3], sub1 = [2, 6, 8], sub2 = [2, 3].
With 4, we can't extend sub1, but we can extend sub2, so sub1 = [2, 6, 8], sub2 = [2, 3, 4].
With 5, we can't extend sub1, but we can extend sub2, so sub1 = [2, 6, 8], sub2 = [2, 3, 4, 5].
With 1, we can't extend neighter sub1 nor sub2, but we need to keep 1, so sub1 = [2, 6, 8], sub2 = [2, 3, 4, 5], sub3 = [1].
Finally, length of longest increase subsequence = len(sub2) = 4.
In the above steps, we need to keep different sub arrays (sub1, sub2..., subk) which causes poor performance. But we notice that we can just keep one sub array, when new number x is not greater than the last element of the subsequence sub, we do binary search to find the smallest element >= x in sub, and replace with number x.
Let's run that example nums = [2, 6, 8, 3, 4, 5, 1] again:
Let pick the first element, sub = [2].
6 is greater than previous number, sub = [2, 6]
8 is greater than previous number, sub = [2, 6, 8]
3 is less than previous number, so we can't extend the subsequence sub. We need to find the smallest number >= 3 in sub, it's 6. Then we overwrite it, now sub = [2, 3, 8].
4 is less than previous number, so we can't extend the subsequence sub. We overwrite 8 by 4, so sub = [2, 3, 4].
5 is greater than previous number, sub = [2, 3, 4, 5].
1 is less than previous number, so we can't extend the subsequence sub. We overwrite 2 by 1, so sub = [1, 3, 4, 5].
Finally, length of longest increase subsequence = len(sub) = 4.

Complexity

Time: O(N * logN), where N <= 2500 is the number of elements in array nums.
Space: O(N), we can achieve O(1) in space by overwriting values of sub into original nums array.

Complexity:

Time: O(N * logN)
Space: O(N)

超级丑数 是一个正整数，并满足其所有质因数都出现在质数数组 primes 中。

给你一个整数 n 和一个整数数组 primes ，返回第 n 个 超级丑数 。

题目数据保证第 n 个 超级丑数 在 32-bit 带符号整数范围内。

示例 1：

输入：n = 12, primes = [2,7,13,19]
输出：32
解释：给定长度为 4 的质数数组 primes = [2,7,13,19]，前 12 个超级丑数序列为：[1,2,4,7,8,13,14,16,19,26,28,32] 。
示例 2：

输入：n = 1, primes = [2,3,5]
输出：1
解释：1 不含质因数，因此它的所有质因数都在质数数组 primes = [2,3,5] 中。
 
提示：

1 <= n <= 105
1 <= primes.length <= 100
2 <= primes[i] <= 1000
题目数据 保证 primes[i] 是一个质数
primes 中的所有值都 互不相同 ，且按 递增顺序 排列

It is actually like how we merge k sorted list:

ugly number                       k sorted list
    1                            2     7    13   19     1 * [2,7,13,19]
    |                            |     |    |    |
    2                            4     14   26   38     2 * [2,7,13,19]
    |                            |     |    |    |
    4                            8     28   52   76     4 * [2,7,13,19]
    |                            |     |    |    |              
    7                            14    49   91   133    7 * [2,7,13,19]
    |                            |     |    |    |
    8                            16    56   ...   ...   8 * [2,7,13,19]
    |                            |     |    |     |
    .                            .     .     .    .
    .                            .     .     .    .
    .                            .     .     .    .
We can see that each prime number in primes[] form a sorted list, and now our job is to merge them and find the nth minimum.

Here we don't have the next pointer for each node to trace the next potential candidate. But as we can see in the graph, we can make use of the ugly number we have produced so far!

给你一个整数数组 coins ，表示不同面额的硬币；以及一个整数 amount ，表示总金额。

计算并返回可以凑成总金额所需的 最少的硬币个数 。如果没有任何一种硬币组合能组成总金额，返回 -1 。

你可以认为每种硬币的数量是无限的。

示例 1：

输入：coins = [1, 2, 5], amount = 11
输出：3
解释：11 = 5 + 5 + 1
示例 2：

输入：coins = [2], amount = 3
输出：-1
示例 3：

输入：coins = [1], amount = 0
输出：0

提示：

1 <= coins.length <= 12
1 <= coins[i] <= 231 - 1
0 <= amount <= 104

If you carefully observe the below 3 codes. You will see that the DP Memoization is dervied from the Recursion code just by changing 3 lines and the DP Tabulation is derived from the DP Memoization.

Recursion
Time: O(2^n)
Space: O(n)

Writing a recursive function is all about find two things:

The base case: Just calculate the output for the smallest possible input
The choice diagram: For any given input, just see what all choices do we have.




DP Memoization
Time: O(n.m)
Space: O(n.m)

In the above recursive case, we were doing repeated work in the form of subproblems. Hence we store the results of those subproblems in a table to reduce the number of recursive calls.


DP Tabulation
Time: O(n.m)
Space: O(n.m)

We have reached the best conceivable run time for this question but since we have recursive calls in the previous algorithm. It might lead to stackoverflow error in the worst case when recursive calls are a lot. Hence we want to totally emit the notion of recursions. To do that, we simply convert the recursion into iterative code.

The below code is bottom up dynamic programming because we are starting from the first element in the 2D array and filling the DP from this first element till the last element. And eventually, the last cell stores our final result.

Step I -- Think naively

At first glance, the problem exhibits the feature of "optimal substructure": if we want to rob maximum amount of money from current binary tree (rooted at root), we surely hope that we can do the same to its left and right subtrees.

So going along this line, let's define the function rob(root) which will return the maximum amount of money that we can rob for the binary tree rooted at root; the key now is to construct the solution to the original problem from solutions to its subproblems, i.e., how to get rob(root) from rob(root.left), rob(root.right), ... etc.

Apparently the analyses above suggest a recursive solution. And for recursion, it's always worthwhile figuring out the following two properties:

Termination condition: when do we know the answer to rob(root) without any calculation? Of course when the tree is empty ---- we've got nothing to rob so the amount of money is zero.

Recurrence relation: i.e., how to get rob(root) from rob(root.left), rob(root.right), ... etc. From the point of view of the tree root, there are only two scenarios at the end: root is robbed or is not. If it is, due to the constraint that "we cannot rob any two directly-linked houses", the next level of subtrees that are available would be the four "grandchild-subtrees" (root.left.left, root.left.right, root.right.left, root.right.right). However if root is not robbed, the next level of available subtrees would just be the two "child-subtrees" (root.left, root.right). We only need to choose the scenario which yields the larger amount of money.


However the solution runs very slowly (1186 ms) and barely got accepted (the time complexity turns out to be exponential, see my comments below).

Step II -- Think one step further

In step I, we only considered the aspect of "optimal substructure", but think little about the possibilities of overlapping of the subproblems. For example, to obtain rob(root), we need rob(root.left), rob(root.right), rob(root.left.left), rob(root.left.right), rob(root.right.left), rob(root.right.right); but to get rob(root.left), we also need rob(root.left.left), rob(root.left.right), similarly for rob(root.right). The naive solution above computed these subproblems repeatedly, which resulted in bad time performance. Now if you recall the two conditions for dynamic programming (DP): "optimal substructure" + "overlapping of subproblems", we actually have a DP problem. A naive way to implement DP here is to use a hash map to record the results for visited subtrees.



Step III -- Think one step back

In step I, we defined our problem as rob(root), which will yield the maximum amount of money that can be robbed of the binary tree rooted at root. This leads to the DP problem summarized in step II.

Now let's take one step back and ask why we have overlapping subproblems. If you trace all the way back to the beginning, you'll find the answer lies in the way how we have defined rob(root). As I mentioned, for each tree root, there are two scenarios: it is robbed or is not. rob(root) does not distinguish between these two cases, so "information is lost as the recursion goes deeper and deeper", which results in repeated subproblems.

If we were able to maintain the information about the two scenarios for each tree root, let's see how it plays out. Redefine rob(root) as a new function which will return an array of two elements, the first element of which denotes the maximum amount of money that can be robbed if root is not robbed, while the second element signifies the maximum amount of money robbed if it is robbed.

Let's relate rob(root) to rob(root.left) and rob(root.right)..., etc. For the 1st element of rob(root), we only need to sum up the larger elements of rob(root.left) and rob(root.right), respectively, since root is not robbed and we are free to rob its left and right subtrees. For the 2nd element of rob(root), however, we only need to add up the 1st elements of rob(root.left) and rob(root.right), respectively, plus the value robbed from root itself, since in this case it's guaranteed that we cannot rob the nodes of root.left and root.right.

As you can see, by keeping track of the information of both scenarios, we decoupled the subproblems and the solution essentially boiled down to a greedy one. Here is the program:

背包问题 后面和回溯问题一起搞定

给你一个 只包含正整数 的 非空 数组 nums 。请你判断是否可以将这个数组分割成两个子集，使得两个子集的元素和相等。

示例 1：

输入：nums = [1,5,11,5]
输出：true
解释：数组可以分割成 [1, 5, 5] 和 [11] 。
示例 2：

输入：nums = [1,2,3,5]
输出：false
解释：数组不能分割成两个元素和相等的子集。

提示：

1 <= nums.length <= 200
1 <= nums[i] <= 100

Approach #1: Brute Force with Initial Character Map [Time Limit Exceeded]
Intuition and Algorithm

In a typical brute force, for all starting indices i of A and j of B, we will check for the longest matching subarray A[i:i+k] == B[j:j+k] of length k. This would look roughly like the following psuedocode:

ans = 0
for i in [0 .. A.length - 1]:
    for j in [0 .. B.length - 1]:
        k = 0
        while (A[i+k] == B[j+k]): k += 1 #and i+k < A.length etc.
        ans = max(ans, k)
Our insight is that in typical cases, most of the time A[i] != B[j]. We could instead keep a hashmap Bstarts[A[i]] = all j such that B[j] == A[i], and only loop through those in our j loop.





Approach #2: Binary Search with Naive Check [Time Limit Exceeded]
Intuition

If there is a length k subarray common to A and B, then there is a length j <= k subarray as well.

Let check(length) be the answer to the question "Is there a subarray with length length, common to A and B?" This is a function with range that must take the form [True, True, ..., True, False, False, ..., False] with at least one True. We can binary search on this function.

Algorithm

Focusing on the binary search, our invariant is that check(hi) will always be False. We'll start with hi = min(len(A), len(B)) + 1; clearly check(hi) is False.

Now we perform our check in the midpoint mi of lo and hi. When it is possible, then lo = mi + 1, and when it isn't, hi = mi. This maintains the invariant. At the end of our binary search, hi == lo and lo is the lowest value such that check(lo) is False, so we want lo - 1.

As for the check itself, we can naively check whether any A[i:i+k] == B[j:j+k] using set structures.



Approach #3: Dynamic Programming [Accepted]
Intuition and Algorithm

Since a common subarray of A and B must start at some A[i] and B[j], let dp[i][j] be the longest common prefix of A[i:] and B[j:]. Whenever A[i] == B[j], we know dp[i][j] = dp[i+1][j+1] + 1. Also, the answer is max(dp[i][j]) over all i, j.

We can perform bottom-up dynamic programming to find the answer based on this recurrence. Our loop invariant is that the answer is already calculated correctly and stored in dp for any larger i, j.

Brief note about Question-

We have to return the total number of continuous subarrays whose sum equals to k.
Let's take an example not given in the question by taking negative numbers
Suppose our arr is arr[]: [-1, -1, 1]  && k = 0

So, the answer should be '1' as their is only one subarray whose sum is 0 i.e (-1 + 1)
Solution - I (Brute force, TLE)-

Since we are very obedient person and don't want to do anything extra from our side.
So, we will try to generate the sum of each subarray and if matches withk , then increment our answer.
Like, this is the most basic thing we can do.
Time Complexity --> O(n ^ 2) // where n is the size of the array
Space Complexity --> O(1) // we are not using anything extra from our side
It paases [ 85 / 89 ] in built test cases

给定一个包含 [0, n] 中 n 个数的数组 nums ，找出 [0, n] 这个范围内没有出现在数组中的那个数。

示例 1：

输入：nums = [3,0,1]
输出：2
解释：n = 3，因为有 3 个数字，所以所有的数字都在范围 [0,3] 内。2 是丢失的数字，因为它没有出现在 nums 中。
示例 2：

输入：nums = [0,1]
输出：2
解释：n = 2，因为有 2 个数字，所以所有的数字都在范围 [0,2] 内。2 是丢失的数字，因为它没有出现在 nums 中。
示例 3：

输入：nums = [9,6,4,2,3,5,7,0,1]
输出：8
解释：n = 9，因为有 9 个数字，所以所有的数字都在范围 [0,9] 内。8 是丢失的数字，因为它没有出现在 nums 中。
示例 4：

输入：nums = [0]
输出：1
解释：n = 1，因为有 1 个数字，所以所有的数字都在范围 [0,1] 内。1 是丢失的数字，因为它没有出现在 nums 中。

提示：

n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
nums 中的所有数字都 独一无二

The basic idea is to use XOR operation. We all know that a^b^b =a, which means two xor operations with the same number will eliminate the number and reveal the original number.
In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing number.

给你一个整数 n ，对于 0 <= i <= n 中的每个 i ，计算其二进制表示中 1 的个数 ，返回一个长度为 n + 1 的数组 ans 作为答案。

示例 1：

输入：n = 2
输出：[0,1,1]
解释：
0 --> 0
1 --> 1
2 --> 10
示例 2：

输入：n = 5
输出：[0,1,1,2,1,2]
解释：
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

提示：

0 <= n <= 105

进阶：

很容易就能实现时间复杂度为 O(n log n) 的解决方案，你可以在线性时间复杂度 O(n) 内用一趟扫描解决此问题吗？
你能不使用任何内置函数解决此问题吗？（如，C++ 中的 __builtin_popcount ）

给你两个整数 a 和 b ，不使用 运算符 + 和 - ​​​​​​​，计算并返回两整数之和。

示例 1：

输入：a = 1, b = 2
输出：3
示例 2：

输入：a = 2, b = 3
输出：5

提示：

-1000 <= a, b <= 1000

Now we can not use the + operator, which means it is obvious that we have to use some sort of bit manupulation. But the real question is how?
The ans lies within the procedure of the addition, which I am going to show you below.

from binary level, how do we exactly add two numbers, let us see -->

         _
a = 2 = 010
b = 3 = 011
------------
c = 5 = 101
In the position, where we have a dash above, there we are generating a carry, which will be carried over to the next bit, and added to that next bit. So, the addition pattern goes like -

  0         0         1         1
+ 0        +1       + 0       + 1
----      ----      ----      ----
  0         1         1         0 (with a carry 1)
Is this pattern similar to you? Have you seen this in the XOR table? Let's see the XOR table quickly -

a b | XOR
- - | - -
0 0 | 0
0 1 | 1
1 0 | 1
1 1 | 0
These are the exact same, hence for addition, we need to use the XOR operator. But what to do with the carry? Hey, we need to add that carry to the next bit, right? That is what we have seen in the implementation of the addition as well. We will do that only, but we can NOT use addition anyway.
But, before that, let's do the XOR for 2 and 3 example.

-- Doing only XOR --
a = 2 = 010
b = 3 = 011
------------
c = 1 = 001
1 is NOT our answer, and in this procedure, we have left the carry out, which is 100. Now what's the pattern for finding the carry? It is after we AND the two numbers, we will LEFT-SHIFT the result by 1. Didn't get it?

-- Doing only AND --
a = 2 = 010
b = 3 = 011
------------
c = 2 = 010
-----------
Doing Left-Shift by 1 (<<1)
-----------
c = 4 = 100
So, we found out the carry as well, and believe me or not, but it is the entire Algorithm. You have to repeat the steps of 1. XOR and 2. AND with Left-Shift, until the step no 2. becomes 0, and you will have your answer.

Example -

a = 2 = 010
b = 3 = 011
-----------
x = 1 = 001 = a
c = 4 = 100 = b
-----------
x = 5 = 101
c = 0 = 000

x = XOR & c = AND with Left-Shift
Since carry becomes 0, hence our answer is the XOR result = 5.

Below is the working code for the same, and you can run this code to find the desired answer.

class Solution {
    public int getSum(int a, int b) {
        while(b != 0){
            int temp = (a&b)<<1;
            a = a ^ b;
            b = temp;
        }
        return a;
    }
}
Time Complexity: O(1)
Space Complexity: O(1)

Time is O(1), because the max and min bounds are 1000 and -1000 respectively, which means the input will NOT be arbitrarily large, and it will be in the limits, hence the time will be constant.

The code for c and c++ will be very similar, and python will be a little different. If you like this approach, then please give me a thumbs up.

Thanks & Happy Coding :)

Solution I:
We use XOR bitwise operatoin to get all the bits that are set either in x or in y, not both.
Then we count the number of such bits and we're done!

class Solution {
public:
    int hammingDistance(int x, int y) {
        int res = 0;
        int num = x^y;
        while (num) {
            res += num % 2;
            num >>= 1;
        }
        return res;
    }
};
Solution II - Without XOR:
We iterate x and y in parallel.
If (x % 2 != y % 2) - only one of the rightmost bits are set - we add one to res.

class Solution {
public:
    int hammingDistance(int x, int y) {
        int res = 0;
        while (x || y) {
            res += (x % 2 != y % 2);
            x >>= 1, y >>= 1;
        }
        return res;
    }
};

✔️ Solution - I (Brute-Force)

We can simply iterate and find the leftmost bit (MSB) that is set. From that point onwards, we flip every bit till we reach the rightmost bit (LSB). To flip a bit, we can use ^ 1 (XOR 1) operation. This will flip a bit to 0 if it is set and flip to 1 if it is unset.

C++

class Solution {
public:
    int findComplement(int num) {
        int i = 31;
        while((num & 1 << i) == 0) i--;   // skip the left 0 bits till we reach the 1st set bit from left
        while(i >= 0)
            num ^= 1 << i--;              // flip all bits by XORing with 1
        return num;
    }
};
Python

class Solution:
    def findComplement(self, num):
        i = 31
        while (num & 1 << i) == 0: 
            i -= 1
        while i >= 0:
            num ^= 1 << i
            i -= 1
        return num
Time Complexity : O(N), where N is the number of bits. In this case, since we are starting from i=31, it should be constant but I am denoting time complexity of this approach as generalized O(N)
Space Complexity : O(1)

✔️ Solution - II (Bit-Manipulation Tricks)

The above method basically finds the leftmost set bit and XORs the remaining bits with 1. A more efficient way to do the same would be to simply XOR num with another number having all bits to the right of nums's 1st set bit as 1 and rest bit to left as 0. This would achieve the same thing as above.

For eg. num = 13
=>  num  = 13  (1101)
=>  mask = 15  (1111)
--------------------
   ^       2   (0010)    We got all the bits flipped
But how to get mask?

A simple way would be to initialize mask = 0 and keep flipping bits of it to 1 starting from the rightmost bit (LSB) till we reach the leftmost set bit of num.
Now, how do we know that we reached the leftmost set bit in num?

We use another variable tmp = num. Each time, we will rightshift tmp essentially removing the rightmost bit. When we remove the leftmost set bit from it, it will become 0. Thus, we loop till tmp is not 0.
C++

class Solution {
public:
    int findComplement(int num) {
        int mask = 0;                                 // all bits in mask are initially 0
        for(int tmp = num; tmp; tmp >>= 1)            // set bits in mask to 1 till we reach leftmost set bit in num
            mask = (mask << 1) | 1;                   // leftshifts and sets the rightmost bit to 1
			
        return mask ^ num;                            // finally XORing with mask will flip all bits in num
    }
};
Python

class Solution:
    def findComplement(self, num):
        mask, tmp = 0, num
        while tmp:
            mask = (mask << 1) | 1
            tmp >>= 1
        return mask ^ num
Another way to do the same would be to the other way around and start with mask with all bits as 1. Then we keep setting rightmost bits in mask to 0 one by one till there are no common set bits left in num and mask (num & mask == 0 when no common set bits are present).

For eg. num = 13 (1101)  and  mask = -1 (111...1111)  [all 32 set bits are set in -1 binary representation]
 
1. num =  000...1101 
   mask = 111...1110   => we still have common set bits
   
2. num =  000...1101 
   mask = 111...1100   => we still have common set bits   
  
3. num =  000...1101
   mask = 111...1000   => we still have common set bits  
   
4. num =  000...1101 
   mask = 111...0000   => no more common bits left
Now what?

Now we can simply flip all bits in mask (using ~ operator). It will now have all rightmost bits set to one starting from leftmost set bit in num, i.e, we now have the same mask that we had in previous approach.
Now, we can XOR it with num and we get the flipped result
C++

class Solution {
public:
    int findComplement(int num) {
        uint mask = -1;                             // -1 is represented in binary as all bits set to 1
        while(mask & num) mask <<= 1;               // remove rightmost bits 1 by 1 till no common bits are left
        return ~mask ^ num;                         // XORs with 1 & flips all bits in num starting from the leftmost set bit 
    }
};
Python

class Solution:
    def findComplement(self, num):
        mask = -1
        while mask & num: mask <<= 1
        return ~mask ^ num
This approach used 1 less operation inside loop (1st approach used 3 operations: right-shift >> on tmp, left-shift << on mask and | with 1 to set rightmost bit. This one uses 2: & to check if there are common bits in mask and num and left-shift << to remove the right set bits one by one)

Time Complexity : O(P) / O(log num), where P is the position of leftmost set bit in num. O(P) ~ O(log2(num))
Space Complexity : O(1)

✔️ Solution - III (Flip Bits from Right to Left)

We can start from the right-end and flip bits one by one. This is somewhat similar to 1st solution in above appraoch. But here we will not use mask but rather a bit starting with i=1 and just directly XOR it with num, then left-shift it & repeat thus flipping the bits in num one by one.

So, how do we know when to stop? We stop as soon i > num. This denotes that we have passed the leftmost set bit in num and thus we have flipped all the bits that needed to be flipped.

C++

class Solution {
public:
    int findComplement(int num) {
        int i = 1;
        while(i <= num)
            num ^= i,
            i <<= 1;
        return num;
    }
};
Python

class Solution():
    def findComplement(self, num):
        i = 1
        while i <= num:
            num ^= i
            i <<= 1
        return num    
Time Complexity : O(P) / O(log num)
Space Complexity : O(1)

给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。

示例 1：

输入：head = [1,2,3,4]
输出：[2,1,4,3]
示例 2：

输入：head = []
输出：[]
示例 3：

输入：head = [1]
输出：[1]

给你一个整数 n ，请你生成并返回所有由 n 个节点组成且节点值从 1 到 n 互不相同的不同 二叉搜索树 。可以按 任意顺序 返回答案。

示例 1：

输入：n = 3
输出：[[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]
示例 2：

输入：n = 1
输出：[[1]]

提示：

1 <= n <= 8

The basic idea is that we can construct the result of n node tree just from the result of n-1 node tree.
Here's how we do it: only 2 conditions: 1) The nth node is the new root, so newroot->left = oldroot;
2) the nth node is not root, we traverse the old tree, every time the node in the old tree has a right child, we can perform: old node->right = nth node, nth node ->left = right child; and when we reach the end of the tree, don't forget we can also add the nth node here.
One thing to notice is that every time we push a TreeNode in our result, I push the clone version of the root, and I recover what I do to the old node immediately. This is because you may use the old tree for several times.

给你一个二叉树的根节点 root ，判断其是否是一个有效的二叉搜索树。

有效 二叉搜索树定义如下：

节点的左子树只包含 小于 当前节点的数。
节点的右子树只包含 大于 当前节点的数。
所有左子树和右子树自身必须也是二叉搜索树。

示例 1：

输入：root = [2,1,3]
输出：true
示例 2：

输入：root = [5,1,4,null,null,3,6]
输出：false
解释：根节点的值是 5 ，但是右子节点的值是 4 。

提示：

树中节点数目范围在[1, 104] 内
-231 <= Node.val <= 231 - 1

For the recursive solution, we set a lower bound and a upper bound for the tree. When we recurse on the left subtree, the upper bound becomes the value of its root. When we recurse on the right subtree, the lower bound becomes the value of its root.

给你一个二叉树的根节点 root ， 检查它是否轴对称。

示例 1：

输入：root = [1,2,2,3,4,4,3]
输出：true
示例 2：

输入：root = [1,2,2,null,3,null,3]
输出：false

提示：

树中节点数目在范围 [1, 1000] 内
-100 <= Node.val <= 100

进阶：你可以运用递归和迭代两种方法解决这个问题吗？

Explanation :

class Solution {
public:
    bool solve(TreeNode * r1, TreeNode * r2)
    {
        // See the tree diagram are r1 and r2 null ? No, so this line dont execute
        if(r1 == NULL && r2 == NULL)
            return true; 
		
        // Is any one of r1 or r2 null ? Or are these values different ? No. Both values are
        // same so this else if wont execute either
        else if(r1 == NULL || r2 == NULL || r1->val != r2->val)
            return false; 
        
        // Now comes the main part, we are calling 2 seperate function calls 
        return solve(r1->left, r2->right) && solve(r1->right, r2->left);
        // First solve() before && will execute
        // r1->left is 3 and r2->right = 3
        // Both values are same , they will by pass both if and else if statement
        // Now again r1->left is null and r2->right is null
        // So they will return true from first if condtion
        // Now the scene is : we have executed first solve() before && and it has
        // returned us True so expression becomes ' return true && solve() '
        // Now solve after && will execute 
        // Similarly it will check for 4 and 4 , it will by pass if else statements
        // next time both will become null, so will return true
        // Thus 2nd solve() at the end will also hold true
        // and we know 'true && true' is true
        // so true will be returned to caller, and thus tree is mirror of itself.
        // Similarly you can check for any testcase, flow of execution will remain same.
        
    }
    
    bool isSymmetric(TreeNode* root) 
    {
        // Imagine a tree:    1
        //                 2     2
        //               3   4 4   3
        // We are standing on root that is 1, function begins
        // and now r1 and r2 points to 2 and 2 respectively.                      
        return solve(root->left, root->right);     
    }
};

给你一个由 ‘1’（陆地）和 ‘0’（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例 1：

输入：grid = [
[“1”,“1”,“1”,“1”,“0”],
[“1”,“1”,“0”,“1”,“0”],
[“1”,“1”,“0”,“0”,“0”],
[“0”,“0”,“0”,“0”,“0”]
]
输出：1
示例 2：

输入：grid = [
[“1”,“1”,“0”,“0”,“0”],
[“1”,“1”,“0”,“0”,“0”],
[“0”,“0”,“1”,“0”,“0”],
[“0”,“0”,“0”,“1”,“1”]
]
输出：3

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 ‘0’ 或 ‘1’

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

示例 1：

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]
示例 2：

输入：head = [1,2]
输出：[2,1]
示例 3：

输入：head = []
输出：[]

提示：

链表中节点的数目范围是 [0, 5000]
-5000 <= Node.val <= 5000

Well, since the head pointer may also be modified, we create a pre that points to it to facilitate the reverse process.

For the example list 1 -> 2 -> 3 -> 4 -> 5 in the problem statement, it will become 0 -> 1 -> 2 -> 3 -> 4 -> 5 (we init pre -> val to be 0). We also set a pointer cur to head. Then we keep inserting cur -> next after pre until cur becomes the last node. This idea uses three pointers (pre, cur and temp). You may implement it as follows.

Well, remember to take advantage of the property of binary search trees, which is, node -> left -> val < node -> val < node -> right -> val. Moreover, both p and q will be the descendants of the root of the subtree that contains both of them. And the root with the largest depth is just the lowest common ancestor. This idea can be turned into the following simple recursive code.

给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个节点 p、q，最近公共祖先表示为一个节点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

示例 1：

输入：root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
输出：3
解释：节点 5 和节点 1 的最近公共祖先是节点 3 。
示例 2：

输入：root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
输出：5
解释：节点 5 和节点 4 的最近公共祖先是节点 5 。因为根据定义最近公共祖先节点可以为节点本身。
示例 3：

输入：root = [1,2], p = 1, q = 2
输出：1

提示：

树中节点数目在范围 [2, 105] 内。
-109 <= Node.val <= 109
所有 Node.val 互不相同 。
p != q
p 和 q 均存在于给定的二叉树中。

Approach 1: Recursive Approach
Intuition

The approach is pretty intuitive. Traverse the tree in a depth first manner. The moment you encounter either of the nodes p or q, return some boolean flag. The flag helps to determine if we found the required nodes in any of the paths. The least common ancestor would then be the node for which both the subtree recursions return a True flag. It can also be the node which itself is one of p or q and for which one of the subtree recursions returns a True flag.

Let us look at the formal algorithm based on this idea.

Algorithm

Start traversing the tree from the root node.
If the current node itself is one of p or q, we would mark a variable mid as True and continue the search for the other node in the left and right branches.
If either of the left or the right branch returns True, this means one of the two nodes was found below.
If at any point in the traversal, any two of the three flags left, right or mid become True, this means we have found the lowest common ancestor for the nodes p and q.

Complexity Analysis

Time Complexity: O(N)O(N), where NN is the number of nodes in the binary tree. In the worst case we might be visiting all the nodes of the binary tree.

Space Complexity: O(N)O(N). This is because the maximum amount of space utilized by the recursion stack would be NN since the height of a skewed binary tree could be NN.





Approach 2: Iterative using parent pointers
Intuition

If we have parent pointers for each node we can traverse back from p and q to get their ancestors. The first common node we get during this traversal would be the LCA node. We can save the parent pointers in a dictionary as we traverse the tree.

Algorithm

Start from the root node and traverse the tree.
Until we find p and q both, keep storing the parent pointers in a dictionary.
Once we have found both p and q, we get all the ancestors for p using the parent dictionary and add to a set called ancestors.
Similarly, we traverse through ancestors for node q. If the ancestor is present in the ancestors set for p, this means this is the first ancestor common between p and q (while traversing upwards) and hence this is the LCA node.

Complexity Analysis

Time Complexity : O(N)O(N), where NN is the number of nodes in the binary tree. In the worst case we might be visiting all the nodes of the binary tree.

Space Complexity : O(N)O(N). In the worst case space utilized by the stack, the parent pointer dictionary and the ancestor set, would be NN each, since the height of a skewed binary tree could be NN.

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]
示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

提示:

1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder 和 inorder 均 无重复 元素
inorder 均出现在 preorder
preorder 保证 为二叉树的前序遍历序列
inorder 保证 为二叉树的中序遍历序列

Solution
You need to understand preorder and postorder traversal first, and then go ahead.

Basic idea is:

preorder[0] is the root node of the tree
preorder[x] is a root node of a sub tree
In in-order traversal
When inorder[index] is an item in the in-order traversal
inorder[0]-inorder[index-1] are on the left branch
inorder[index+1]-inorder[size()-1] are on the right branch
if there is nothing on the left, that means the left child of the node is NULL
if there is nothing on the right, that means the right child of the node is NULL
Algorithm:

Start from rootIdx 0
Find preorder[rootIdx] from inorder, let's call the index pivot
Create a new node with inorder[pivot]
Create its left child recursively
Create its right child recursively
Return the created node.
The implementation is self explanatory. Have a look :)

给定一个单链表的头节点  head ，其中的元素 按升序排序 ，将其转换为高度平衡的二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差不超过 1。

示例 1:

输入: head = [-10,-3,0,5,9]
输出: [0,-3,9,-10,null,5]
解释: 一个可能的答案是[0，-3,9，-10,null,5]，它表示所示的高度平衡的二叉搜索树。
示例 2:

输入: head = []
输出: []

提示:

head 中的节点数在[0, 2 * 104] 范围内
-105 <= Node.val <= 105

Recursively build tree.

find midpoint by fast/slow method, use middle node as root.
build left child by first half of the list
build right child by second half of the list (head is midpoint->next)

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]
示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]
示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

提示：

树中节点总数在范围 [0, 5000] 内
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000

Idea

DFS from the root down to it's descendants:
We need to keep current path (which stores elements in the path) so far.
We need to keep the remain targetSum so far (after minus value of elements in the path).
If we already reach into leaf node
Check if targetSum == 0 then we found a valid path from root to leaf node which sum equal to targetSum, so add current path to the answer.
Else dfs on left node and on the right node.


Complexity

Time: O(N^2), where N <= 5000 is the number of elements in the binary tree.

First, we think the time complexity is O(N) because we only visit each node once.
But we forgot to calculate the cost to copy the current path when we found a valid path, which in the worst case can cost O(N^2), let see the following example for more clear.

Extra Space (without counting output as space): O(H), where H is height of the binary tree. This is the space for stack recursion or keeping path so far.

Approach 1: Recursive Solution
The current solution is very simple. We make use of a function construct(nums, l, r), which returns the maximum binary tree consisting of numbers within the indices ll and rr in the given numsnums array(excluding the r^{th}r 
th
  element).

The algorithm consists of the following steps:

Start with the function call construct(nums, 0, n). Here, nn refers to the number of elements in the given numsnums array.

Find the index, max_imax 
i
​
 , of the largest element in the current range of indices (l:r-1)(l:r−1). Make this largest element, nums[max\_i]nums[max_i] as the local root node.

Determine the left child using construct(nums, l, max_i). Doing this recursively finds the largest element in the subarray left to the current largest element.

Similarly, determine the right child using construct(nums, max_i + 1, r).

Return the root node to the calling function.

'''
class Solution {
public:

int dia=0;

int diameterOfBinaryTree(TreeNode* root) {
    
    //base case 1:
    if(!root)
        return 0;
    
    //recursively, call the helper function with the root
    helper(root);
    
    //return the diameter value
    return dia;
}

int helper(TreeNode* node){
    
    //base case 2
    if(!node)
        return 0;
    
    //initialize two pointers, left & right ones
    int leftNode=helper(node->left);
    int rightNode=helper(node->right);
    
    //updated diameter value is the max path between two nodes-> path between two leaves
    dia=max(dia, leftNode+rightNode);
    
    //for each node-> add the diameter by one, as you pass an edge
    return 1+max(leftNode, rightNode);
}
};

给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。

示例 1：

输入：root = [1,null,2,3]
输出：[1,3,2]
示例 2：

输入：root = []
输出：[]
示例 3：

输入：root = [1]
输出：[1]

提示：

树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100

给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[9,20],[15,7]]
示例 2：

输入：root = [1]
输出：[[1]]
示例 3：

输入：root = []
输出：[]

提示：

树中节点数目在范围 [0, 2000] 内
-1000 <= Node.val <= 1000

What is level in our binary tree? It is set of nodes, for which distance between root and these nodes are constant. And if we talk about distances, it can be a good idea to use bfs.

We put our root into queue, now we have level 0 in our queue.
On each step extract all nodes from queue and put their children to to opposite end of queue. In this way we will have full level in the end of each step and our queue will be filled with nodes from the next level.
In the end we just return result.
Complexity
Time complexity is O(n): we perform one bfs on our tree. Space complexity is also O(n), because we have answer of this size.

Code

给你二叉树的根节点 root ，返回其节点值的 锯齿形层序遍历 。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[20,9],[15,7]]
示例 2：

输入：root = [1]
输出：[[1]]
示例 3：

输入：root = []
输出：[]

提示：

树中节点数目在范围 [0, 2000] 内
-100 <= Node.val <= 100

Assuming after traversing the 1st level, nodes in queue are {9, 20, 8}, And we are going to traverse 2nd level, which is even line and should print value from right to left [8, 20, 9].

We know there are 3 nodes in current queue, so the vector for this level in final result should be of size 3.
Then, queue [i] -> goes to -> vector[queue.size() - 1 - i]
i.e. the ith node in current queue should be placed in (queue.size() - 1 - i) position in vector for that line.

For example, for node(9), it's index in queue is 0, so its index in vector should be (3-1-0) = 2.

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]
示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

提示:

1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder 和 inorder 均 无重复 元素
inorder 均出现在 preorder
preorder 保证 为二叉树的前序遍历序列
inorder 保证 为二叉树的中序遍历序列

olution
You need to understand preorder and postorder traversal first, and then go ahead.

Basic idea is:

preorder[0] is the root node of the tree
preorder[x] is a root node of a sub tree
In in-order traversal
When inorder[index] is an item in the in-order traversal
inorder[0]-inorder[index-1] are on the left branch
inorder[index+1]-inorder[size()-1] are on the right branch
if there is nothing on the left, that means the left child of the node is NULL
if there is nothing on the right, that means the right child of the node is NULL
Algorithm:

Start from rootIdx 0
Find preorder[rootIdx] from inorder, let's call the index pivot
Create a new node with inorder[pivot]
Create its left child recursively
Create its right child recursively
Return the created node.
The implementation is self explanatory. Have a look :)

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。

示例 1：

输入：root = [1,null,2,3]
输出：[1,2,3]
示例 2：

输入：root = []
输出：[]
示例 3：

输入：root = [1]
输出：[1]
示例 4：

输入：root = [1,2]
输出：[1,2]
示例 5：

输入：root = [1,null,2]
输出：[1,2]

提示：

树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100

There are three solutions to this problem.

Iterative solution using stack --- O(n) time and O(n) space;
Recursive solution --- O(n) time and O(n) space (function call stack);
Morris traversal --- O(n) time and O(1) space.

给定一个二叉树的 根节点 root，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。

示例 1:

输入: [1,2,3,null,5,null,4]
输出: [1,3,4]
示例 2:

输入: [1,null,3]
输出: [1,3]
示例 3:

输入: []
输出: []

提示:

二叉树的节点个数的范围是 [0,100]
-100 <= Node.val <= 100

To invert a binary tree, we swap the left and right subtrees and invert them recursively/iteratively.

给定一个二叉搜索树的根节点 root ，和一个整数 k ，请你设计一个算法查找其中第 k 个最小元素（从 1 开始计数）。

给你二叉树的根结点 root ，请你将它展开为一个单链表：

展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
展开后的单链表应该与二叉树 先序遍历 顺序相同。

示例 1：

输入：root = [1,2,5,3,4,null,6]
输出：[1,null,2,null,3,null,4,null,5,null,6]
示例 2：

输入：root = []
输出：[]
示例 3：

输入：root = [0]
输出：[0]

提示：

树中结点数在范围 [0, 2000] 内
-100 <= Node.val <= 100