前后缀操作：求出前后缀上的最小值数组，然后枚举 $j$

class Solution {
public:
    int minimumSum(vector<int> &nums) {
        int n = nums.size();
        vector<int> l(n), r(n);//l[i]=min{nums[0],...,nums[i]}, r[i]=min{nums[i],...,nums[n-1]}
        l[0] = nums[0];
        for (int i = 1; i < n; i++)
            l[i] = min(l[i - 1], nums[i]);
        r[n - 1] = nums[n - 1];
        for (int i = n - 2; i >= 0; i--)
            r[i] = min(r[i + 1], nums[i]);

        int res = INT32_MAX;
        for (int j = 1; j < n - 1; j++)
            if (l[j - 1] < nums[j] && r[j + 1] < nums[j])
                res = min(res, l[j - 1] + nums[j] + r[j + 1]);
        return res == INT32_MAX ? -1 : res;
    }
};

同 A …

class Solution {
public:
    int minimumSum(vector<int> &nums) {
        int n = nums.size();
        vector<int> l(n), r(n);//l[i]=min{nums[0],...,nums[i]}, r[i]=min{nums[i],...,nums[n-1]}
        l[0] = nums[0];
        for (int i = 1; i < n; i++)
            l[i] = min(l[i - 1], nums[i]);
        r[n - 1] = nums[n - 1];
        for (int i = n - 2; i >= 0; i--)
            r[i] = min(r[i + 1], nums[i]);

        int res = INT32_MAX;
        for (int j = 1; j < n - 1; j++)
            if (l[j - 1] < nums[j] && r[j + 1] < nums[j])
                res = min(res, l[j - 1] + nums[j] + r[j + 1]);
        return res == INT32_MAX ? -1 : res;
    }
};

枚举：枚举分组中有最多下标的组的下标数 $i$ ，设一个数 $val$ 在 $nums$ 的出现次数为 $ci$ ，若 $val$ 可以满足条件的被分为若干组，则有 $ci=a\times i+b\times (i-1),a\ge 0,b\ge 0$，等价于 $ci=ab\times i-b,ab\ge b\ge 0$，即 $\lceil \frac{ci}{i}\rceil \ge \lceil \frac{ci}{i}\rceil \times i-ci$

class Solution {
public:
    int minGroupsForValidAssignment(vector<int> &nums) {
        unordered_map<int, int> cnt;
        for (auto x: nums)
            cnt[x]++;
        unordered_map<int, int> can;
        int m = cnt.size();//nums中不同的数val的个数
        for (auto &[_, ci]: cnt) {
            for (int i = 1; i <= ci + 1; i++) {
                int ab = (ci - 1) / i + 1;
                int b = ab * i - ci;
                if (ab >= b)
                    can[i]++;
            }
        }
        int mx = 1;
        for (auto &[gs, ci]: can)
            if (ci == m)//对nums中每个数val都可以划分为若干大小为gs-1和若干大小为g的组
                mx = max(mx, gs);
        int res = 0;
        for (auto &[_, ci]: cnt)
            res += (ci - 1) / mx + 1;
        return res;
    }
};

动态规划：首先需要预处理求出 $cost$ 数组（ $cost[i][j]$ 为将字符串 $s[i,j]$ 修改为半回文串的最少修改次数），然后设 $p[i][j]$ 为使 $s[0,i]$ 可分为 $j$ 个半回文串的最少操作数，枚举其第 $j$ 个半回文串可能的长度来进行状态转移

class Solution {
public:
    vector<vector<int>> cost;
    string s;
    int n;

    void comp_cost(int i, int j, int d) {//以d为模数(题目描述中的d)更新cost[i][j]
        int g = (j - i + 1) / d;//分组大小
        int tot = 0;
        for (int ind = 0; ind < d; ind++) {//枚举d个分组，若s[i,j]是半回文串则每个分组都是回文串
            for (int l = i + ind, r = i + ind + d * (g - 1); l <= r; l += d, r -= d)
                if (s[l] != s[r])
                    tot++;
        }
        cost[i][j] = min(cost[i][j], tot);//更新cost[i][j]
    }

    int minimumChanges(string s, int k) {
        this->s = s;
        n = s.size();
        cost = vector<vector<int>>(n, vector<int>(n, INT32_MAX));
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {//s[i,j]
                int len = j - i + 1;
                for (int d = 1; d * d <= len; d++)
                    if (len % d == 0) {//len的因子d
                        comp_cost(i, j, d);
                        if (d != len / d && d != 1)//len的因子len/d
                            comp_cost(i, j, len / d);
                    }
            }
        }
        int p[n][n + 1];// loc, k
        for (int i = 0; i < n; i++)
            for (int j = 0; j <= n; j++)
                p[i][j] = INT32_MAX;//无效状态标志
        for (int i = 1; i < n; i++) {
            for (int j = 1; j <= (i + 1) / 2; j++) {
                if (j == 1)
                    p[i][j] = cost[0][i];
                for (int pre = 0; i - pre > 1; pre++)
                    if (p[pre][j - 1] != INT32_MAX)
                        p[i][j] = min(p[i][j], p[pre][j - 1] + cost[pre + 1][i]);//考虑最后一个半回文串为s[pre+1][i]的情况
            }
        }
        return p[n - 1][k];
    }
};