1312-minimum-insertion-steps-to-make-a-string-palindrome.cpp

class Solution {
public:
    int minInsertions(string s) {
        int n = s.length();
        vector<vector<int>> dp(n + 1, vector<int>(n + 1));
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j)
                dp[i + 1][j + 1] = s[i] == s[n - 1 - j] ? dp[i][j] + 1 : max(dp[i][j + 1], dp[i + 1][j]);
        return n - dp[n][n];
    }
};