28_MinInsertion_toMakeString_palindrome.cpp

// min deletion to make string palindrome = min insertion to make string palindrome
// Same as question 24

// Question Link :- https://www.geeksforgeeks.org/problems/minimum-number-of-deletions4610/1
// Minimum number of deletions to make string Palindrome

// Question Link :- https://leetcode.com/problems/minimum-insertion-steps-to-make-a-string-palindrome/
// Minimum Insertion Steps to Make a String Palindrome


// Brute Force
// T.C = O(2^n)
// S.C = O(n)


// Better Solution (Tabulation)
// T.C = O(n^2)
// S.C = O(n^2)
class Solution {
public:
    int LCS(string X, string Y, int n, int m) {
        int dp[n + 1][m + 1];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                if (i == 0 || j == 0) {
                    dp[i][j] = 0;
                }
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (X[i - 1] == Y[j - 1]) {    // when last character is same
                    dp[i][j] = 1 + dp[i - 1][j - 1];
                } else {    // when last character is not same -> pick max
                    dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
                }
            }
        }
        return dp[n][m];
    }

    int minInsertions(string s) {
        int n = s.length();
        string rev_s = s;
        reverse(rev_s.begin(), rev_s.end());
        
        return n - LCS(s, rev_s, n, n);
    }
};