Min # of deletion or insertions to make palindrome

// longest common subsequence
// [1 ,2 ,3, 4]
// subsequnce (order matters but maybe not continuous)
// substring or subarray .. (need to be contigous)
// subset same as subsequence ..
int solve(string s1){
    string s2 = s1;
    reverse(s2.begin(),s2.end());
    int sa = s1.length();
    int sb = sa;
    vector<vector<int>> dp (sa+1 ,vector<int>(sb+1,0));
    // a represents the ath element .. b represents the knapsack present

    for(int a=1;a<=sa;a++){
        for(int b=1;b<=sb;b++){
            if(s1[a-1] == s2[b-1]){
            dp[a][b] = 1 + dp[a-1][b-1]; //here we can choose ith element again
            }else{
            dp[a][b] = max(dp[a-1][b] , dp[a][b-1]); // cant choose the ath element becuase its weight is more
            }
        }
    }
    return (sa - dp[sa][sa]);

}