1312. Minimum Insertion Steps to Make a String Palindrome

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+1;i++){
        for(int j=0;j<m+1;j++){
            if(i*j==0){
                dp[i][j]=0;
            }
            else if(x[i-1]==y[j-1]){
                dp[i][j]= 1+dp[i-1][j-1];
            }
            else
            dp[i][j]= max(dp[i-1][j],dp[i][j-1]);
        }
    }
    return dp[n][m];

}
    int minInsertions(string s) {
        string s1=s;
        int n= s.length();
        reverse(s1.begin(),s1.end());
        return n-lcs(s,s1,n,n);
    }
};