C++ Solution with longest common subsequence logic

#1

ALGORITHM
1.Initialize another string B= A(given string)
2.Reverse String B
3. Find common subsequence between A and B.

int Solution::solve(string A) {
    int n=A.length();
    string B=A;
    reverse(B.begin(),B.end());
    int dp[n+1][n+1];
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=n;j++)
        {
            if(i==0 || j==0)
            dp[i][j]=0;
            else if(A[i-1]==B[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][n];
    
}