Implementing the longest palindromic subsequence algorithm in C++

#include <iostream>
#include <algorithm> // to use reverse() function
using namespace std;
// LCS is used to find the LPS of a string:
// just find the LCS of the original and reversed string
// to get the LPS.
string LCS(string original, string reversed) 
{ 
    int m = original.length(), n = reversed.length(); 
  
    int memo[m+1][n+1]; // create a 2D array to serve as a memo
    // Populate the memo using the mathematical equation for LCS.
    // Note that a cell, memo[i][j], denotes the 
    // length of LCS/LPS of original[0..i-1] and reversed[0..j-1].
    for (int i = 0; i <= m; i++) 
    { 
        for (int j = 0; j <= n; j++) 
        { 
            if (i == 0 || j == 0) 
                memo[i][j] = 0; 
            else if (original[i-1] == reversed[j-1]) 
                memo[i][j] = memo[i-1][j-1] + 1; 
            else
                memo[i][j] = max(memo[i-1][j], memo[i][j-1]); 
        } 
    } 
      
    // create an empty string to store the LPS.
    string LPS = "";
  
    // Start backtracking from the right-bottom most
    // cell of the memo.
    int i = m, j = n; 
    while (i > 0 && j > 0) 
    { 
        // if the current character of original and
        // reversed are the same, then that character is 
        // a part of the LPS. Move diagonally to the top-left
        // cell in this case.
        if (original[i-1] == reversed[j-1]) 
        { 
            LPS += original[i-1];
            i--; 
            j--;   
        } 
        // If not same, then find the larger of 
        // two and go in the direction of larger 
        // valued cell.
        else if (memo[i-1][j] > memo[i][j-1]) 
            i--; 
        else
            j--; 
    } 
    return LPS; 
} 
// driver code
int main() 
{ 
    string original = "ABBCAD"; 
    cout << "Original string is: " << original << endl;
    string reversed = original; 
    reverse(reversed.begin(), reversed.end());   
    cout << "Longest Palindromic Subsequence is: " << 
            LCS(original, reversed) << endl; 
    return 0; 
}