16 Longest Common Subsequence Bottom Up(DP).cpp

#include <bits/stdc++.h>
using namespace std;

int LCS(string X, string Y, int n, int m) {
	int dp[n + 1][m + 1]; // DP - matrix

	// initialize 1st row and of dp matrix with 0 according to the base condition of recursion // base case of recursion --> for initialization of dp - matrix
	for (int i = 0; i <= n; i++)
		for (int j = 0; j <= m; j++)
			if (i == 0 || j == 0)
				dp[i][j] = 0;
// choise diagram is used to fill rest of the matrix 
	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]; // last row and last column element will give the length of the LCS;
}

int main() {
	string X, Y; cin >> X >> Y;
	int n = X.length(), m = Y.length();

	cout << LCS(X, Y, n, m) << endl;
	return 0;
}

// https://www.geeksforgeeks.org/longest-common-subsequence-dp-4/#:~:text=Time%20complexity%20of%20the%20above,length%20of%20LCS%20is%200.&text=In%20the%20above%20partial%20recursion,%E2%80%9D)%20is%20being%20solved%20twice.