O(mn)

Author: skjha1

src/Algorithms/Dynamic Programming/16 Longest Common Subsequence Bottom Up(DP).cpp

@@ -0,0 +1,32 @@
+
+#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.