O(N * M)

Author: skjha1

src/Algorithms/Dynamic Programming/15 Longest Common Subsequence Top down(Memoization).cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+
+int dp[1001][1001];
+
+int LCS(string X, string Y, int n, int m) {
+	// base case
+	if (n == 0 || m == 0)
+		dp[n][m] = 0;
+
+	if (dp[n][m] != -1) // when table is not having -1 then return the value which is preseent in that block
+		return dp[n][m];
+
+  // choice diagram 
+	// when last character is same
+	if (X[n - 1] == Y[m - 1])
+		dp[n][m] = 1 + LCS(X, Y, n - 1, m - 1);// count the number and decreament the both's string length // store the value in particular block 
+	// when last character is not same -> pick max
+	else
+		dp[n][m] = max(LCS(X, Y, n - 1, m), LCS(X, Y, n, m - 1)); // one take full and another by leaving last char and vice versa // store the value in particular block 
+
+	return dp[n][m];
+}
+
+int main() {
+	string X, Y; cin >> X >> Y;
+	int n = X.length(), m = Y.length();
+
+	memset(dp, -1, sizeof(dp)); // intialize the whole dp matrix with -1; // from memset we can initialise either -1 and zero;
+
+	cout << LCS(X, Y, n, m) << endl;
+	return 0;
+}