Create LongestCommonSubsequence.cpp

LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”.

Algorithms/Dynamic Programming/C++/LongestCommonSubsequence.cpp

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+/* Dynamic Programming C++ implementation of LCS problem */
+#include<bits/stdc++.h> 
+using namespace std; 
+
+int max(int a, int b); 
+
+/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
+int lcs( char *X, char *Y, int m, int n ) 
+{ 
+	int L[m + 1][n + 1]; 
+	int i, j; 
+
+	/* Following steps build L[m+1][n+1] in 
+	bottom up fashion. Note that L[i][j] 
+	contains length of LCS of X[0..i-1] 
+	and Y[0..j-1] */
+	for (i = 0; i <= m; i++) 
+	{ 
+		for (j = 0; j <= n; j++) 
+		{ 
+		if (i == 0 || j == 0) 
+			L[i][j] = 0; 
+
+		else if (X[i - 1] == Y[j - 1]) 
+			L[i][j] = L[i - 1][j - 1] + 1; 
+
+		else
+			L[i][j] = max(L[i - 1][j], L[i][j - 1]); 
+		} 
+	} 
+
+	/* L[m][n] contains length of LCS 
+	for X[0..n-1] and Y[0..m-1] */
+	return L[m][n]; 
+} 
+
+/* Utility function to get max of 2 integers */
+int max(int a, int b) 
+{ 
+	return (a > b)? a : b; 
+} 
+
+// Driver Code 
+int main() 
+{ 
+	char X[] = "AGGTAB"; 
+	char Y[] = "GXTXAYB"; 
+
+	int m = strlen(X); 
+	int n = strlen(Y); 
+
+	cout << "Length of LCS is "
+		<< lcs( X, Y, m, n ); 
+
+	return 0; 
+} 
+