C++ implementation added for Longest Common Subsequence in between tw…

C++/Knapsack_0-1.cpp

@@ -0,0 +1,52 @@
+// Problem Statement 
+/*
+Given weights and values of n items, put these items in a knapsack of capacity W to
+get the maximum total value in the knapsack. In other words, given two integer arrays
+val[0..n-1] and wt[0..n-1] which represent values and weights associated with n items
+respectively. Also given an integer W which represents knapsack capacity, 
+find out the maximum value subset of val[] such that sum of the weights 
+of this subset is smaller than or equal to W. You cannot break an item, 
+either pick the complete item or don’t pick it (0-1 property).
+
+*/
+
+// Algorithm :
+
+// A Dynamic Programming based 
+// solution for 0-1 Knapsack problem 
+#include<bits/stdc++.h>
+using namespace std ;
+
+// Returns the maximum value that 
+// can be put in a knapsack of capacity W 
+int knapSack(int W, int wt[], int val[], int n) 
+{ 
+	int i, w; 
+	int K[n + 1][W + 1]; 
+
+	// Build table K[][] in bottom up manner 
+	for (i = 0; i <= n; i++) { 
+		for (w = 0; w <= W; w++) { 
+			if (i == 0 || w == 0) 
+				K[i][w] = 0; 
+			else if (wt[i - 1] <= w) 
+				K[i][w] = max( 
+					val[i - 1] + K[i - 1][w - wt[i - 1]], 
+					K[i - 1][w]); 
+			else
+				K[i][w] = K[i - 1][w]; 
+		} 
+	} 
+
+	return K[n][W]; 
+} 
+
+int main() 
+{ 
+	int val[] = { 60, 100, 120 }; 
+	int wt[] = { 10, 20, 30 }; 
+	int W = 50; 
+	int n = sizeof(val) / sizeof(val[0]); 
+	out<< knapSack(W, wt, val, n)) <<'\n'; 
+	return 0; 
+} 

C++/Longest_Common_Subsequence.cpp

@@ -0,0 +1,38 @@
+#include<bits/stdc++.h>
+using namespace std ;
+
+int LCS(string a,string b){
+    int m = a.length() ;
+    int n = b.length() ;
+
+    int lcs[m + 1][n + 1];  
+
+    for (int i = 0; i <= m; i++)  
+    {  
+        for (int j = 0; j <= n; j++)  
+        {  
+        if (i == 0 || j == 0)  
+            lcs[i][j] = 0;  
+
+        else if (X[i - 1] == Y[j - 1])  
+            lcs[i][j] = lcs[i - 1][j - 1] + 1;  
+
+        else
+            lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1]);  
+        }  
+    }  
+
+    /* lcs[m][n] contains length of LCS  
+    for X[0..n-1] and Y[0..m-1] */
+    return lcs[m][n];
+}
+
+int main(){
+    string str1,str2 ; 
+    cout << "Take two strings to find their LCS\n";
+    cin >> str1 >> str2 ; 
+    cout << "Length of LCS is " 
+         << LCS( str1 , str2 ); 
+
+    return 0 ;     
+}