Added Dynamic Programming and Tree Folder

Data Structures/Dynamic Programming/0:1 Knapsack/0:1Knapsack.cpp

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+// 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]); 
+	printf("%d\n", knapSack(W, wt, val, n)); 
+	return 0; 
+} 

Data Structures/Dynamic Programming/0:1 Knapsack/equal_partition.cpp

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+// A Dynamic Programming based 
+// C++ program to partition problem 
+#include <bits/stdc++.h> 
+using namespace std; 
+
+// Returns true if arr[] can be partitioned 
+// in two subsets of equal sum, otherwise false 
+bool findPartiion (int arr[], int n) 
+{ 
+	int sum = 0; 
+	int i, j; 
+
+	// Calculate sum of all elements 
+	for (i = 0; i < n; i++) 
+	sum += arr[i]; 
+
+	if (sum % 2 != 0) 
+	return false; 
+
+	bool part[sum / 2 + 1][n + 1]; 
+
+	// initialize top row as true 
+	for (i = 0; i <= n; i++) 
+	part[0][i] = true; 
+
+	// initialize leftmost column, 
+	// except part[0][0], as 0 
+	for (i = 1; i <= sum / 2; i++) 
+	part[i][0] = false; 
+
+	// Fill the partition table in botton up manner 
+	for (i = 1; i <= sum / 2; i++) 
+	{ 
+		for (j = 1; j <= n; j++) 
+		{ 
+			part[i][j] = part[i][j - 1]; 
+			if (i >= arr[j - 1]) 
+			part[i][j] = part[i][j] || 
+						part[i - arr[j - 1]][j - 1]; 
+		}	 
+	} 
+
+	/* // uncomment this part to print table 
+	for (i = 0; i <= sum/2; i++) 
+	{ 
+	for (j = 0; j <= n; j++) 
+		cout<<part[i][j]; 
+	cout<<endl; 
+	} */
+
+	return part[sum / 2][n]; 
+} 
+
+// Driver Code 
+int main() 
+{ 
+	int arr[] = {3, 1, 1, 2, 2, 1}; 
+	int n = sizeof(arr) / sizeof(arr[0]); 
+	if (findPartiion(arr, n) == true) 
+		cout << "Can be divided into two subsets of equal sum"; 
+	else
+		cout << "Can not be divided into"
+			<< " two subsets of equal sum"; 
+	return 0; 
+} 
+
+// This code is contributed by rathbhupendra 

Data Structures/Dynamic Programming/0:1 Knapsack/minimum_subset_sum_difference.cpp

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+// A Recursive C program to solve minimum sum partition 
+// problem. 
+#include <bits/stdc++.h> 
+using namespace std; 
+
+// Returns the minimum value of the difference of the two sets. 
+int findMin(int arr[], int n) 
+{ 
+	// Calculate sum of all elements 
+	int sum = 0; 
+	for (int i = 0; i < n; i++) 
+		sum += arr[i]; 
+
+	// Create an array to store results of subproblems 
+	bool dp[n+1][sum+1]; 
+
+	// Initialize first column as true. 0 sum is possible 
+	// with all elements. 
+	for (int i = 0; i <= n; i++) 
+		dp[i][0] = true; 
+
+	// Initialize top row, except dp[0][0], as false. With 
+	// 0 elements, no other sum except 0 is possible 
+	for (int i = 1; i <= sum; i++) 
+		dp[0][i] = false; 
+
+	// Fill the partition table in bottom up manner 
+	for (int i=1; i<=n; i++) 
+	{ 
+		for (int j=1; j<=sum; j++) 
+		{ 
+			// If i'th element is excluded 
+			dp[i][j] = dp[i-1][j]; 
+
+			// If i'th element is included 
+			if (arr[i-1] <= j) 
+				dp[i][j] |= dp[i-1][j-arr[i-1]]; 
+		} 
+	} 
+
+	// Initialize difference of two sums. 
+	int diff = INT_MAX; 
+
+	// Find the largest j such that dp[n][j] 
+	// is true where j loops from sum/2 t0 0 
+	for (int j=sum/2; j>=0; j--) 
+	{ 
+		// Find the 
+		if (dp[n][j] == true) 
+		{ 
+			diff = sum-2*j; 
+			break; 
+		} 
+	} 
+	return diff; 
+} 
+
+// Driver program to test above function 
+int main() 
+{ 
+	int arr[] = {3, 1, 4, 2, 2, 1}; 
+	int n = sizeof(arr)/sizeof(arr[0]); 
+	cout << "The minimum difference between 2 sets is "
+		<< findMin(arr, n); 
+	return 0; 
+} 

Data Structures/Dynamic Programming/0:1 Knapsack/subset_sum.cpp

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+// A Dynamic Programming solution 
+// for subset sum problem 
+#include <bits/stdc++.h>
+
+using namespace std;
+
+// Returns true if there is a subset of set[] 
+// with sun equal to given sum 
+bool isSubsetSum(int set[], int n, int sum) 
+{ 
+	// The value of subset[i][j] will be true if 
+	// there is a subset of set[0..j-1] with sum 
+	// equal to i 
+	bool subset[n + 1][sum + 1]; 
+
+	// If sum is 0, then answer is true 
+	for (int i = 0; i <= n; i++) 
+		subset[i][0] = true; 
+
+	// If sum is not 0 and set is empty, 
+	// then answer is false 
+	for (int i = 1; i <= sum; i++) 
+		subset[0][i] = false; 
+
+	// Fill the subset table in botton up manner 
+	for (int i = 1; i <= n; i++) { 
+		for (int j = 1; j <= sum; j++) { 
+			if (j < set[i - 1]) 
+				subset[i][j] = subset[i - 1][j]; 
+			if (j >= set[i - 1]) 
+				subset[i][j] = subset[i - 1][j] 
+							|| subset[i - 1][j - set[i - 1]]; 
+		} 
+	} 
+
+	/* // uncomment this code to print table 
+	for (int i = 0; i <= n; i++) 
+	{ 
+	for (int j = 0; j <= sum; j++) 
+		printf ("%4d", subset[i][j]); 
+	printf("\n"); 
+	}*/
+
+	return subset[n][sum]; 
+} 
+
+// Driver program to test above function 
+int main() 
+{ 
+	int set[] = { 3, 34, 4, 12, 5, 2 }; 
+	int sum = 9; 
+	int n = sizeof(set) / sizeof(set[0]); 
+	if (isSubsetSum(set, n, sum) == true) 
+		printf("Found a subset with given sum"); 
+	else
+		printf("No subset with given sum"); 
+	return 0; 
+} 
+// This code is contributed by Arjun Tyagi. 

Data Structures/Dynamic Programming/0:1 Knapsack/subsets_with_sum_equal_to.cpp

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+// A Dynamic Programming solution 
+// for subset sum problem 
+#include <bits/stdc++.h>
+
+using namespace std;
+
+// Returns true if there is a subset of set[] 
+// with sun equal to given sum 
+int isSubsetSum(int set[], int n, int sum) 
+{ 
+	// The value of subset[i][j] will be true if 
+	// there is a subset of set[0..j-1] with sum 
+	// equal to i 
+	int subset[n + 1][sum + 1]; 
+
+	// If sum is 0, then answer is true 
+	for (int i = 0; i <= n; i++) 
+		subset[i][0] = true; 
+
+	// If sum is not 0 and set is empty, 
+	// then answer is false 
+	for (int i = 1; i <= sum; i++) 
+		subset[0][i] = false; 
+
+	// Fill the subset table in botton up manner 
+	for (int i = 1; i <= n; i++) { 
+		for (int j = 1; j <= sum; j++) { 
+			if (j < set[i - 1]) 
+				subset[i][j] = subset[i - 1][j]; 
+			if (j >= set[i - 1]) 
+				subset[i][j] = subset[i - 1][j] 
+							+ subset[i - 1][j - set[i - 1]]; 
+		} 
+	} 
+
+	/* // uncomment this code to print table 
+	for (int i = 0; i <= n; i++) 
+	{ 
+	for (int j = 0; j <= sum; j++) 
+		printf ("%4d", subset[i][j]); 
+	printf("\n"); 
+	}*/
+
+	return subset[n][sum]; 
+} 
+
+// Driver program to test above function 
+int main() 
+{ 
+	int set[] = { 2, 3, 5, 8, 10}; 
+	int sum = 10; 
+	int n = sizeof(set) / sizeof(set[0]); 
+	cout<<isSubsetSum(set, n, sum)<<endl; 
+
+
+	return 0; 
+} 
+// This code is contributed by Arjun Tyagi. 

Data Structures/Dynamic Programming/Longest Common Subsequence/LCS.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( string X, string 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() 
+{ 
+	string X = "AGGTAB"; 
+	string Y = "GXTXAYB"; 
+
+	int m = X.length(); 
+	int n = Y.length(); 
+
+	cout << "Length of LCS is "
+		<< lcs( X, Y, m, n ); 
+
+	return 0; 
+} 
+
+// This code is contributed by rathbhupendra