DP partition problem added.

Author: 96mohitm

DP/PartitionProblem/partition.cpp

@@ -0,0 +1,62 @@
+/*
+Given a set of positive integers, find if it can be divided into two subsets with equal sum.
+Partition problem is special case of Subset Sum Problem which itself is a special case of the Knapsack Problem.
+The idea is to calculate sum of all elements in the set. 
+*/
+
+#include <iostream>
+#include <string>
+using namespace std;
+
+// Return true if there exists a subset of arr[] with given sum
+bool subsetSum(int arr[], int n, int sum)
+{
+    // return true if sum becomes 0 (subset found)
+    if (sum == 0)
+        return true;
+
+    // base case: no items left or sum becomes negative
+    if (n < 0 || sum < 0)
+        return false;
+
+    // Case 1. include current item in the subset (arr[n]) and recurse
+    // for remaining items (n - 1) with remaining sum (sum - arr[n])
+    bool include = subsetSum(arr, n - 1, sum - arr[n]);
+
+    // Case 2. exclude current item n from subset and recurse for
+    // remaining items (n - 1)
+    bool exclude = subsetSum(arr, n - 1, sum);
+
+    // return true if we get subset by including or excluding current item
+    return include || exclude;
+}
+
+// Return true if given array arr[0..n-1] can be divided into two
+// subsets with equal sum
+bool partition(int arr[], int n)
+{
+    int sum = 0;
+    for (int i = 0; i < n; i++)
+       sum += arr[i];
+
+    // return true if sum is even and array can can be divided into
+    // two subsets with equal sum
+    return !(sum & 1) && subsetSum(arr, n - 1, sum/2);
+}
+
+// main function to solve partition problem
+int main()
+{
+    // Input: set of items
+    int arr[] = { 7, 3, 1, 5, 4, 8 };
+
+    // number of items
+    int n = sizeof(arr) / sizeof(arr[0]);
+
+    if (partition(arr, n))
+        cout << "Yes";
+    else
+        cout << "No";
+
+    return 0;
+}