Given a set of integers, the task is to divide it into two sets S1 and S2 such that the absolute difference between their sums is minimum.

If there is a set S with n elements, then if we assume Subset1 has m elements, Subset2 must have n-m elements and the value of abs(sum(Subset1) – sum(Subset2)) should be minimum.

Example:

Input: arr[] = {1, 6, 11, 5} Output: 1 Explanation: Subset1 = {1, 5, 6}, sum of Subset1 = 12 Subset2 = {11}, sum of Subset2 = 11

This problem is mainly an extension to the Dynamic Programming| Set 18 (Partition Problem).

**Recursive Solution**

The recursive approach is to generate all possible sums from all the values of array and to check which solution is the most optimal one.

To generate sums we either include the i’th item in set 1 or don’t include, i.e., include in set 2.

## C

// A Recursive C program to solve minimum sum partition // problem. #include <bits/stdc++.h> using namespace std; // Function to find the minimum sum int findMinRec(int arr[], int i, int sumCalculated, int sumTotal) { // If we have reached last element. Sum of one // subset is sumCalculated, sum of other subset is // sumTotal-sumCalculated. Return absolute difference // of two sums. if (i==0) return abs((sumTotal-sumCalculated) - sumCalculated); // For every item arr[i], we have two choices // (1) We do not include it first set // (2) We include it in first set // We return minimum of two choices return min(findMinRec(arr, i-1, sumCalculated+arr[i-1], sumTotal), findMinRec(arr, i-1, sumCalculated, sumTotal)); } // Returns minimum possible difference between sums // of two subsets int findMin(int arr[], int n) { // Compute total sum of elements int sumTotal = 0; for (int i=0; i<n; i++) sumTotal += arr[i]; // Compute result using recursive function return findMinRec(arr, n, 0, sumTotal); } // 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 two sets is " << findMin(arr, n); return 0; }

## Java

// JAVA code to partition a set into two subsets // such that the difference of subset sums // is minimum import java.util.*; class GFG { // Function to find the minimum sum public static int findMinRec(int arr[], int i, int sumCalculated, int sumTotal) { // If we have reached last element. // Sum of one subset is sumCalculated, // sum of other subset is sumTotal- // sumCalculated. Return absolute // difference of two sums. if (i == 0) return Math.abs((sumTotal-sumCalculated) - sumCalculated); // For every item arr[i], we have two choices // (1) We do not include it first set // (2) We include it in first set // We return minimum of two choices return Math.min(findMinRec(arr, i - 1, sumCalculated + arr[i-1], sumTotal), findMinRec(arr, i-1, sumCalculated, sumTotal)); } // Returns minimum possible difference between // sums of two subsets public static int findMin(int arr[], int n) { // Compute total sum of elements int sumTotal = 0; for (int i = 0; i < n; i++) sumTotal += arr[i]; // Compute result using recursive function return findMinRec(arr, n, 0, sumTotal); } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = {3, 1, 4, 2, 2, 1}; int n = arr.length; System.out.print("The minimum difference"+ " between two sets is " + findMin(arr, n)); } } // This code is contributed by Arnav Kr. Mandal.

Output:

The minimum difference between two sets is 1

**Time Complexity:**

All the sums can be generated by either (1) including that element in set 1. (2) without including that element in set 1. So possible combinations are :- arr[0] (1 or 2) -> 2 values arr[1] (1 or 2) -> 2 values . . . arr[n] (2 or 2) -> 2 values So time complexity will be 2*2*..... *2 (For n times), that is O(2^n).

**Dynamic Programming**

The problem can be solved using dynamic programming when the sum of the elements is not too big. We can create a 2D array dp[n+1][sum+1] where n is number of elements in given set and sum is sum of all elements. We can construct the solution in bottom up manner.

The task is to divide the set into two parts. We will consider the following factors for dividing it. Let dp[n+1][sum+1] = {1 if some subset from 1st to i'th has a sum equal to j 0 otherwise} i ranges from {1..n} j ranges from {0..(sum of all elements)} So dp[n+1][sum+1] will be 1 if 1) The sum j is achieved including i'th item 2) The sum j is achieved excluding i'th item. Let sum of all the elements be S. To find Minimum sum difference, w have to find j such that Min{sum - j*2 : dp[n][j] == 1 } where j varies from 0 to sum/2 The idea is, sum of S1 is j and it should be closest to sum/2, i.e., 2*j should be closest to sum.

Below is C++ implementation of above code.

// 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; }

Output:

The minimum difference between 2 sets is 1

Time Complexity = O(n*sum) where n is number of elements and sum is sum of all elements.

Note that the above solution is in Pseudo Polynomial Time (time complexity is dependent on numeric value of input).

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This article is contributed by **Abhiraj Smit**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.