Print equal sum sets of Array (Partition Problem) using Dynamic Programming
Last Updated :
23 Apr, 2025
Given an array arr[]. Determine whether it is possible to split the array into two sets such that the sum of elements in both sets is equal. If it is possible, then print both sets. If it is not possible then output -1.
Examples :
Input : arr = {5, 5, 1, 11}
Output : Set 1 = {5, 5, 1}, Set 2 = {11}
Sum of both the sets is 11 and equal.
Input : arr = {1, 5, 3}
Output : -1
No partitioning results in equal sum sets.Prerequisite: Partition Problem
Approach:
In the previous post, a solution using recursion is discussed. In this post, a solution using Dynamic Programming is explained.
The idea is to declare two sets set 1 and set 2. To recover the solution, traverse the boolean dp table backward starting from the final result dp[n][k], where n = number of elements and k = sum/2. Set 1 will consist of elements that contribute to sum k and other elements that do not contribute are added to set 2. Follow these steps at each position to recover the solution.
- Check if dp[i-1][sum] is true or not. If it is true, then the current element does not contribute to sum k. Add this element to set 2. Update index i by i-1 and sum remains unchanged.
- If dp[i-1][sum] is false, then current element contribute to sum k. Add current element to set 1. Update index i by i-1 and sum by sum-arr[i-1].
Repeat the above steps until each index position is traversed.
Implementation:
C++
// CPP program to print equal sum sets of array.
#include <bits/stdc++.h>
using namespace std;
// Function to print equal sum
// sets of array.
void printEqualSumSets(int arr[], int n)
{
int i, currSum;
// Finding sum of array elements
int sum = accumulate(arr, arr+n, 0);
// Check sum is even or odd. If odd
// then array cannot be partitioned.
// Print -1 and return.
if (sum & 1) {
cout << "-1";
return;
}
// Divide sum by 2 to find
// sum of two possible subsets.
int k = sum >> 1;
// Boolean DP table to store result
// of states.
// dp[i][j] = true if there is a
// subset of elements in first i elements
// of array that has sum equal to j.
bool dp[n + 1][k + 1];
// If number of elements are zero, then
// no sum can be obtained.
for (i = 1; i <= k; i++)
dp[0][i] = false;
// Sum 0 can be obtained by not selecting
// any element.
for (i = 0; i <= n; i++)
dp[i][0] = true;
// Fill the DP table in bottom up manner.
for (i = 1; i <= n; i++) {
for (currSum = 1; currSum <= k; currSum++) {
// Excluding current element.
dp[i][currSum] = dp[i - 1][currSum];
// Including current element
if (arr[i - 1] <= currSum)
dp[i][currSum] = dp[i][currSum] |
dp[i - 1][currSum - arr[i - 1]];
}
}
// Required sets set1 and set2.
vector<int> set1, set2;
// If partition is not possible print
// -1 and return.
if (!dp[n][k]) {
cout << "-1\n";
return;
}
// Start from last element in dp table.
i = n;
currSum = k;
while (i > 0 && currSum >= 0) {
// If current element does not
// contribute to k, then it belongs
// to set 2.
if (dp[i - 1][currSum]) {
i--;
set2.push_back(arr[i]);
}
// If current element contribute
// to k then it belongs to set 1.
else if (dp[i - 1][currSum - arr[i - 1]]) {
i--;
currSum -= arr[i];
set1.push_back(arr[i]);
}
}
// Print elements of both the sets.
cout << "Set 1 elements: ";
for (i = 0; i < set1.size(); i++)
cout << set1[i] << " ";
cout << "\nSet 2 elements: ";
for (i = 0; i < set2.size(); i++)
cout << set2[i] << " ";
}
// Driver program.
int main()
{
int arr[] = { 5, 5, 1, 11 };
int n = sizeof(arr) / sizeof(arr[0]);
printEqualSumSets(arr, n);
return 0;
}
// Java program to print
// equal sum sets of array.
import java.io.*;
import java.util.*;
class GFG
{
// Function to print equal
// sum sets of array.
static void printEqualSumSets(int []arr,
int n)
{
int i, currSum, sum = 0;
// Finding sum of array elements
for (i = 0; i < arr.length; i++)
sum += arr[i];
// Check sum is even or odd.
// If odd then array cannot
// be partitioned. Print -1
// and return.
if ((sum & 1) == 1)
{
System.out.print("-1");
return;
}
// Divide sum by 2 to find
// sum of two possible subsets.
int k = sum >> 1;
// Boolean DP table to store
// result of states.
// dp[i,j] = true if there is a
// subset of elements in first i
// elements of array that has sum
// equal to j.
boolean [][]dp = new boolean[n + 1][k + 1];
// If number of elements are zero,
// then no sum can be obtained.
for (i = 1; i <= k; i++)
dp[0][i] = false;
// Sum 0 can be obtained by
// not selecting any element.
for (i = 0; i <= n; i++)
dp[i][0] = true;
// Fill the DP table
// in bottom up manner.
for (i = 1; i <= n; i++)
{
for (currSum = 1;
currSum <= k;
currSum++)
{
// Excluding current element.
dp[i][currSum] = dp[i - 1][currSum];
// Including current element
if (arr[i - 1] <= currSum)
dp[i][currSum] = dp[i][currSum] |
dp[i - 1][currSum - arr[i - 1]];
}
}
// Required sets set1 and set2.
List<Integer> set1 = new ArrayList<Integer>();
List<Integer> set2 = new ArrayList<Integer>();
// If partition is not possible
// print -1 and return.
if (!dp[n][k])
{
System.out.print("-1\n");
return;
}
// Start from last
// element in dp table.
i = n;
currSum = k;
while (i > 0 && currSum >= 0)
{
// If current element does
// not contribute to k, then
// it belongs to set 2.
if (dp[i - 1][currSum])
{
i--;
set2.add(arr[i]);
}
// If current element contribute
// to k then it belongs to set 1.
else if (dp[i - 1][currSum - arr[i - 1]])
{
i--;
currSum -= arr[i];
set1.add(arr[i]);
}
}
// Print elements of both the sets.
System.out.print("Set 1 elements: ");
for (i = 0; i < set1.size(); i++)
System.out.print(set1.get(i) + " ");
System.out.print("\nSet 2 elements: ");
for (i = 0; i < set2.size(); i++)
System.out.print(set2.get(i) + " ");
}
// Driver Code
public static void main(String args[])
{
int []arr = new int[]{ 5, 5, 1, 11 };
int n = arr.length;
printEqualSumSets(arr, n);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
# Python3 program to print equal sum
# sets of array.
import numpy as np
# Function to print equal sum
# sets of array.
def printEqualSumSets(arr, n) :
# Finding sum of array elements
sum_array = sum(arr)
# Check sum is even or odd. If odd
# then array cannot be partitioned.
# Print -1 and return.
if (sum_array & 1) :
print("-1")
return
# Divide sum by 2 to find
# sum of two possible subsets.
k = sum_array >> 1
# Boolean DP table to store result
# of states.
# dp[i][j] = true if there is a
# subset of elements in first i elements
# of array that has sum equal to j.
dp = np.zeros((n + 1, k + 1))
# If number of elements are zero, then
# no sum can be obtained.
for i in range(1, k + 1) :
dp[0][i] = False
# Sum 0 can be obtained by not
# selecting any element.
for i in range(n + 1) :
dp[i][0] = True
# Fill the DP table in bottom up manner.
for i in range(1, n + 1) :
for currSum in range(1, k + 1) :
# Excluding current element.
dp[i][currSum] = dp[i - 1][currSum]
# Including current element
if (arr[i - 1] <= currSum) :
dp[i][currSum] = (dp[i][currSum] or
dp[i - 1][currSum - arr[i - 1]])
# Required sets set1 and set2.
set1, set2 = [], []
# If partition is not possible print
# -1 and return.
if ( not dp[n][k]) :
print("-1")
return
# Start from last element in dp table.
i = n
currSum = k
while (i > 0 and currSum >= 0) :
# If current element does not
# contribute to k, then it belongs
# to set 2.
if (dp[i - 1][currSum]) :
i -= 1
set2.append(arr[i])
# If current element contribute
# to k then it belongs to set 1.
elif (dp[i - 1][currSum - arr[i - 1]]) :
i -= 1
currSum -= arr[i]
set1.append(arr[i])
# Print elements of both the sets.
print("Set 1 elements:", end = " ")
for i in range(len(set1)) :
print(set1[i], end = " ")
print("\nSet 2 elements:", end = " ")
for i in range(len(set2)) :
print(set2[i], end = " ")
# Driver Code
if __name__ == "__main__" :
arr = [ 5, 5, 1, 11 ]
n = len(arr)
printEqualSumSets(arr, n)
# This code is contributed by Ryuga
// C# program to print
// equal sum sets of array.
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
// Function to print equal
// sum sets of array.
static void printEqualSumSets(int []arr,
int n)
{
int i, currSum, sum = 0;
// Finding sum of array elements
for (i = 0; i < arr.Length; i++)
sum += arr[i];
// Check sum is even or odd.
// If odd then array cannot
// be partitioned. Print -1
// and return.
if ((sum & 1) == 1)
{
Console.Write("-1");
return;
}
// Divide sum by 2 to find
// sum of two possible subsets.
int k = sum >> 1;
// Boolean DP table to store
// result of states.
// dp[i,j] = true if there is a
// subset of elements in first i
// elements of array that has sum
// equal to j.
bool [,]dp = new bool[n + 1, k + 1];
// If number of elements are zero,
// then no sum can be obtained.
for (i = 1; i <= k; i++)
dp[0, i] = false;
// Sum 0 can be obtained by
// not selecting any element.
for (i = 0; i <= n; i++)
dp[i, 0] = true;
// Fill the DP table
// in bottom up manner.
for (i = 1; i <= n; i++)
{
for (currSum = 1; currSum <= k; currSum++)
{
// Excluding current element.
dp[i, currSum] = dp[i - 1, currSum];
// Including current element
if (arr[i - 1] <= currSum)
dp[i, currSum] = dp[i, currSum] |
dp[i - 1, currSum - arr[i - 1]];
}
}
// Required sets set1 and set2.
List<int> set1 = new List<int>();
List<int> set2 = new List<int>();
// If partition is not possible
// print -1 and return.
if (!dp[n, k])
{
Console.Write("-1\n");
return;
}
// Start from last
// element in dp table.
i = n;
currSum = k;
while (i > 0 && currSum >= 0)
{
// If current element does
// not contribute to k, then
// it belongs to set 2.
if (dp[i - 1, currSum])
{
i--;
set2.Add(arr[i]);
}
// If current element contribute
// to k then it belongs to set 1.
else if (dp[i - 1, currSum - arr[i - 1]])
{
i--;
currSum -= arr[i];
set1.Add(arr[i]);
}
}
// Print elements of both the sets.
Console.Write("Set 1 elements: ");
for (i = 0; i < set1.Count; i++)
Console.Write(set1[i] + " ");
Console.Write("\nSet 2 elements: ");
for (i = 0; i < set2.Count; i++)
Console.Write(set2[i] + " ");
}
// Driver Code.
static void Main()
{
int []arr = { 5, 5, 1, 11 };
int n = arr.Length;
printEqualSumSets(arr, n);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
<script>
// Javascript program to print equal sum sets of array.
// Function to print equal sum
// sets of array.
function printEqualSumSets(arr, n)
{
var i, currSum;
// Finding sum of array elements
var sum = 0;
for(var i =0; i< arr.length; i++)
{
sum+=arr[i];
}
// Check sum is even or odd. If odd
// then array cannot be partitioned.
// Print -1 and return.
if (sum & 1) {
document.write( "-1");
return;
}
// Divide sum by 2 to find
// sum of two possible subsets.
var k = sum >> 1;
// Boolean DP table to store result
// of states.
// dp[i][j] = true if there is a
// subset of elements in first i elements
// of array that has sum equal to j.
var dp = Array.from(Array(n+1), ()=> Array(k+1));
// If number of elements are zero, then
// no sum can be obtained.
for (i = 1; i <= k; i++)
dp[0][i] = false;
// Sum 0 can be obtained by not selecting
// any element.
for (i = 0; i <= n; i++)
dp[i][0] = true;
// Fill the DP table in bottom up manner.
for (i = 1; i <= n; i++) {
for (currSum = 1; currSum <= k; currSum++) {
// Excluding current element.
dp[i][currSum] = dp[i - 1][currSum];
// Including current element
if (arr[i - 1] <= currSum)
dp[i][currSum] = dp[i][currSum] |
dp[i - 1][currSum - arr[i - 1]];
}
}
// Required sets set1 and set2.
var set1 = [], set2=[];
// If partition is not possible print
// -1 and return.
if (!dp[n][k]) {
document.write( "-1<br>");
return;
}
// Start from last element in dp table.
i = n;
currSum = k;
while (i > 0 && currSum >= 0) {
// If current element does not
// contribute to k, then it belongs
// to set 2.
if (dp[i - 1][currSum]) {
i--;
set2.push(arr[i]);
}
// If current element contribute
// to k then it belongs to set 1.
else if (dp[i - 1][currSum - arr[i - 1]]) {
i--;
currSum -= arr[i];
set1.push(arr[i]);
}
}
// Print elements of both the sets.
document.write( "Set 1 elements: ");
for (i = 0; i < set1.length; i++)
document.write( set1[i] + " ");
document.write( "<br>Set 2 elements: ");
for (i = 0; i < set2.length; i++)
document.write( set2[i] + " ");
}
// Driver program.
var arr = [ 5, 5, 1, 11 ];
var n = arr.length;
printEqualSumSets(arr, n);
</script>
OutputSet 1 elements: 1 5 5
Set 2 elements: 11
Complexity Analysis:
- Time Complexity: O(n*k), where k = sum(arr) / 2
- Auxiliary Space: O(n*k)
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