Check if it possible to partition in k subarrays with equal sum
Last Updated :
11 Jul, 2025
Given an array A of size N, and a number K. Task is to find out if it is possible to partition the array A into K contiguous subarrays such that the sum of elements within each of these subarrays is the same.
Prerequisite: Count the number of ways to divide an array into three contiguous parts having equal sum
Examples :
Input : arr[] = { 1, 4, 2, 3, 5 } K = 3
Output : Yes
Explanation :
Three possible partitions which have equal sum :
(1 + 4), (2 + 3) and (5)
Input : arr[] = { 1, 1, 2, 2 } K = 2
Output : No
Approach :
Can be solved by using Prefix Sums. Firstly, note that total sum of all elements in the array should be divisible by K to create K partitions each having equal sum. If it is divisible then, check each partition have an equal sum by doing :
- For a particular K, each subarray should have a required sum = total_sum / K.
- Starting from the 0th index, start comparing prefix sum, as soon as it is equal to the sum, it implies the end of one subarray (let's say at index j).
- From (j + 1)th index, find another suitable i whose sum (prefix_sum[i] - prefix_sum[j]) gets equal to the required sum. And the process goes until required number of contiguous subarrays i.e. K is found.
- If at any index, any subarray sum becomes greater than required sum, break out from loop since each subarray should contain that an equal sum.
Following is the implementation for above Approach
C++
// CPP Program to check if array
// can be split into K contiguous
// subarrays each having equal sum
#include <bits/stdc++.h>
using namespace std;
// function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
bool KpartitionsPossible(int arr[], int n, int K)
{
// Creating and filling prefix sum array
int prefix_sum[n];
prefix_sum[0] = arr[0];
for (int i = 1; i < n; i++)
prefix_sum[i] = prefix_sum[i - 1] + arr[i];
// return false if total_sum is not
// divisible by K
int total_sum = prefix_sum[n-1];
if (total_sum % K != 0)
return false;
// a temporary variable to check
// there are exactly K partitions
int temp = 0;
int pos = -1;
for (int i = 0; i < n; i++)
{
// find suitable i for which first
// partition have the required sum
// and then find next partition and so on
if (prefix_sum[i] - (pos == -1 ? 0 :
prefix_sum[pos]) == total_sum / K)
{
pos = i;
temp++;
}
// if it becomes greater than
// required sum break out
else if (prefix_sum[i] - prefix_sum[pos] >
total_sum / K)
break;
}
// check if temp has reached to K
return (temp == K);
}
// Driver Code
int main()
{
int arr[] = { 4, 4, 3, 5, 6, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
int K = 3;
if (KpartitionsPossible(arr, n, K))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java Program to check if an array
// can be split into K contiguous
// subarrays each having equal sum
public class GfG{
// Function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
static boolean KpartitionsPossible(int arr[], int n, int K)
{
// Creating and filling prefix sum array
int prefix_sum[] = new int[n];
prefix_sum[0] = arr[0];
for (int i = 1; i < n; i++)
prefix_sum[i] = prefix_sum[i - 1] + arr[i];
// return false if total_sum is not divisible by K
int total_sum = prefix_sum[n-1];
if (total_sum % K != 0)
return false;
// a temporary variable to check
// there are exactly K partitions
int temp = 0, pos = -1;
for (int i = 0; i < n; i++)
{
// find suitable i for which first
// partition have the required sum
// and then find next partition and so on
if (prefix_sum[i] - (pos == -1 ? 0 :
prefix_sum[pos]) == total_sum / K)
{
pos = i;
temp++;
}
// if it becomes greater than
// required sum break out
else if (prefix_sum[i] - (pos == -1 ? 0 :
prefix_sum[pos]) > total_sum / K)
break;
}
// check if temp has reached to K
return (temp == K);
}
public static void main(String []args){
int arr[] = { 4, 4, 3, 5, 6, 2 };
int n = arr.length;
int K = 3;
if (KpartitionsPossible(arr, n, K))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Rituraj Jain
# Python 3 Program to check if array
# can be split into K contiguous
# subarrays each having equal sum
# function returns true to it is possible to
# create K contiguous partitions each having
# equal sum, otherwise false
def KpartitionsPossible(arr, n, K):
# Creating and filling prefix sum array
prefix_sum = [0 for i in range(n)]
prefix_sum[0] = arr[0]
for i in range(1, n, 1):
prefix_sum[i] = prefix_sum[i - 1] + arr[i]
# return false if total_sum is not
# divisible by K
total_sum = prefix_sum[n - 1]
if (total_sum % K != 0):
return False
# a temporary variable to check
# there are exactly K partitions
temp = 0
pos = -1
for i in range(0, n, 1):
# find suitable i for which first
# partition have the required sum
# and then find next partition and so on
if (pos == -1):
sub = 0
else:
sub = prefix_sum[pos]
if (prefix_sum[i] - sub == total_sum / K) :
pos = i
temp += 1
# if it becomes greater than
# required sum break out
elif (prefix_sum[i] -
prefix_sum[pos] > total_sum / K):
break
# check if temp has reached to K
return (temp == K)
# Driver Code
if __name__ =='__main__':
arr = [4, 4, 3, 5, 6, 2]
n = len(arr)
K = 3
if (KpartitionsPossible(arr, n, K)):
print("Yes")
else:
print("No")
# This code is contributed by
# Shashank_Sharma
// C# Program to check if an array
// can be split into K contiguous
// subarrays each having equal sum
using System;
class GfG
{
// Function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
static bool KpartitionsPossible(int[] arr, int n, int K)
{
// Creating and filling prefix sum array
int[] prefix_sum = new int[n];
prefix_sum[0] = arr[0];
for (int i = 1; i < n; i++)
prefix_sum[i] = prefix_sum[i - 1] + arr[i];
// return false if total_sum is not divisible by K
int total_sum = prefix_sum[n-1];
if (total_sum % K != 0)
return false;
// a temporary variable to check
// there are exactly K partitions
int temp = 0, pos = -1;
for (int i = 0; i < n; i++)
{
// find suitable i for which first
// partition have the required sum
// and then find next partition and so on
if (prefix_sum[i] - (pos == -1 ? 0 :
prefix_sum[pos]) == total_sum / K)
{
pos = i;
temp++;
}
// if it becomes greater than
// required sum break out
else if (prefix_sum[i] - (pos == -1 ? 0 :
prefix_sum[pos]) > total_sum / K)
break;
}
// check if temp has reached to K
return (temp == K);
}
// Driver code
public static void Main()
{
int[] arr = { 4, 4, 3, 5, 6, 2 };
int n = arr.Length;
int K = 3;
if (KpartitionsPossible(arr, n, K))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by ChitraNayal
<?php
// PHP Program to check if array
// can be split into K contiguous
// subarrays each having equal sum
// function returns true to
// it is possible to create
// K contiguous partitions
// each having equal sum,
// otherwise false
function KpartitionsPossible($arr,
$n, $K)
{
// Creating and filling
// prefix sum array
$prefix_sum = Array();
$prefix_sum[0] = $arr[0];
for ($i = 1; $i < $n; $i++)
$prefix_sum[$i] = $prefix_sum[$i - 1] +
$arr[$i];
// return false if total_sum
// is not divisible by K
$total_sum = $prefix_sum[$n - 1];
if ($total_sum % $K != 0)
return false;
// a temporary variable to
// check there are exactly
// K partitions
$temp = 0;
$pos = -1;
for ($i = 0; $i < $n; $i++)
{
// find suitable i for which
// first partition have the
// required sum and then find
// next partition and so on
if ($prefix_sum[$i] - ($pos == -1 ? 0 :
$prefix_sum[$pos]) ==
(int)$total_sum / $K)
{
$pos = $i;
$temp++;
}
}
// check if temp has
// reached to K
return ($temp == $K);
}
// Driver Code
$arr = array (4, 4, 3,
5, 6, 2);
$n = sizeof($arr) ;
$K = 3;
if (KpartitionsPossible($arr, $n, $K))
echo "Yes";
else
echo "No";
// This code is contributed by m_kit
?>
<script>
// Javascript Program to check if an array
// can be split into K contiguous
// subarrays each having equal sum
// Function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
function KpartitionsPossible(arr, n, K)
{
// Creating and filling prefix sum array
let prefix_sum = new Array(n);
prefix_sum[0] = arr[0];
for (let i = 1; i < n; i++)
prefix_sum[i] = prefix_sum[i - 1] +
arr[i];
// return false if total_sum is
// not divisible by K
let total_sum = prefix_sum[n-1];
if (total_sum % K != 0)
return false;
// a temporary variable to check
// there are exactly K partitions
let temp = 0, pos = -1;
for (let i = 0; i < n; i++)
{
// find suitable i for which first
// partition have the required sum
// and then find next partition and so on
if (prefix_sum[i] - (pos == -1 ? 0 :
prefix_sum[pos]) ==
parseInt(total_sum / K, 10))
{
pos = i;
temp++;
}
// if it becomes greater than
// required sum break out
else if (prefix_sum[i] - (pos == -1 ? 0 :
prefix_sum[pos]) >
parseInt(total_sum / K, 10))
break;
}
// check if temp has reached to K
return (temp == K);
}
let arr = [ 4, 4, 3, 5, 6, 2 ];
let n = arr.length;
let K = 3;
if (KpartitionsPossible(arr, n, K))
document.write("Yes");
else
document.write("No");
</script>
Time Complexity: O(N), where N is the size of array.
Auxiliary Space: O(N), where N is the size of array.
We can further reduce the space complexity to O(1).
Since the array will be divided to k sub arrays and all the sub arrays will be continuous. So idea is to calculate the count of sub arrays whose sum is equal to sum of whole array divided by k.
if count == k print Yes else print No.
Steps to solve the problem:
- Initialize sum as 0 and count as 0.
- Calculate the sum of all elements in the array arr.
- Check if the sum is divisible by k or not. If not, return 0 as it is not possible to divide the array into k subarrays of equal sum.
- Update the sum as divide the sum by k.
- Initialize ksum as 0.
- Iterate through the array arr and add the current element to ksum.
- If ksum is equal to the target sum, it means that we have found a subarray, so we reset ksum to 0 and increment the count by 1.
- If the count is equal to k, it means that we have successfully divided the array into k subarrays of equal sum, so return 1. Otherwise, return 0.
Following is the implementation for above Approach
C++
// CPP Program to check if array
// can be split into K contiguous
// subarrays each having equal sum
#include <bits/stdc++.h>
using namespace std;
// function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
int KpartitionsPossible(int arr[], int n, int k)
{
int sum = 0;
int count = 0;
// calculate the sum of the array
for(int i = 0; i < n; i++)
sum = sum + arr[i];
if(sum % k != 0)
return 0;
sum = sum / k;
int ksum = 0;
// ksum denotes the sum of each subarray
for(int i = 0; i < n; i++)
{
ksum=ksum + arr[i];
// one subarray is found
if(ksum == sum)
{
// to locate another
ksum = 0;
count++;
}
}
if(count == k)
return 1;
else
return 0;
}
// Driver code
int main() {
int arr[] = { 1, 1, 2, 2};
int k = 2;
int n = sizeof(arr) / sizeof(arr[0]);
if (KpartitionsPossible(arr, n, k) == 0)
cout << "Yes";
else
cout<<"No";
return 0;
}
//Java Program to check if array
// can be split into K contiguous
// subarrays each having equal sum
public class GFG {
// function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
static int KpartitionsPossible(int arr[], int n, int k) {
int sum = 0;
int count = 0;
// calculate the sum of the array
for (int i = 0; i < n; i++) {
sum = sum + arr[i];
}
if (sum % k != 0) {
return 0;
}
sum = sum / k;
int ksum = 0;
// ksum denotes the sum of each subarray
for (int i = 0; i < n; i++) {
ksum = ksum + arr[i];
// one subarray is found
if (ksum == sum) {
// to locate another
ksum = 0;
count++;
}
}
if (count == k) {
return 1;
} else {
return 0;
}
}
// Driver Code
public static void main(String[] args) {
int arr[] = {1, 1, 2, 2};
int k = 2;
int n = arr.length;
if (KpartitionsPossible(arr, n, k) == 0) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
/*This code is contributed by PrinciRaj1992*/
# Python3 Program to check if array
# can be split into K contiguous
# subarrays each having equal sum
# Function returns true to it is possible
# to create K contiguous partitions each
# having equal sum, otherwise false
def KpartitionsPossible(arr, n, k) :
sum = 0
count = 0
# calculate the sum of the array
for i in range(n) :
sum = sum + arr[i]
if(sum % k != 0) :
return 0
sum = sum // k
ksum = 0
# ksum denotes the sum of each subarray
for i in range(n) :
ksum = ksum + arr[i]
# one subarray is found
if(ksum == sum) :
# to locate another
ksum = 0
count += 1
if(count == k) :
return 1
else :
return 0
# Driver code
if __name__ == "__main__" :
arr = [ 1, 1, 2, 2]
k = 2
n = len(arr)
if (KpartitionsPossible(arr, n, k) == 0) :
print("Yes")
else :
print("No")
# This code is contributed by Ryuga
// C# Program to check if array
// can be split into K contiguous
// subarrays each having equal sum
using System;
public class GFG{
// function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
static int KpartitionsPossible(int []arr, int n, int k) {
int sum = 0;
int count = 0;
// calculate the sum of the array
for (int i = 0; i < n; i++) {
sum = sum + arr[i];
}
if (sum % k != 0) {
return 0;
}
sum = sum / k;
int ksum = 0;
// ksum denotes the sum of each subarray
for (int i = 0; i < n; i++) {
ksum = ksum + arr[i];
// one subarray is found
if (ksum == sum) {
// to locate another
ksum = 0;
count++;
}
}
if (count == k) {
return 1;
} else {
return 0;
}
}
// Driver Code
public static void Main() {
int []arr = {1, 1, 2, 2};
int k = 2;
int n = arr.Length;
if (KpartitionsPossible(arr, n, k) == 0) {
Console.Write("Yes");
} else {
Console.Write("No");
}
}
}
/*This code is contributed by PrinciRaj1992*/
<?php
// PHP Program to check if array
// can be split into K contiguous
// subarrays each having equal sum
// function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
function KpartitionsPossible($arr, $n, $k)
{
$sum = 0;
$count = 0;
// calculate the sum of the array
for($i = 0; $i < $n; $i++)
$sum = $sum + $arr[$i];
if($sum % $k != 0)
return 0;
$sum = $sum / $k;
$ksum = 0;
// ksum denotes the sum of each subarray
for( $i = 0; $i < $n; $i++)
{
$ksum = $ksum + $arr[$i];
// one subarray is found
if($ksum == $sum)
{
// to locate another
$ksum = 0;
$count++;
}
}
if($count == $k)
return 1;
else
return 0;
}
// Driver code
$arr = array(1, 1, 2, 2);
$k = 2;
$n = count($arr);
if (KpartitionsPossible($arr, $n, $k) == 0)
echo "Yes";
else
echo "No";
// This code is contributed by
// Rajput-Ji
?>
<script>
// Javascript program to check if array
// can be split into K contiguous
// subarrays each having equal sum
// Function returns true to it is possible to
// create K contiguous partitions each having
// equal sum, otherwise false
function KpartitionsPossible(arr, n, k)
{
let sum = 0;
let count = 0;
// Calculate the sum of the array
for(let i = 0; i < n; i++)
sum = sum + arr[i];
if (sum % k != 0)
return 0;
sum = parseInt(sum / k, 10);
let ksum = 0;
// ksum denotes the sum of each subarray
for(let i = 0; i < n; i++)
{
ksum = ksum + arr[i];
// One subarray is found
if (ksum == sum)
{
// To locate another
ksum = 0;
count++;
}
}
if (count == k)
return 1;
else
return 0;
}
// Driver code
let arr = [ 1, 1, 2, 2 ];
let k = 2;
let n = arr.length;
if (KpartitionsPossible(arr, n, k) == 0)
document.write("Yes");
else
document.write("No");
// This code is contributed by mukesh07
</script>
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