Check if array can be divided into two sub-arrays such that their absolute difference is K

Given an array arr[] and an integer K, the task is to find whether the array can be divided into two sub-arrays such that the absolute difference of the sum of the elements of both the sub-arrays is K.

Examples:

Input: arr[] = {2, 4, 5, 1}, K = 0
Output: Yes
{2, 4} and {5, 1} are the two possible sub-arrays.
|(2 + 4) – (5 + 1)| = |6 – 6| = 0

Input: arr[] = {2, 4, 1, 5}, K = 2
Output: No

Approach:

  • Assume there exists an answer, let the sum of elements of the sub-array (with smaller sum) is S.
  • Sum of the elements of the second array will be S + K.
  • And, S + S + K must be equal to sum of all the elements of the array say totalSum = 2 *S + K.
  • S = (totalSum – K) / 2
  • Now, traverse the array till we achieve a sum of S starting from the first element and if its not possible then print No.
  • Else print Yes.

Below is the implementation of the above approach:

C++

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#include <bits/stdc++.h>
using namespace std;
  
// Function that return true if it is possible
// to divide the array into sub-arrays
// that satisfy the given condition
bool solve(int array[], int size, int k)
{
    // To store the sum of all the elements
    // of the array
    int totalSum = 0;
    for (int i = 0; i < size; i++)
        totalSum += array[i];
  
    // Sum of any sub-array cannot be
    // a floating point value
    if ((totalSum - k) % 2 == 1)
        return false;
  
    // Required sub-array sum
    int S = (totalSum - k) / 2;
  
    int sum = 0;
    for (int i = 0; i < size; i++) {
        sum += array[i];
        if (sum == S)
            return true;
    }
  
    return false;
}
  
// Driver Code
int main()
{
    int array[] = { 2, 4, 1, 5 };
    int k = 2;
    int size = sizeof(array) / sizeof(array[0]);
    if (solve(array, size, k))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

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Java

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/*package whatever //do not write package name here */
  
import java.io.*;
  
class GFG 
{
      
// Function that return true if it is possible
// to divide the array into sub-arrays
// that satisfy the given condition
static boolean solve(int array[], int size, int k)
{
    // To store the sum of all the elements
    // of the array
    int totalSum = 0;
    for (int i = 0; i < size; i++)
        totalSum += array[i];
  
    // Sum of any sub-array cannot be
    // a floating point value
    if ((totalSum - k) % 2 == 1)
        return false;
  
    // Required sub-array sum
    int S = (totalSum - k) / 2;
  
    int sum = 0;
    for (int i = 0; i < size; i++) 
    {
        sum += array[i];
        if (sum == S)
            return true;
    }
    return false;
}
  
    // Driver Code
    public static void main (String[] args)
    {
        int array[] = { 2, 4, 1, 5 };
        int k = 2;
        int size = array.length;
          
        if (solve(array, size, k))
            System.out.println ("Yes");
        else
            System.out.println ("No" );
    }
}
  
// This Code is contributed by akt_mit 

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Python3

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# Function that return true if it is possible
# to divide the array into sub-arrays
# that satisfy the given condition
def solve(array,size,k):
    # To store the sum of all the elements
    # of the array
    totalSum = 0
    for i in range (0,size):
        totalSum += array[i]
  
    # Sum of any sub-array cannot be
    # a floating point value
    if ((totalSum - k) % 2 == 1):
        return False
  
    # Required sub-array sum
    S = (totalSum - k) / 2
  
    sum = 0;
    for i in range (0,size):
        sum += array[i]
        if (sum == S):
            return True
      
  
    return False
  
  
# Driver Code
array= [2, 4, 1, 5]
k = 2
n = 4
if (solve(array, n, k)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by iAyushRaj.

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C#

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using System;
class GFG
{
  
// Function that return true if it is possible
// to divide the array into sub-arrays
// that satisfy the given condition
public static bool solve(int[] array, int size, int k)
{
    // To store the sum of all the elements
    // of the array
    int totalSum = 0;
    for (int i = 0; i < size; i++)
        totalSum += array[i];
  
    // Sum of any sub-array cannot be
    // a floating point value
    if ((totalSum - k) % 2 == 1)
        return false;
  
    // Required sub-array sum
    int S = (totalSum - k) / 2;
  
    int sum = 0;
    for (int i = 0; i < size; i++) 
    {
        sum += array[i];
        if (sum == S)
            return true;
    }
  
    return false;
}
  
// Driver Code
public static void Main()
{
    int[] array = { 2, 4, 1, 5 };
    int k = 2;
    int size = 4;
      
    if (solve(array, size, k))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by iAyushRaj.

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PHP

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<?php
// Function that return true if it is possible
// to divide the array into sub-arrays
// that satisfy the given condition
function solve($array, $size,$k)
{
    // To store the sum of all the elements
    // of the array
    $totalSum = 0;
    for ($i = 0; $i < $size; $i++)
        $totalSum += $array[$i];
  
    // Sum of any sub-array cannot be
    // a floating point value
    if (($totalSum - $k) % 2 == 1)
        return false;
  
    // Required sub-array sum
    $S = ($totalSum - $k) / 2;
  
    $sum = 0;
    for ($i = 0; $i < $size; $i++) 
    {
        $sum += $array[$i];
        if ($sum == $S)
            return true;
    }
  
    return false;
}
  
// Driver Code
$array = array( 2, 4, 1, 5 );
$k = 2;
$size = sizeof($array);
if (solve($array, $size, $k))
    echo "Yes";
else
    echo "No";
  
// This code is contributed by iAyushRaj.
?>

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Output:

No


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Improved By : jit_t, iAyushRaj