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| = 0Input: 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++
#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 conditionbool 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 Codeint 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;} |
Java
/*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 conditionstatic 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 |
Python3
# Function that return true if it is possible# to divide the array into sub-arrays# that satisfy the given conditiondef 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 Codearray= [2, 4, 1, 5]k = 2n = 4if (solve(array, n, k)): print("Yes")else: print("No")# This code is contributed by iAyushRaj. |
C#
using System;class GFG{// Function that return true if it is possible// to divide the array into sub-arrays// that satisfy the given conditionpublic 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 Codepublic 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. |
PHP
<?php// Function that return true if it is possible// to divide the array into sub-arrays// that satisfy the given conditionfunction 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.?> |
Javascript
<script>// Javascript program to illustrate // the above problem // Function that return true if it is possible// to divide the array into sub-arrays// that satisfy the given conditionfunction solve(array, size, k){ // To store the sum of all the elements // of the array let totalSum = 0; for (let 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 let S = (totalSum - k) / 2; let sum = 0; for (let i = 0; i < size; i++) { sum += array[i]; if (sum == S) return true; } return false;}// Driver Code let array = [ 2, 4, 1, 5 ]; let k = 2; let size = array.length; if (solve(array, size, k)) document.write("Yes"); else document.write ("No" );</script> |
Output
No
Complexity Analysis:
- Time Complexity: O(n) where n is the size of the array.
- Auxiliary Space: O(1)
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