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 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;
} |
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 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 |
Python3
# 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. |
C#
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. |
PHP
<?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. ?> |
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 condition function 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)