Find the minimum value to be added so that array becomes balanced
Last Updated :
19 Sep, 2023
Given an array of even size, task is to find minimum value that can be added to an element so that array become balanced. An array is balanced if the sum of the left half of the array elements is equal to the sum of right half. Suppose, we have an array 1 3 1 2 4 3. The Sum of first three elements is 1 + 3 + 1 = 5 and sum of last three elements is 2 + 4 + 3 = 9
So this is unbalanced, to make it balanced the minimum number we can add is 4 to any element in first half.
Examples :
Input : 1 2 1 2 1 3
Output : 2
Sum of first 3 elements is 1 + 2 + 1 = 4,
sum of last three elements is 2 + 1 + 3 = 6
To make the array balanced you can add 2.
Input : 20 10
Output : 10
The idea is simple, we compute sums of first and second halves. Once these sums are computed, we return absolute difference of these two values.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int minValueToBalance( int a[], int n)
{
int sum1 = 0;
for ( int i = 0; i < n/2; i++)
sum1 += a[i];
int sum2 = 0;
for ( int i = n/2; i < n; i++)
sum2 += a[i];
return abs (sum1 - sum2);
}
int main()
{
int arr[] = {1, 7, 1, 1, 3, 1};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << minValueToBalance(arr, n)<<endl;
return 0;
}
|
Java
class Minimum
{
public static int minValueToBalance( int a[],
int n)
{
int sum1 = 0 ;
for ( int i = 0 ; i < n / 2 ; i++)
sum1 += a[i];
int sum2 = 0 ;
for ( int i = n/ 2 ; i < n; i++)
sum2 += a[i];
return Math.abs(sum1 - sum2);
}
public static void main(String[] args)
{
int arr[] = { 1 , 7 , 1 , 1 , 3 , 1 };
int n = 6 ;
System.out.print(minValueToBalance(arr, n));
}
}
|
Python3
def minValueToBalance(a, n):
sum1 = 0
for i in range ( int (n / 2 )):
sum1 + = a[i]
sum2 = 0 ;
i = int (n / 2 )
while i < n:
sum2 + = a[i]
i = i + 1
return abs (sum1 - sum2)
arr = [ 1 , 7 , 1 , 1 , 3 , 1 ]
n = len (arr)
print (minValueToBalance(arr, n))
|
C#
using System;
class Minimum {
public static int minValueToBalance( int []a,
int n)
{
int sum1 = 0;
for ( int i = 0; i < n / 2; i++)
sum1 += a[i];
int sum2 = 0;
for ( int i = n / 2; i < n; i++)
sum2 += a[i];
return Math.Abs(sum1 - sum2);
}
public static void Main()
{
int []arr = {1, 7, 1, 1, 3, 1};
int n = 6;
Console.Write(minValueToBalance(arr, n));
}
}
|
PHP
<?php
function minValueToBalance( $a , $n )
{
$sum1 = 0;
for ( $i = 0; $i < $n / 2; $i ++)
$sum1 += $a [ $i ];
$sum2 = 0;
for ( $i = $n / 2; $i < $n ; $i ++)
$sum2 += $a [ $i ];
return abs ( $sum1 - $sum2 );
}
$arr = array (1, 7, 1, 1, 3, 1);
$n = sizeof( $arr ) / sizeof( $arr [0]);
echo minValueToBalance( $arr , $n );
?>
|
Javascript
<script>
function minValueToBalance(a, n)
{
let sum1 = 0;
for (let i = 0; i < Math.floor(n/2); i++)
sum1 += a[i];
let sum2 = 0;
for (let i = Math.floor(n/2); i < n; i++)
sum2 += a[i];
return Math.abs(sum1 - sum2);
}
let arr = [1, 7, 1, 1, 3, 1];
let n = arr.length;
document.write(minValueToBalance(arr, n) + "<br>" );
</script>
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Time complexity: O(n), where n is the size of the input array.
Space complexity: O(1), as the space used is constant irrespective of the input size.
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