Find maximum array sum after making all elements same with repeated subtraction
Last Updated :
12 Sep, 2023
Given an array of n elements, find out the maximum possible sum of all elements such that all the elements are equal. Only operation allowed is choosing any two elements and replacing the larger of them by the absolute difference of the two.
Examples:
Input : 9 12 3 6
Output : 12
Explanation :
9 12 3 6
replace a2 = 12 with a2-a4 = 12 - 6 => 6
i.e, 9 6 3 6
replace a4 = 6 with a4-a3 = 6 - 3 => 3
i.e, 9 6 3 3
replace a1 = 9 with a1-a2 = 9 - 6 => 3
i.e, 3 6 3 3
replace a2 = 6 with a2-a4 = 6 - 3 => 3
i,e. 3 3 3 3
Now, at this point we have all the elements equal,
hence we can return our answer from here.
Input : 4 8 6 10
Output : 8
Explanation :
Resultant array formed will be:
4 8 6 10
replace a4 = 10 with a4-a1 = 10 - 4 => 6
i.e, 4 8 6 6
replace a3 = 6 with a3-a1 = 6 - 4 => 2
i.e, 4 8 2 6
replace a2 = 8 with a2-a4 = 8 - 6 => 2
i.e, 4 2 2 6
replace a4 = 6 with a4-a1 = 6 - 4 => 2
i,e. 4 2 2 2
replace a1 = 4 with a1-a2 = 4 - 2 => 2
i,e. 2 2 2 2
Now, at this point we have all the elements equal,
hence we can return our answer from here.
By analyzing the given operation i.e,
ai = ai - aj where ai > aj
We see that this is similar to finding GCD through Euclidean algorithm as:
GCD(a, b) = GCD(b, a - b)
And also, the order of rearrangement does not matter, we can proceed by taking any two elements and replace the larger value by the absolute difference of the two, and repeat among them till the difference comes out to be zero[both the elements be same]. That is, taking out the GCD of any two numbers. The reason for this to work is, GCD is associative and commutative.
So the idea is the take the GCD of all the elements at once and replace all the elements by that result.
Implementation:
C++
#include<bits/stdc++.h>
using namespace std;
int GCD( int a, int b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
int findMaxSumUtil( int arr[], int n)
{
int finalGCD = arr[0];
for ( int i = 1; i < n; i++)
finalGCD = GCD(arr[i], finalGCD);
return finalGCD;
}
int findMaxSum( int arr[], int n)
{
int maxElement = findMaxSumUtil(arr, n);
return (maxElement * n);
}
int main()
{
int arr[] = {8, 20, 12, 36};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << findMaxSum(arr, n) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
static int GCD( int a, int b)
{
if (b == 0 )
return a;
return GCD(b, a % b);
}
static int findMaxSumUtil( int arr[], int n)
{
int finalGCD = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
finalGCD = GCD(arr[i], finalGCD);
return finalGCD;
}
static int findMaxSum( int arr[], int n)
{
int maxElement = findMaxSumUtil(arr, n);
return (maxElement * n);
}
public static void main (String[] args) {
int arr[] = { 8 , 20 , 12 , 36 };
int n = arr.length;
System.out.println(findMaxSum(arr, n));
}
}
|
Python3
def GCD(a, b):
if (b = = 0 ): return a
return GCD(b, a % b)
def findMaxSumUtil(arr, n):
finalGCD = arr[ 0 ]
for i in range ( 1 , n):
finalGCD = GCD(arr[i], finalGCD)
return finalGCD
def findMaxSum(arr, n):
maxElement = findMaxSumUtil(arr, n)
return (maxElement * n)
arr = [ 8 , 20 , 12 , 36 ]
n = len (arr)
print (findMaxSum(arr, n))
|
C#
using System;
class GFG {
static int GCD( int a, int b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
static int findMaxSumUtil( int []arr, int n)
{
int finalGCD = arr[0];
for ( int i = 1; i < n; i++)
finalGCD = GCD(arr[i], finalGCD);
return finalGCD;
}
static int findMaxSum( int []arr, int n)
{
int maxElement = findMaxSumUtil(arr, n);
return (maxElement * n);
}
public static void Main () {
int []arr = {8, 20, 12, 36};
int n = arr.Length;
Console.WriteLine(findMaxSum(arr, n));
}
}
|
PHP
<?php
function GCD( $a , $b )
{
if ( $b == 0)
return $a ;
return GCD( $b , $a % $b );
}
function findMaxSumUtil( $arr , $n )
{
$finalGCD = $arr [0];
for ( $i = 1; $i < $n ; $i ++)
$finalGCD = GCD( $arr [ $i ], $finalGCD );
return $finalGCD ;
}
function findMaxSum( $arr , $n )
{
$maxElement = findMaxSumUtil( $arr , $n );
return ( $maxElement * $n );
}
$arr = array (8, 20, 12, 36);
$n = count ( $arr );
echo findMaxSum( $arr , $n ) ;
?>
|
Javascript
<script>
function GCD(a, b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
function findMaxSumUtil(arr, n)
{
let finalGCD = arr[0];
for (let i = 1; i < n; i++)
finalGCD = GCD(arr[i], finalGCD);
return finalGCD;
}
function findMaxSum(arr, n)
{
let maxElement = findMaxSumUtil(arr, n);
return (maxElement * n);
}
let arr = [8, 20, 12, 36];
let n = arr.length;
document.write(findMaxSum(arr, n));
</script>
|
Time Complexity: O(N * log max(a, b)), where N is the size of the given array and a & b are the elements of the array of which gcd is to be done.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Share your thoughts in the comments
Please Login to comment...