Minimum operations to make GCD of array a multiple of k
Given an array and k, we need to find the minimum operations needed to make GCD of the array equal or multiple of k. Here an operation means either increment or decrements an array element by 1.
Examples:
Input : a = { 4, 5, 6 }, k = 5
Output : 2
Explanation : We can increase 4 by 1 so that it becomes 5 and decrease 6 by 1 so that it becomes 5. Hence minimum operation will be 2.
Input : a = { 4, 9, 6 }, k = 5
Output : 3
Explanation : Just like the previous example we can increase and decrease 4 and 6 by 1 and increase 9 by 1 so that it becomes 10. Now each element has GCD 5. Hence minimum operation will be 3.
Here we have to make the gcd of the array equal or multiple to k, which means there will be cases in which some elements are near k or to some of its multiple. So, to solve this we just have to make each array value equal to or multiple to K. By doing this we will achieve our solution as if every element is multiple of k then it’s GCD will be at least K. Now our next target is to convert the array elements in the minimum operation i.e. minimum number of increment and decrement. This minimum value of increment or decrement can be known only by taking the remainder of each number from K i.e. either we have to take the remainder value or (k-remainder) value, whichever is minimum among them.
Below is implementation of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
int MinOperation( int a[], int n, int k)
{
int result = 0;
for ( int i = 0; i < n; ++i) {
if (a[i] != 1 && a[i] > k)
result = result + min(a[i] % k, k - a[i] % k);
else
result = result + k - a[i];
}
return result;
}
int main()
{
int arr[] = { 4, 5, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
int k = 5;
cout << MinOperation(arr, n, k);
return 0;
}
|
C
#include <stdio.h>
int min( int num1, int num2)
{
return (num1 > num2) ? num2 : num1;
}
int MinOperation( int a[], int n, int k)
{
int result = 0;
for ( int i = 0; i < n; ++i) {
if (a[i] != 1 && a[i] > k)
result = result + min(a[i] % k, k - a[i] % k);
else
result = result + k - a[i];
}
return result;
}
int main()
{
int arr[] = { 4, 5, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
int k = 5;
printf ( "%d" , MinOperation(arr, n, k));
return 0;
}
|
Java
import java.io.*;
class GFG {
static int MinOperation( int a[], int n, int k)
{
int result = 0 ;
for ( int i = 0 ; i < n; ++i) {
if (a[i] != 1 && a[i] > k)
result = result + Math.min(a[i] % k, k - a[i] % k);
else
result = result + k - a[i];
}
return result;
}
public static void main(String[] args)
{
int arr[] = { 4 , 5 , 6 };
int n = arr.length;
int k = 5 ;
System.out.println(MinOperation(arr, n, k));
}
}
|
Python3
def MinOperation(a, n, k):
result = 0
for i in range (n) :
if (a[i] ! = 1 and a[i] > k) :
result = (result + min (a[i] % k,
k - a[i] % k))
else :
result = result + k - a[i]
return result
if __name__ = = "__main__" :
arr = [ 4 , 5 , 6 ]
n = len (arr)
k = 5
print (MinOperation(arr, n, k))
|
C#
using System;
public class GFG{
static int MinOperation( int []a,
int n, int k)
{
int result = 0;
for ( int i = 0; i < n; ++i)
{
if (a[i] != 1 && a[i] > k)
{
result = result +
Math.Min(a[i] % k,
k - a[i] % k);
}
else
{
result = result + k - a[i];
}
}
return result;
}
static public void Main (){
int []arr = {4, 5, 6};
int n = arr.Length;
int k = 5;
Console.WriteLine(MinOperation(arr, n, k));
}
}
|
PHP
<?php
function MinOperation( $a , $n , $k )
{
$result = 0;
for ( $i = 0; $i < $n ; ++ $i )
{
if ( $a [ $i ] != 1 && $a [ $i ] > $k )
{
$result = $result + min( $a [ $i ] %
$k , $k -
$a [ $i ] % $k );
}
else
{
$result = $result +
$k - $a [ $i ];
}
}
return $result ;
}
$arr = array (4, 5, 6);
$n = sizeof( $arr ) /
sizeof( $arr [0]);
$k = 5;
echo MinOperation( $arr , $n , $k );
?>
|
Javascript
<script>
function MinOperation(a, n, k)
{
let result = 0;
for (let i = 0; i < n; ++i)
{
if (a[i] != 1 && a[i] > k)
{
result = result + Math.min(a[i] %
k, k -
a[i] % k);
}
else
{
result = result +
k - a[i];
}
}
return result;
}
let arr = [4, 5, 6];
let n = arr.length;
let k = 5;
document.write(MinOperation(arr, n, k));
</script>
|
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
Last Updated :
17 Aug, 2022
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