Given an array arr[] and an integer K, the task is to find the minimum number of operations to make all the elements of the array arr[] equal. In one operation, K is subtracted from an element and this K is added into other element. If it is not possible then print -1.
Examples:
Input: arr[] = {5, 8, 11}, K = 3
Output: 1
Explanation:
Operation 1: Subtract 3 from arr[2](=11) and add 3 to arr[0](=5).
Now, all the elements of the array arr[] are equal.
Input: arr[] = {1, 2, 3, 4}, K = 2
Output: -1
Approach: This problem can be solved using the Greedy algorithm. First check the sum of all elements of the array arr[] is divisible by N or not. If it is not divisible that means it is impossible to make all elements of the array arr[] equal. Otherwise, try to add or subtract the value of K to make every element equal to sum of arr[] / N. Follow the steps below to solve this problem:
- Initialize a variable sum as 0 to store sum of all elements of the array arr[].
- Iterate in the range [0, N-1] using the variable i and update sum as sum + arr[i].
- If sum % N is not equal to 0, then print -1 and return.
- Initialize a variable valueAfterDivision as sum/N to store that this much value, each element of array arr[] is store and count as 0 to store the minimum number of operation needed to make all array element equal.
- Iterate in the range [0, N-1] using the variable i:
- If abs of valueAfterDivision – arr[i] % K is not equal to 0, then print -1 and return.
- Update count as count + abs of valueAfterDivision – arr[i] / K.
- After completing the above steps, print count/2 as the answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void miniOperToMakeAllEleEqual(int arr[], int n, int k)
{
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
}
if (sum % n) {
cout << -1;
return;
}
int valueAfterDivision = sum / n;
int count = 0;
for (int i = 0; i < n; i++) {
if (abs(valueAfterDivision - arr[i]) % k != 0) {
cout << -1;
return;
}
count += abs(valueAfterDivision - arr[i]) / k;
}
cout << count / 2 << endl;
}
int main()
{
int n = 3, k = 3;
int arr[3] = { 5, 8, 11 };
miniOperToMakeAllEleEqual(arr, n, k);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void miniOperToMakeAllEleEqual(int arr[], int n, int k)
{
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
}
if (sum % n != 0) {
System.out.println(-1);
return;
}
int valueAfterDivision = sum / n;
int count = 0;
for (int i = 0; i < n; i++) {
if (Math.abs(valueAfterDivision - arr[i]) % k != 0) {
System.out.println(-1);
return;
}
count += Math.abs(valueAfterDivision - arr[i]) / k;
}
System.out.println((int)count / 2);
}
public static void main (String[] args)
{
int n = 3, k = 3;
int arr[] = { 5, 8, 11 };
miniOperToMakeAllEleEqual(arr, n, k);
}
}
|
Python3
def miniOperToMakeAllEleEqual(arr, n, k):
sum = 0
for i in range(n):
sum += arr[i]
if (sum % n):
print(-1)
return
valueAfterDivision = sum // n
count = 0
for i in range(n):
if (abs(valueAfterDivision - arr[i]) % k != 0):
print(-1)
return
count += abs(valueAfterDivision - arr[i]) // k
print(count // 2)
if __name__ == '__main__':
n = 3
k = 3
arr = [ 5, 8, 11 ]
miniOperToMakeAllEleEqual(arr, n, k)
|
C#
using System;
class GFG
{
static void miniOperToMakeAllEleEqual(int[] arr, int n,
int k)
{
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
}
if (sum % n != 0) {
Console.WriteLine(-1);
return;
}
int valueAfterDivision = sum / n;
int count = 0;
for (int i = 0; i < n; i++) {
if (Math.Abs(valueAfterDivision - arr[i]) % k
!= 0) {
Console.WriteLine(-1);
return;
}
count += Math.Abs(valueAfterDivision - arr[i])
/ k;
}
Console.WriteLine((int)count / 2);
}
static void Main()
{
int n = 3, k = 3;
int[] arr = { 5, 8, 11 };
miniOperToMakeAllEleEqual(arr, n, k);
}
}
|
Javascript
<script>
function miniOperToMakeAllEleEqual(arr, n, k)
{
let sum = 0;
for (let i = 0; i < n; i++) {
sum += arr[i];
}
if (sum % n) {
document.write(-1);
return;
}
let valueAfterDivision = sum / n;
let count = 0;
for (let i = 0; i < n; i++) {
if (Math.abs(valueAfterDivision - arr[i]) % k != 0) {
document.write(-1);
return;
}
count += Math.abs(valueAfterDivision - arr[i]) / k;
}
document.write(Math.floor(count / 2));
}
let n = 3, k = 3;
let arr = [5, 8, 11];
miniOperToMakeAllEleEqual(arr, n, k);
</script>
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Time Complexity: O(N)
Auxiliary Space: O(1)