Given an array of N integers, the task is to find the number of operations required to make all elements in the array equal. In one operation we can distribute equal weights from the maximum element to the rest of the array elements. If it is not possible to make the array elements equal after performing the above operations then print -1.
Examples:
Input: arr[] = [1, 6, 1, 1, 1];
Output: 4
Explanation: Since arr becomes [2, 2, 2, 2, 2] after distribution from max element.
Input : arr[] = [2, 2, 3];
Output : -1
Explanation: Here arr becomes [3, 3, 1] after distribution.
Algorithm:
- Declare temporary variable to store number of times operation is performed.
- Find maximum element of the given array and store its index value.
- Check if all the elements are equal to the maximum element after n subtractions.
- Again check that each element is equal to other elements and return n.
Below is the implementation of above approach:
C++
#include<bits/stdc++.h>
using namespace std;
int find_n(int a[],int n)
{
int j = 0, k = 0, s = 0;
int x = *max_element(a, a + n);
int y = *min_element(a, a + n);
for (int i = 0; i < n; i++)
{
if (a[i] == x)
{
s = i;
break;
}
}
for (int i =0;i<n;i++)
{
if (a[i] != x && a[i] <= y && a[i] != 0)
{
a[j] += 1;
a[s] -= 1;
x -= 1;
k += 1;
j += 1;
}
else if (a[i] != 0)
{
j += 1;
}
}
for (int i = 0; i < n; i++)
{
if (a[i] != x)
{
k = -1;
break;
}
}
return k;
}
int main()
{
int a[] = {1, 6, 1, 1, 1};
int n = sizeof(a)/sizeof(a[0]);
cout << (find_n(a, n));
return 0;
}
|
Java
import java.util.Arrays;
class GFG {
static int find_n(int[] a) {
int j = 0, k = 0, s = 0;
int x = Arrays.stream(a).max().getAsInt();
int y = Arrays.stream(a).min().getAsInt();
for (int i : a) {
if (a[i] == x) {
s = i;
break;
}
}
for (int i : a) {
if (i != x && i <= y && i != 0) {
a[j] += 1;
a[s] -= 1;
x -= 1;
k += 1;
j += 1;
} else if (i != 0) {
j += 1;
}
}
for (int i : a) {
if (a[i] != x) {
k = -1;
break;
}
}
return k;
}
public static void main(String[] args) {
int[] a = {1, 6, 1, 1, 1};
System.out.println(find_n(a));
}
}
|
Python3
def find_n(a):
j, k = 0, 0
x = max(a)
for i in range(len(a)):
if(a[i] == x):
s = i
break
for i in a:
if(i != x and i <= min(a) and i !='\0'):
a[j] += 1
a[s] -= 1
x -= 1
k += 1
j += 1
elif(i != '\0'):
j += 1
for i in range(len(a)):
if(a[i] != x):
k = -1
break
return k
a = [1, 6, 1, 1, 1]
print (find_n(a))
|
C#
using System;
using System.Linq;
class GFG
{
static int find_n(int []a)
{
int j = 0, k = 0, s = 0;
int x = a.Max();
int y = a.Min();
foreach(int i in a)
{
if (a[i] == x)
{
s = i;
break;
}
}
foreach (int i in a)
{
if (i != x && i <= y && i != 0)
{
a[j] += 1;
a[s] -= 1;
x -= 1;
k += 1;
j += 1;
}
else if (i != 0)
{
j += 1;
}
}
foreach (int i in a)
{
if (a[i] != x)
{
k = -1;
break;
}
}
return k;
}
public static void Main()
{
int[] a = {1, 6, 1, 1, 1};
Console.Write(find_n(a));
}
}
|
PHP
<?php
function find_n(&$a)
{
$j = 0;
$k = 0;
$x = max($a);
for ($i = 0; $i < sizeof($a); $i++)
{
if($a[$i] == $x)
{
$s = $i;
break;
}
}
for($i = 0; $i < sizeof($a); $i++)
{
if($a[$i] != $x and $a[$i] <= min($a) and
$a[$i] !=0)
{
$a[$j] += 1;
$a[$s] -= 1;
$x -= 1;
$k += 1;
$j += 1;
}
else if($a[$i] != 0)
$j += 1;
}
for($i = 0; $i < sizeof($a); $i++)
{
if($a[$i] != $x)
{
$k = -1;
break;
}
}
return $k;
}
$a = array(1, 6, 1, 1, 1);
echo(find_n($a));
?>
|
Javascript
<script>
function find_n(a) {
var j = 0, k = 0, s = 0;
var x = Math.max.apply(Math, a);
var y = Math.min.apply(Math, a);
for (var i = 0; i < n; i++)
{
if (a[i] == x)
{
s = i;
break;
}
}
for (var i =0;i<n;i++)
{
if (a[i] != x && a[i] <= y && a[i] != 0)
{
a[j] += 1;
a[s] -= 1;
x -= 1;
k += 1;
j += 1;
}
else if (a[i] != 0)
{
j += 1;
}
}
for (var i = 0; i < n; i++)
{
if (a[i] != x)
{
k = -1;
break;
}
}
return k;
}
var a = [1, 6, 1, 1, 1];
var n = a.length;
document.write(find_n(a,n));
</script>
|
Complexity Analysis:
- Time complexity: O(n), where n represents the size of the given array.
- Auxiliary Space: O(1), no extra space is required, so it is a constant.