Make array elements equal with minimum cost
Given an array of integers arr[], the task is to find the minimum number of steps to make array elements equal by following two operations –
- Add or subtract 2 from the element with 0 cost
- Add or subtract 1 from the element with 1 cost
Examples:
Input: arr[] = {4, 3, 2, 1}
Output: 2
Explanation:
As in the above example all the array elements can be equal to the 4 or 3 with 2 as minimum cost
Cost of Making Array Elements equal to 4
Index 0: As index 0 is already equal to 4, the cost is 0
Index 1: 4 – 3 = 1, Adding 1 to the element will cost 1
Index 2: 4 – 2 = 2, Adding 2 to the element will cost 0
Index 3: 4 – 1 = 3, Add 2 and 1 once in the array element, which will cost 1, because 2 can be added any number of times with cost 0.
Total Cost = 1 + 1 = 2
Input: arr[] = {3, 2, 1}
Output: 1
Explanation:
As in the above example all the array elements can be made equal to 3 or 1 with 1 as minimum cost
Cost of Making Array Elements equal to 3
Index 0: As index 0 is already equal to 3, the cost is 0
Index 1: 3 – 2 = 1, Adding 1 to the element will cost 1
Index 2: 3 – 1 = 2, Adding 2 to the element will cost 0
Total Cost = 1
Approach: The idea is to use the fact that adding 2 to any element of the array will cost 0. Also the second observation is that adding 2 to any odd number will result in an Odd number and similarly adding 2 to an even number will result in an even number only. Therefore, any element can be made equal to a nearest integer at zero cost but with the same property that it will remain odd if it is odd as initial value or it will remain even if it is even as initial. So, to find the minimum cost we can find the minimum count of the odd or even elements in the array.
Algorithm:
- Find the count of the odd or even elements in the array (say count).
- Find the minimum value between count and (length – count) where length is the length of the array.
- The minimum value will be desired cost to make all array elements equal.
Explanation with Example:
Given Array be - {4, 2, 3, 1}
Count of Odd Elements - 2 ({3, 1})
Count of Even Elements - 2 ({4, 2})
Hence, the minimum cost required is 2 and
the array elements can be made equal
to any integer of even or odd value
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void makearrayequal( int arr[], int n)
{
int x = 0;
for ( int i = 0; i < n; i++) {
x += arr[i] & 1;
}
cout << min(x, n - x) << endl;
}
int main()
{
int arr[] = { 4, 3, 2, 1 };
int n = sizeof (arr) / sizeof (arr[0]);
makearrayequal(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void makearrayequal( int arr[], int n)
{
int x = 0 ;
for ( int i = 0 ; i < n; i++) {
x += (arr[i] & 1 );
}
System.out.println(Math.min(x, n - x));
}
public static void main (String[] args)
{
int arr[] = { 4 , 3 , 2 , 1 };
int n = arr.length;
makearrayequal(arr, n);
}
}
|
Python3
def makearrayequal(arr, n) :
x = 0 ;
for i in range (n) :
x + = arr[i] & 1 ;
print ( min (x, n - x));
if __name__ = = "__main__" :
arr = [ 4 , 3 , 2 , 1 ];
n = len (arr);
makearrayequal(arr, n);
|
C#
using System;
class GFG {
static void makearrayequal( int []arr, int n)
{
int x = 0;
for ( int i = 0; i < n; i++) {
x += (arr[i] & 1);
}
Console.WriteLine(Math.Min(x, n - x));
}
public static void Main ( string [] args)
{
int []arr = { 4, 3, 2, 1 };
int n = arr.Length;
makearrayequal(arr, n);
}
}
|
Javascript
<script>
function makearrayequal(arr , n)
{
var x = 0;
for (i = 0; i < n; i++)
{
x += (arr[i] & 1);
}
document.write(Math.min(x, n - x));
}
var arr = [ 4, 3, 2, 1 ];
var n = arr.length;
makearrayequal(arr, n);
</script>
|
Performance Analysis:
- Time Complexity: O(N).
- Auxiliary Space: O(1)
Last Updated :
14 Dec, 2022
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