Minimum operations required to change the array such that |arr[i] – M| <= 1
Last Updated :
29 Dec, 2022
Given an array arr[] of integers, the task is to find the minimum number of operations required to change the array elements such that for any positive integer M, |arr[i] – M| ? 1 for all valid i.
In a single operation, any element of the array can either be incremented or decremented by 1.
Examples:
Input: arr[] = {10, 1, 4}
Output: 7
If we change 1 into 2 and 10 into 4 with count of operations being |1 – 2| + |10 – 4| = 7
After changing, array becomes {4, 2, 4} where every element’s absolute difference with M = 3 is ? 1
Input: arr[] = {5, 7, 4, 1, 4}
Output: 4
Approach: Starting from the minimum element of the array to the maximum element of the array say num, calculate the count of operations required to change every element such that its absolute difference with num is ? 1. The minimum among all possible operations is the required answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int changeTheArray( int arr[], int n)
{
int minEle = *(std::min_element(arr, arr + n));
int maxEle = *(std::max_element(arr, arr + n));
int minOperations = INT_MAX;
for ( int num = minEle; num <= maxEle; num++) {
int operations = 0;
for ( int i = 0; i < n; i++) {
if (arr[i] != num) {
operations += ( abs (num - arr[i]) - 1);
}
}
minOperations = min(minOperations, operations);
}
return minOperations;
}
int main()
{
int arr[] = { 10, 1, 4 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << changeTheArray(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG {
static int changeTheArray( int arr[], int n)
{
int minEle = Arrays.stream(arr).min().getAsInt();
int maxEle = Arrays.stream(arr).max().getAsInt();
int minOperations = Integer.MAX_VALUE;
for ( int num = minEle; num <= maxEle; num++) {
int operations = 0 ;
for ( int i = 0 ; i < n; i++) {
if (arr[i] != num) {
operations += (Math.abs(num - arr[i]) - 1 );
}
}
minOperations = Math.min(minOperations, operations);
}
return minOperations;
}
public static void main(String args[])
{
int arr[] = { 10 , 1 , 4 };
int n = arr.length;
System.out.println(changeTheArray(arr, n));
}
}
|
Python3
import math
import sys
def changeTheArray(arr, n):
minEle = min (arr)
maxEle = max (arr)
minOperations = sys.maxsize
for num in range (minEle, maxEle + 1 ):
operations = 0
for i in range (n):
if arr[i] ! = num:
operations + = ( abs (num - arr[i]) - 1 )
minOperations = min (minOperations, operations)
return minOperations
if __name__ = = '__main__' :
arr = [ 10 , 1 , 4 ]
n = len (arr)
print (changeTheArray(arr, n))
|
C#
using System;
using System.Linq;
class GFG
{
static int changeTheArray( int []arr, int n)
{
int minEle = arr.Min();
int maxEle = arr.Max();
int minOperations = int .MaxValue;
for ( int num = minEle; num <= maxEle; num++)
{
int operations = 0;
for ( int i = 0; i < n; i++)
{
if (arr[i] != num)
{
operations += (Math.Abs(num - arr[i]) - 1);
}
}
minOperations = Math.Min(minOperations, operations);
}
return minOperations;
}
public static void Main(String []args)
{
int []arr = { 10, 1, 4 };
int n = arr.Length;
Console.WriteLine(changeTheArray(arr, n));
}
}
|
Javascript
<script>
function changeTheArray(arr, n)
{
let minEle = Math.min(...arr);
let maxEle = Math.max(...arr);
let minOperations = Number.MAX_VALUE;
for (let num = minEle; num <= maxEle; num++) {
let operations = 0;
for (let i = 0; i < n; i++) {
if (arr[i] != num) {
operations += (Math.abs(num - arr[i]) - 1);
}
}
minOperations = Math.min(minOperations, operations);
}
return minOperations;
}
let arr = [ 10, 1, 4 ];
let n = arr.length;
document.write(changeTheArray(arr, n));
</script>
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Time Complexity: O((maxEle-minEle)*n)
Auxiliary Space: O(1), as no extra space has been taken.
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