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++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the minimum
// number of operations required
int changeTheArray(int arr[], int n)
{
// Minimum and maximum elements from the array
int minEle = *(std::min_element(arr, arr + n));
int maxEle = *(std::max_element(arr, arr + n));
// To store the minimum number of
// operations required
int minOperations = INT_MAX;
for (int num = minEle; num <= maxEle; num++) {
// To store the number of operations required
// to change every element to either
// (num - 1), num or (num + 1)
int operations = 0;
for (int i = 0; i < n; i++) {
// If current element is not already num
if (arr[i] != num) {
// Add the count of operations
// required to change arr[i]
operations += (abs(num - arr[i]) - 1);
}
}
// Update the minimum operations so far
minOperations = min(minOperations, operations);
}
return minOperations;
}
// Driver code
int main()
{
int arr[] = { 10, 1, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << changeTheArray(arr, n);
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG {
// Function to return the minimum
// number of operations required
static int changeTheArray(int arr[], int n)
{
// Minimum and maximum elements from the array
int minEle = Arrays.stream(arr).min().getAsInt();
int maxEle = Arrays.stream(arr).max().getAsInt();
// To store the minimum number of
// operations required
int minOperations = Integer.MAX_VALUE;
for (int num = minEle; num <= maxEle; num++) {
// To store the number of operations required
// to change every element to either
// (num - 1), num or (num + 1)
int operations = 0;
for (int i = 0; i < n; i++) {
// If current element is not already num
if (arr[i] != num) {
// Add the count of operations
// required to change arr[i]
operations += (Math.abs(num - arr[i]) - 1);
}
}
// Update the minimum operations so far
minOperations = Math.min(minOperations, operations);
}
return minOperations;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 10, 1, 4 };
int n = arr.length;
System.out.println(changeTheArray(arr, n));
}
}
// This code has been contributed by 29AjayKumar
# Python3 implementation of the approach
import math
import sys
# Function to return the minimum
# number of operations required
def changeTheArray(arr, n):
# Minimum and maximum elements
# from the array
minEle = min(arr)
maxEle = max(arr)
# To store the minimum number of
# operations required
minOperations = sys.maxsize
for num in range(minEle, maxEle + 1):
# To store the number of operations required
# to change every element to either
# (num - 1), num or (num + 1)
operations = 0
for i in range(n):
# If current element is not already num
if arr[i] != num:
operations += (abs(num - arr[i]) - 1)
# Update the minimum operations so far
minOperations = min(minOperations, operations)
return minOperations
# Driver code
if __name__=='__main__':
arr = [10, 1, 4]
n = len(arr)
print(changeTheArray(arr, n))
# This code is contributed by Vikash Kumar 37
// C# implementation of the approach
using System;
using System.Linq;
class GFG
{
// Function to return the minimum
// number of operations required
static int changeTheArray(int []arr, int n)
{
// Minimum and maximum elements from the array
int minEle = arr.Min();
int maxEle = arr.Max();
// To store the minimum number of
// operations required
int minOperations = int.MaxValue;
for (int num = minEle; num <= maxEle; num++)
{
// To store the number of operations required
// to change every element to either
// (num - 1), num or (num + 1)
int operations = 0;
for (int i = 0; i < n; i++)
{
// If current element is not already num
if (arr[i] != num)
{
// Add the count of operations
// required to change arr[i]
operations += (Math.Abs(num - arr[i]) - 1);
}
}
// Update the minimum operations so far
minOperations = Math.Min(minOperations, operations);
}
return minOperations;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 10, 1, 4 };
int n = arr.Length;
Console.WriteLine(changeTheArray(arr, n));
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript implementation of the approach
// Function to return the minimum
// number of operations required
function changeTheArray(arr, n)
{
// Minimum and maximum elements from the array
let minEle = Math.min(...arr);
let maxEle = Math.max(...arr);
// To store the minimum number of
// operations required
let minOperations = Number.MAX_VALUE;
for (let num = minEle; num <= maxEle; num++) {
// To store the number of operations required
// to change every element to either
// (num - 1), num or (num + 1)
let operations = 0;
for (let i = 0; i < n; i++) {
// If current element is not already num
if (arr[i] != num) {
// Add the count of operations
// required to change arr[i]
operations += (Math.abs(num - arr[i]) - 1);
}
}
// Update the minimum operations so far
minOperations = Math.min(minOperations, operations);
}
return minOperations;
}
// Driver code
let arr = [ 10, 1, 4 ];
let n = arr.length;
document.write(changeTheArray(arr, n));
</script>
Time Complexity: O((maxEle-minEle)*n)
Auxiliary Space: O(1), as no extra space has been taken.
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