# Minimum count of array elements that must be changed such that difference between maximum and minimum array element is N – 1

• Difficulty Level : Easy
• Last Updated : 27 Oct, 2021

Given an array arr[] consisting of N integers, the task is to find the minimum number of array elements that must be changed to any arbitrary integers such that the difference between maximum and minimum array element is (N – 1) and all array elements must be distinct.

Examples:

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Input: arr[] = {1, 2, 3, 5, 6}
Output: 1
Explanation:
Change 6->4, the final array will be {1, 2, 3, 5, 4} and the difference is equal to 5 – 1 = 4.

Input: arr[] = {1, 10, 100, 1000}
Output: 3

Approach: The given problem can be solved by using the Sorting and Sliding Window Technique. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the minimum changes``// needed to make difference of maximum``// and minimum element as (N - 1)``int` `minOperations(vector<``int``>& A)``{``    ``int` `N = A.size();` `    ``// Stores the resultant count``    ``int` `ans = N;` `    ``// Maintain a pointer j that will``    ``// denote the rightmost position of``    ``// the valid array``    ``int` `j = 0;` `    ``// Sort the array``    ``sort(begin(A), end(A));` `    ``// Only keep unique elements``    ``A.erase(unique(begin(A), end(A)),``            ``end(A));` `    ``// Store the new size of the array``    ``// after removing duplicates``    ``int` `M = A.size();``  ` `    ``// IterM;ate over the range``    ``for` `(``int` `i = 0; i < M; ++i) {``        ``while` `(j < M && A[j] <= A[i] + N - 1) {``            ``++j;``        ``}` `        ``// Check minimum over this``        ``// starting point``        ``ans = min(ans, N - j + i);` `        ``// The length of this subarray``        ``// is `j - i`. Replace `N - j + i```        ``// elements to make it good``    ``}` `    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``vector<``int``> arr = { 1, 10, 100, 1000 };``    ``cout << minOperations(arr);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.Arrays;``import` `java.util.*;` `class` `GFG {` `    ``// Function to find the minimum changes``    ``// needed to make difference of maximum``    ``// and minimum element as (N - 1)``    ``public` `static` `int` `minOperations(``int``[] A) {``        ``int` `N = A.length;` `        ``// Stores the resultant count``        ``int` `ans = N;` `        ``// Maintain a pointer j that will``        ``// denote the rightmost position of``        ``// the valid array``        ``int` `j = ``0``;` `        ``// Sort the array``        ``Arrays.sort(A);` `        ``// Only keep unique elements``        ``removeDups(A);` `        ``// Store the new size of the array``        ``// after removing duplicates``        ``int` `M = A.length;` `        ``// IterM;ate over the range``        ``for` `(``int` `i = ``0``; i < M; ++i) {``            ``while` `(j < M && A[j] <= A[i] + N - ``1``) {``                ``++j;``            ``}` `            ``// Check minimum over this``            ``// starting point``            ``ans = Math.min(ans, N - j + i);` `            ``// The length of this subarray``            ``// is `j - i`. Replace `N - j + i```            ``// elements to make it good``        ``}` `        ``return` `ans;``    ``}` `    ``public` `static` `void` `removeDups(``int``[] a) {` `        ``LinkedHashSet set = ``new` `LinkedHashSet();` `        ``// adding elements to LinkedHashSet``        ``for` `(``int` `i = ``0``; i < a.length; i++)``            ``set.add(a[i]);` `    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String args[]) {``        ``int``[] arr = { ``1``, ``10``, ``100``, ``1000` `};``        ``System.out.println(minOperations(arr));` `    ``}` `}` `// This code is contributed by saurabh_jaiswal.`

## Python3

 `# Python 3 program for the above approach` `# Function to find the minimum changes``# needed to make difference of maximum``# and minimum element as (N - 1)``def` `minOperations(A):``    ``N ``=` `len``(A)` `    ``# Stores the resultant count``    ``ans ``=` `N` `    ``# Maintain a pointer j that will``    ``# denote the rightmost position of``    ``# the valid array``    ``j ``=` `0` `    ``# Sort the array``    ``A.sort()` `    ``# Only keep unique elements``    ``A ``=` `list``(``set``(A))` `    ``# Store the new size of the array``    ``# after removing duplicates``    ``A.sort()``    ``M ``=` `len``(A)``  ` `    ``# IterM;ate over the range``    ``for` `i ``in` `range``(M):``        ``while` `(j < M ``and` `A[j] <``=` `A[i] ``+` `N ``-` `1``):``            ``j ``+``=` `1` `        ``# Check minimum over this``        ``# starting point``        ``ans ``=` `min``(ans, N ``-` `j ``+` `i)` `        ``# The length of this subarray``        ``# is `j - i`. Replace `N - j + i```        ``# elements to make it good` `    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``10``, ``100``, ``1000``]``    ``print``(minOperations(arr))``    ` `    ``# This code is contributed by ipg2016107.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG {` `    ``// Function to find the minimum changes``    ``// needed to make difference of maximum``    ``// and minimum element as (N - 1)``    ``public` `static` `int` `minOperations(``int``[] A) {``        ``int` `N = A.Length;` `        ``// Stores the resultant count``        ``int` `ans = N;` `        ``// Maintain a pointer j that will``        ``// denote the rightmost position of``        ``// the valid array``        ``int` `j = 0;` `        ``// Sort the array``        ``Array.Sort(A);` `        ``// Only keep unique elements``        ``removeDups(A);` `        ``// Store the new size of the array``        ``// after removing duplicates``        ``int` `M = A.Length;` `        ``// IterM;ate over the range``        ``for` `(``int` `i = 0; i < M; ++i) {``            ``while` `(j < M && A[j] <= A[i] + N - 1) {``                ``++j;``            ``}` `            ``// Check minimum over this``            ``// starting point``            ``ans = Math.Min(ans, N - j + i);` `            ``// The length of this subarray``            ``// is `j - i`. Replace `N - j + i```            ``// elements to make it good``        ``}` `        ``return` `ans;``    ``}` `    ``public` `static` `void` `removeDups(``int``[] a) {` `        ``HashSet<``int``> ``set` `= ``new` `HashSet<``int``>();` `        ``// adding elements to LinkedHashSet``        ``for` `(``int` `i = 0; i < a.Length; i++)``            ``set``.Add(a[i]);` `    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main() {``        ``int``[] arr = { 1, 10, 100, 1000 };``        ``Console.Write(minOperations(arr));``    ``}` `}` `// This code is contributed by saurabh_jaiswal.`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

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