# Find minimum changes required in an array for it to contain k distinct elements

Given an array arr of size N and a number K. The task is to find the minimum elements to be replaced in the array with any number such that the array consists of K distinct elements.
Note: The array might consist of repeating elements.
Examples:

Input : arr[]={1, 2, 2, 8}, k = 1
Output :
The elements to be changed are 1, 8
Input : arr[]={1, 2, 7, 8, 2, 3, 2, 3}, k = 2
Output :
The elements to be changed are 1, 7, 8

Approach: Since the task is to replace minimum elements from the array so we won’t replace elements which have more frequency in the array. So just define an array freq[] which stores the frequency of each number present in the array arr, then sort freq in descending order. So, first k elements of freq array don’t need to be replaced.
Below is the implementation of the above approach :

## C++

 `// CPP program to minimum changes required ` `// in an array for k distinct elements.` `#include ` `using` `namespace` `std;`   `#define MAX 100005`   `// Function to minimum changes required ` `// in an array for k distinct elements.` `int` `Min_Replace(``int` `arr[], ``int` `n, ``int` `k)` `{` `    ``sort(arr, arr + n);`   `    ``// Store the frequency of each element` `    ``int` `freq[MAX];` `    `  `    ``memset``(freq, 0, ``sizeof` `freq);` `    `  `    ``int` `p = 0;` `    ``freq[p] = 1;` `    `  `    ``// Store the frequency of elements` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``if` `(arr[i] == arr[i - 1])` `            ``++freq[p];` `        ``else` `            ``++freq[++p];` `    ``}`   `    ``// Sort frequencies in descending order` `    ``sort(freq, freq + n, greater<``int``>());` `    `  `    ``// To store the required answer` `    ``int` `ans = 0;` `    ``for` `(``int` `i = k; i <= p; i++)` `        ``ans += freq[i];` `        `  `    ``// Return the required answer` `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 7, 8, 2, 3, 2, 3 };` `    `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    `  `    ``int` `k = 2;` `    `  `    ``cout << Min_Replace(arr, n, k);` `    `  `    ``return` `0;` `}`

## Java

 `// C# program to minimum changes required ` `// in an array for k distinct elements.` `import` `java.util.*;`   `class` `GFG` `{` `    ``static` `int` `MAX = ``100005``;` `    `  `    ``// Function to minimum changes required ` `    ``// in an array for k distinct elements.` `    ``static` `int` `Min_Replace(``int` `[] arr, ` `                           ``int` `n, ``int` `k)` `    ``{` `        ``Arrays.sort(arr);` `    `  `        ``// Store the frequency of each element` `        ``Integer [] freq = ``new` `Integer[MAX];` `        ``Arrays.fill(freq, ``0``);` `        ``int` `p = ``0``;` `        ``freq[p] = ``1``;` `        `  `        ``// Store the frequency of elements` `        ``for` `(``int` `i = ``1``; i < n; i++)` `        ``{` `            ``if` `(arr[i] == arr[i - ``1``])` `                ``++freq[p];` `            ``else` `                ``++freq[++p];` `        ``}` `    `  `        ``// Sort frequencies in descending order` `        ``Arrays.sort(freq, Collections.reverseOrder());` `        `  `        ``// To store the required answer` `        ``int` `ans = ``0``;` `        ``for` `(``int` `i = k; i <= p; i++)` `            ``ans += freq[i];` `            `  `        ``// Return the required answer` `        ``return` `ans;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main (String []args)` `    ``{` `        ``int` `[] arr = { ``1``, ``2``, ``7``, ``8``, ``2``, ``3``, ``2``, ``3` `};` `        `  `        ``int` `n = arr.length;` `        `  `        ``int` `k = ``2``;` `        `  `        ``System.out.println(Min_Replace(arr, n, k));` `    ``}` `}`   `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python 3 program to minimum changes required ` `# in an array for k distinct elements.` `MAX` `=` `100005`   `# Function to minimum changes required ` `# in an array for k distinct elements.` `def` `Min_Replace(arr, n, k):` `    ``arr.sort(reverse ``=` `False``)`   `    ``# Store the frequency of each element` `    ``freq ``=` `[``0` `for` `i ``in` `range``(``MAX``)]` `    `  `    ``p ``=` `0` `    ``freq[p] ``=` `1` `    `  `    ``# Store the frequency of elements` `    ``for` `i ``in` `range``(``1``, n, ``1``):` `        ``if` `(arr[i] ``=``=` `arr[i ``-` `1``]):` `            ``freq[p] ``+``=` `1` `        ``else``:` `            ``p ``+``=` `1` `            ``freq[p] ``+``=` `1`   `    ``# Sort frequencies in descending order` `    ``freq.sort(reverse ``=` `True``)` `    `  `    ``# To store the required answer` `    ``ans ``=` `0` `    ``for` `i ``in` `range``(k, p ``+` `1``, ``1``):` `        ``ans ``+``=` `freq[i]` `        `  `    ``# Return the required answer` `    ``return` `ans`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``1``, ``2``, ``7``, ``8``, ``2``, ``3``, ``2``, ``3``]` `    `  `    ``n ``=` `len``(arr)` `    `  `    ``k ``=` `2` `    `  `    ``print``(Min_Replace(arr, n, k))` `    `  `# This code is contributed by` `# Surendra_Gangwar`

## C#

 `// C# program to minimum changes required ` `// in an array for k distinct elements.` `using` `System;`   `class` `GFG` `{` `    ``static` `int` `MAX = 100005;` `    `  `    ``// Function to minimum changes required ` `    ``// in an array for k distinct elements.` `    ``static` `int` `Min_Replace(``int` `[] arr, ` `                           ``int` `n, ``int` `k)` `    ``{` `        ``Array.Sort(arr);` `    `  `        ``// Store the frequency of each element` `        ``int` `[] freq = ``new` `int``[MAX];` `        `  `        ``int` `p = 0;` `        ``freq[p] = 1;` `        `  `        ``// Store the frequency of elements` `        ``for` `(``int` `i = 1; i < n; i++)` `        ``{` `            ``if` `(arr[i] == arr[i - 1])` `                ``++freq[p];` `            ``else` `                ``++freq[++p];` `        ``}` `    `  `        ``// Sort frequencies in descending order` `        ``Array.Sort(freq);` `        ``Array.Reverse(freq);` `        `  `        ``// To store the required answer` `        ``int` `ans = 0;` `        ``for` `(``int` `i = k; i <= p; i++)` `            ``ans += freq[i];` `            `  `        ``// Return the required answer` `        ``return` `ans;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main ()` `    ``{` `        ``int` `[] arr = { 1, 2, 7, 8, 2, 3, 2, 3 };` `        `  `        ``int` `n = arr.Length;` `        `  `        ``int` `k = 2;` `        `  `        ``Console.WriteLine(Min_Replace(arr, n, k));` `    ``}` `}`   `// This code is contributed by ihritik`

## Javascript

 ``

Output:

`3`

Time Complexity : O(NlogN)

Auxiliary Space: O(1) because it is using constant size freq array

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