# 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 : 2
The elements to be changed are 1, 8

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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); ` `     `  `    ``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 `

Output:

```3
```

Time Complexity : O(NlogN)

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