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
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 <bits/stdc++.h>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 codeint 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 codeif __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
<script>// Javascript program to minimum changes required // in an array for k distinct elements.var MAX = 100005;// Function to minimum changes required // in an array for k distinct elements.function Min_Replace(arr, n, k){ arr.sort((a,b)=>a-b) // Store the frequency of each element var freq = Array(MAX).fill(0); var p = 0; freq[p] = 1; // Store the frequency of elements for (var i = 1; i < n; i++) { if (arr[i] == arr[i - 1]) ++freq[p]; else ++freq[++p]; } // Sort frequencies in descending order freq.sort((a,b)=>b-a); // To store the required answer var ans = 0; for (var i = k; i <= p; i++) ans += freq[i]; // Return the required answer return ans;}// Driver codevar arr = [1, 2, 7, 8, 2, 3, 2, 3];var n = arr.length;var k = 2;document.write( Min_Replace(arr, n, k));</script> |
Output:
3
Time Complexity : O(NlogN)
Auxiliary Space: O(1) because it is using constant size freq array
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
