Minimum removals required to make frequency of each array element equal to its value
Given an array arr[] of size N, the task is to find the minimum count of array elements required to be removed such that frequency of each array element is equal to its value
Examples:
Input: arr[] = { 2, 4, 1, 4, 2 }
Output: 2
Explanation:
Removing arr[1] from the array modifies arr[] to { 2, 1, 4, 2 }
Removing arr[2] from the array modifies arr[] to { 2, 1, 2 }
Distinct elements in the array are: { 1, 2 } with frequencies 1 and 2 respectively.
Therefore, the required output is 2.Input: arr[] = { 2, 7, 1, 8, 2, 8, 1, 8 }
Output: 5
Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:
- Initialize a map say, mp to store the frequency of each distinct element of the array.
- Traverse the array and store the frequency of each distinct element of the array.
- Initialize a variable say, cntMinRem to store the minimum count of array elements required to be removed such that the frequency of arr[i] is equal to arr[i].
- Traverse the map using key value of the map as i and check the following conditions:
- If mp[i] < i, then update the value of cntMinRem += mp[i].
- If mp[i] > i, then update the value of cntMinRem += (mp[i] – i).
- Finally, print the value of cntMinRem.
Below is the implementation of the above approach:
C++14
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum count of// elements required to be removed such// that frequency of arr[i] equal to arr[i]int min_elements(int arr[], int N){ // Stores frequency of each // element of the array unordered_map<int, int> mp; // Traverse the array for (int i = 0; i < N; i++) { // Update frequency // of arr[i] mp[arr[i]]++; } // Stores minimum count of removals int cntMinRem = 0; // Traverse the map for (auto it : mp) { // Stores key value // of the map int i = it.first; // If frequency of i is // less than i if (mp[i] < i) { // Update cntMinRem cntMinRem += mp[i]; } // If frequency of i is // greater than i else if (mp[i] > i) { // Update cntMinRem cntMinRem += (mp[i] - i); } } return cntMinRem;}// Driver Codeint main(){ int arr[] = { 2, 4, 1, 4, 2 }; int N = sizeof(arr) / sizeof(arr[0]); cout << min_elements(arr, N); return 0;} |
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Java
// Java program to implement// the above approachimport java.util.*;class GFG{ // Function to find the minimum count of// elements required to be removed such// that frequency of arr[i] equal to arr[i]public static int min_elements(int arr[], int N){ // Stores frequency of each // element of the array Map<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Traverse the array for(int i = 0; i < N; i++) { // Update frequency // of arr[i] mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1); } // Stores minimum count of removals int cntMinRem = 0; // Traverse the map for(int key : mp.keySet()) { // Stores key value // of the map int i = key; int val = mp.get(i); // If frequency of i is // less than i if (val < i) { // Update cntMinRem cntMinRem += val; } // If frequency of i is // greater than i else if (val > i) { // Update cntMinRem cntMinRem += (val - i); } } return cntMinRem;}// Driver Codepublic static void main(String[] args){ int arr[] = { 2, 4, 1, 4, 2 }; System.out.println(min_elements( arr, arr.length));}}// This code is contributed by grand_master |
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Python3
# Python3 program to implement# the above approach# Function to find the minimum count of# elements required to be removed such# that frequency of arr[i] equal to arr[i]def min_elements(arr, N) : # Stores frequency of each # element of the array mp = {}; # Traverse the array for i in range(N) : # Update frequency # of arr[i] if arr[i] in mp : mp[arr[i]] += 1; else : mp[arr[i]] = 1; # Stores minimum count of removals cntMinRem = 0; # Traverse the map for it in mp : # Stores key value # of the map i = it; # If frequency of i is # less than i if (mp[i] < i) : # Update cntMinRem cntMinRem += mp[i]; # If frequency of i is # greater than i elif (mp[i] > i) : # Update cntMinRem cntMinRem += (mp[i] - i); return cntMinRem;# Driver Codeif __name__ == "__main__" : arr = [ 2, 4, 1, 4, 2 ]; N = len(arr); print(min_elements(arr, N)); # This code is contributed by AnkThon |
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C#
// C# program to implement// the above approachusing System;using System.Collections.Generic; class GFG { // Function to find the minimum count of // elements required to be removed such // that frequency of arr[i] equal to arr[i] static int min_elements(int[] arr, int N) { // Stores frequency of each // element of the array Dictionary<int, int> mp = new Dictionary<int, int>(); // Traverse the array for (int i = 0; i < N; i++) { // Update frequency // of arr[i] if(mp.ContainsKey(arr[i])) { mp[arr[i]]++; } else { mp[arr[i]] = 1; } } // Stores minimum count of removals int cntMinRem = 0; // Traverse the map foreach (KeyValuePair<int, int> it in mp) { // Stores key value // of the map int i = it.Key; // If frequency of i is // less than i if (mp[i] < i) { // Update cntMinRem cntMinRem += mp[i]; } // If frequency of i is // greater than i else if (mp[i] > i) { // Update cntMinRem cntMinRem += (mp[i] - i); } } return cntMinRem; } // Driver code static void Main() { int[] arr = { 2, 4, 1, 4, 2 }; int N = arr.Length; Console.Write(min_elements(arr, N)); }}// This code is contributed by divyesh072019 |
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Output:
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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