Given an array arr[] of size N, the task is to find the minimum number of array elements required to be removed such that the frequency of the remaining array elements become equal.
Examples :
Input: arr[] = {2, 4, 3, 2, 5, 3}
Output: 2
Explanation: Following two possibilities exists:
1) Either remove an occurrence of 2 and 3. The array arr[] modifies to {2, 4, 3, 5}. Therefore, frequency of all the elements become equal.
2) Or, remove an occurrence of 4 and 5. The array arr[] modifies to {2, 3, 2, 3}. Therefore, frequency of all the elements become equal.Input: arr[] = {1, 1, 2, 1}
Output: 1
Naive Approach: We count the frequency of each element in an array. Then for each value v in frequency map, we traverse the frequency map and check whether this current value is less than v, if it is true then we add this current value to our result and if it is false then we add the difference between the current value and v to our result. After each traversal, store minimum of current result and previous result.
Algorithm:
Step 1: Create a function called “minDelete” that accepts the arguments arr, an integer array of size n, and the function name.
Step 2: Create an unordered map called “freq” to keep track of each element’s frequency in the input array. Set each element’s frequency to 0 at the beginning.
Step 3: Increase the frequency of each element in the “freq” map as you traverse the input array.
Step 4: Set two variables, “tempans” and “res,” to their respective initial values of 0 and INT MAX.
Step 5: Use the iterator “itr” to navigate the “freq” map. Repeat the traversal of the map for each element using an additional iterator called “j”.
a. Increase “tempans” by the frequency of the element pointed by “j” if its frequency is lower than that of the element pointed by “itr”.
b. If the frequency of the element pointed by “j” is greater than or equal to that of the element pointed by “itr”, increment “tempans” by the difference between the frequency of the element pointed by “j” and that of the element pointed by “itr”.
Step 6: Update “res” with the minimum value between “res” and “tempans”.
Step 7: Return res.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to get minimum removals required// to make frequency of all remaining elements equalint minDelete(int arr[], int n){ // Create an hash map and store frequencies of all // array elements in it using element as key and // frequency as value unordered_map<int, int> freq; for (int i = 0; i < n; i++) freq[arr[i]]++; // Initialize the result to store the minimum deletions int tempans, res = INT_MAX; // Find deletions required for each element and store // the minimum deletions in result for (auto itr = freq.begin(); itr != freq.end(); itr++) { tempans = 0; for (auto j = freq.begin(); j != freq.end(); j++) { if (j->second < itr->second) { tempans = tempans + j->second; } else { tempans = tempans + (j->second - itr->second); } } res = min(res, tempans); } return res;}// Driver program to run the caseint main(){ int arr[] = {2, 4, 3, 2, 5, 3}; int n = sizeof(arr) / sizeof(arr[0]); cout << minDelete(arr, n); return 0;} |
Java
import java.util.HashMap;import java.util.Map;// Java program for the above approachclass Main { // Function to get minimum removals required // to make frequency of all remaining elements equal static int minDelete(int arr[], int n) { // Create an hash map and store frequencies of all // array elements in it using element as key and // frequency as value Map<Integer, Integer> freq = new HashMap<>(); for (int i = 0; i < n; i++) freq.put(arr[i], freq.getOrDefault(arr[i], 0) + 1); // Initialize the result to store the minimum deletions int tempans, res = Integer.MAX_VALUE; // Find deletions required for each element and store // the minimum deletions in result for (Map.Entry<Integer, Integer> itr : freq.entrySet()) { tempans = 0; for (Map.Entry<Integer, Integer> j : freq.entrySet()) { if (j.getValue() < itr.getValue()) { tempans = tempans + j.getValue(); } else { tempans = tempans + (j.getValue() - itr.getValue()); } } res = Math.min(res, tempans); } return res; } // Driver program to run the case public static void main(String args[]) { int arr[] = { 2, 4, 3, 2, 5, 3 }; int n = arr.length; System.out.println(minDelete(arr, n)); }}// This code is contributed by phasing17. |
Python3
# Python program for the above approach# Function to get minimum removals required# to make frequency of all remaining elements equalimport sysdef minDelete(arr, n): # Create an hash map and store frequencies of all # array elements in it using element as key and # frequency as value freq = {} for i in range(n): if(arr[i] in freq): freq[arr[i]] = freq[arr[i]]+1 else: freq[arr[i]] = 1 # Initialize the result to store the minimum deletions tempans, res = sys.maxsize,sys.maxsize # Find deletions required for each element and store # the minimum deletions in result for [key,value] in freq.items(): tempans = 0 for [key1,value1] in freq.items(): if (value1 < value): tempans = tempans + value1 else: tempans = tempans + (value1 - value) res = min(res, tempans) return res# Driver program to run the casearr = [2, 4, 3, 2, 5, 3]n = len(arr)print(minDelete(arr, n))# This code is contributed by shinjanpatra. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to get minimum removals required // to make frequency of all remaining elements equal static int MinDelete(int[] arr, int n) { // Create a dictionary and store frequencies of all // array elements in it using element as key and // frequency as value Dictionary<int, int> freq = new Dictionary<int, int>(); for (int i = 0; i < n; i++) { if (freq.ContainsKey(arr[i])) { freq[arr[i]]++; } else { freq[arr[i]] = 1; } } // Initialize the result to store the minimum deletions int tempans, res = int.MaxValue; // Find deletions required for each element and store // the minimum deletions in result foreach (KeyValuePair<int, int> itr in freq) { tempans = 0; foreach (KeyValuePair<int, int> j in freq) { if (j.Value < itr.Value) { tempans = tempans + j.Value; } else { tempans = tempans + (j.Value - itr.Value); } } res = Math.Min(res, tempans); } return res; } // Driver program to run the case static void Main(string[] args) { int[] arr = { 2, 4, 3, 2, 5, 3 }; int n = arr.Length; Console.WriteLine(MinDelete(arr, n)); }}// This code is contributed by phasing17 |
Javascript
<script>// JavaScript program for the above approach// Function to get minimum removals required// to make frequency of all remaining elements equalfunction minDelete(arr, n){ // Create an hash map and store frequencies of all // array elements in it using element as key and // frequency as value let freq = new Map(); for (let i = 0; i < n; i++){ if(freq.has(arr[i])){ freq.set(arr[i],freq.get(arr[i])+1); } else freq.set(arr[i],1); } // Initialize the result to store the minimum deletions let tempans, res = Number.MAX_VALUE; // Find deletions required for each element and store // the minimum deletions in result for (let [key,value] of freq){ let tempans = 0; for (let [key1,value1] of freq) { if (value1 < value) { tempans = tempans + value1; } else { tempans = tempans + (value1 - value); } } res = Math.min(res, tempans); } return res;}// Driver program to run the caselet arr = [2, 4, 3, 2, 5, 3];let n = arr.length;document.write(minDelete(arr, n));// This code is contributed by shinjanpatra.</script> |
Output
2
Efficient Approach: Follow the steps below to solve the problem:
- Initialize an ordered map, say freq, to store the frequency of all array elements.
- Traverse the array arr[] and store the respective frequencies of array elements.
- Initialize a vector, say v, to store all the frequencies stored in the map.
- Sort the vector v.
- Initialize a variable, say ans, to store the final count.
- Traverse the vector v and for each frequency, count the number of elements required to be removed to make the frequency of the remaining elements equal.
- Print the minimum of all the count of removals obtained.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count the minimum// removals required to make frequency// of all array elements equalint minDeletions(int arr[], int N){ // Stores frequency of // all array elements map<int, int> freq; // Traverse the array for (int i = 0; i < N; i++) { freq[arr[i]]++; } // Stores all the frequencies vector<int> v; // Traverse the map for (auto z : freq) { v.push_back(z.second); } // Sort the frequencies sort(v.begin(), v.end()); // Count of frequencies int size = v.size(); // Stores the final count int ans = N - (v[0] * size); // Traverse the vector for (int i = 1; i < v.size(); i++) { // Count the number of removals // for each frequency and update // the minimum removals required if (v[i] != v[i - 1]) { int safe = v[i] * (size - i); ans = min(ans, N - safe); } } // Print the final count cout << ans;}// Driver Codeint main(){ // Given array int arr[] = { 2, 4, 3, 2, 5, 3 }; // Size of the array int N = sizeof(arr) / sizeof(arr[0]); // Function call to print the minimum // number of removals required minDeletions(arr, N);} |
Java
/*package whatever //do not write package name here */import java.io.*;import java.util.HashMap;import java.util.Map;import java.util.ArrayList;import java.util.Collections;class GFG { public static void minDeletions(int arr[], int N) { // Stores frequency of // all array elements HashMap<Integer, Integer> map = new HashMap<>(); ; // Traverse the array for (int i = 0; i < N; i++) { Integer k = map.get(arr[i]); map.put(arr[i], (k == null) ? 1 : k + 1); } // Stores all the frequencies ArrayList<Integer> v = new ArrayList<>(); // Traverse the map for (Map.Entry<Integer, Integer> e : map.entrySet()) { v.add(e.getValue()); } // Sort the frequencies Collections.sort(v); // Count of frequencies int size = v.size(); // Stores the final count int ans = N - (v.get(0) * size); // Traverse the vector for (int i = 1; i < v.size(); i++) { // Count the number of removals // for each frequency and update // the minimum removals required if (v.get(i) != v.get(i - 1)) { int safe = v.get(i) * (size - i); ans = Math.min(ans, N - safe); } } // Print the final count System.out.println(ans); } // Driver code public static void main(String[] args) { // Given array int arr[] = { 2, 4, 3, 2, 5, 3 }; // Size of the array int N = 6; // Function call to print the minimum // number of removals required minDeletions(arr, N); }}// This code is contributed by aditya7409. |
Python3
# Python 3 program for the above approachfrom collections import defaultdict# Function to count the minimum# removals required to make frequency# of all array elements equaldef minDeletions(arr, N): # Stores frequency of # all array elements freq = defaultdict(int) # Traverse the array for i in range(N): freq[arr[i]] += 1 # Stores all the frequencies v = [] # Traverse the map for z in freq.keys(): v.append(freq[z]) # Sort the frequencies v.sort() # Count of frequencies size = len(v) # Stores the final count ans = N - (v[0] * size) # Traverse the vector for i in range(1, len(v)): # Count the number of removals # for each frequency and update # the minimum removals required if (v[i] != v[i - 1]): safe = v[i] * (size - i) ans = min(ans, N - safe) # Print the final count print(ans)# Driver Codeif __name__ == "__main__": # Given array arr = [2, 4, 3, 2, 5, 3] # Size of the array N = len(arr) # Function call to print the minimum # number of removals required minDeletions(arr, N) # This code is contributed by chitranayal. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to count the minimum // removals required to make frequency // of all array elements equal static void minDeletions(int []arr, int N) { // Stores frequency of // all array elements Dictionary<int,int> freq = new Dictionary<int,int>(); // Traverse the array for (int i = 0; i < N; i++) { if(freq.ContainsKey(arr[i])) freq[arr[i]]++; else freq[arr[i]] = 1; } // Stores all the frequencies List<int> v = new List<int>(); // Traverse the map foreach (var z in freq) { v.Add(z.Value); } // Sort the frequencies int sz = v.Count; int []temp = new int[sz]; for(int i = 0; i < v.Count; i++) temp[i] = v[i]; Array.Sort(temp); for(int i = 0; i < v.Count; i++) v[i] = temp[i]; // Count of frequencies int size = v.Count; // Stores the final count int ans = N - (v[0] * size); // Traverse the vector for (int i = 1; i < v.Count; i++) { // Count the number of removals // for each frequency and update // the minimum removals required if (v[i] != v[i - 1]) { int safe = v[i] * (size - i); ans = Math.Min(ans, N - safe); } } // Print the final count Console.WriteLine(ans); } // Driver Code public static void Main() { // Given array int []arr = { 2, 4, 3, 2, 5, 3 }; // Size of the array int N = arr.Length; // Function call to print the minimum // number of removals required minDeletions(arr, N); }}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script>// Javascript program for the above approach// Function to count the minimum// removals required to make frequency// of all array elements equalfunction minDeletions(arr, N){ // Stores frequency of // all array elements var freq = new Map(); // Traverse the array for (var i = 0; i < N; i++) { if(freq.has(arr[i])) { freq.set(arr[i], freq.get(arr[i])+1); } else { freq.set(arr[i], 1); } } // Stores all the frequencies var v = []; // Traverse the map freq.forEach((value, key) => { v.push(value); }); // Sort the frequencies v.sort(); // Count of frequencies var size = v.length; // Stores the final count var ans = N - (v[0] * size); // Traverse the vector for (var i = 1; i < v.length; i++) { // Count the number of removals // for each frequency and update // the minimum removals required if (v[i] != v[i - 1]) { var safe = v[i] * (size - i); ans = Math.min(ans, N - safe); } } // Print the final count document.write( ans);}// Driver Code// Given arrayvar arr = [ 2, 4, 3, 2, 5, 3 ];// Size of the arrayvar N = arr.length;// Function call to print the minimum// number of removals requiredminDeletions(arr, N);</script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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