Given an array arr[] of positive integers, the task is to select an element from the array and delete all its occurrences, such that the number of elements selected are minimum and the array size becomes atleast half of its original size.
Note: Size of the given array is always even.
Example:
Input: arr[] = {2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3}
Output: 2
Explanation: First we select 2 and delete all its occurrences after that arr[] becomes – {4, 4, 6, 7, 7, 3, 3, 3} with size = 8. As the size is still greater than half, we select 3 and delete all its occurrences, after that arr[] becomes – {4, 4, 6, 7, 7} with size = 5.Input: arr[] = {3, 3, 3, 3, 3}
Output: 1
Explanation : select 3 and remove all its occurrences.Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 4
Explanation : Since there is no duplicate elements, hence any 4 elements can be removed to make the array length at least half.
Approach: The task can be easily achieved by removing the elements with maximum frequency, as soon as the array size becomes at least half of the actual size, we return the number of unique elements deleted till now.
Follow the steps to solve the problem:
- Use Hash-map to store frequency of the elements in array present.
- Store the frequencies in a list.
- Sort the list and traverse it from the back.
- Select the largest frequency element and decrement it from the array size and increment the count of unique elements deleted.
- If new array size becomes at-least half of the original array size, return the number of unique elements till now.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std; // Function to calculate the minimum // elements removed int reduceArrSize(int arr[],int n) { unordered_map<int,int> hm; // Making frequency map of elements for (int i = 0; i < n; i++) { hm[arr[i]]++; } // Storing frequencies in a list vector<int> freq; for(auto it = hm.begin(); it != hm.end(); it++) { freq.push_back(it->second); } // Sorting the list sort(freq.begin(), freq.end()); int size = n; int idx = freq.size() - 1; int count = 0; // Counting number of elements to be deleted while (size > n/ 2) { size -= freq[idx--]; count++; } return count; } // Driver Code int main() { int arr[] = { 2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3 }; int n = sizeof(arr)/sizeof(arr[0]); int count = reduceArrSize(arr, n); cout<<(count); return 0; }// This code is contributed by Potta Lokesh |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to calculate the minimum // elements removed public static int reduceArrSize(int[] arr) { HashMap<Integer, Integer> hm = new HashMap<>(); // Making frequency map of elements for (int i = 0; i < arr.length; i++) { hm.put(arr[i], hm.getOrDefault(arr[i], 0) + 1); } // Storing frequencies in a list ArrayList<Integer> freq = new ArrayList<Integer>(hm.values()); // Sorting the list Collections.sort(freq); int size = arr.length; int idx = freq.size() - 1; int count = 0; // Counting number of elements to be deleted while (size > arr.length / 2) { size -= freq[idx--]; count++; } return count; } // Driver Code public static void main(String[] args) { int[] arr = { 2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3 }; int count = reduceArrSize(arr); System.out.println(count); }} |
Python3
# python 3 Program to implement# the above approach# Function to calculate the minimum# elements removeddef reduceArrSize(arr,n): hm = {} # Making frequency map of elements for i in range(n): if arr[i] in hm: hm[arr[i]] += 1 else: hm[arr[i]] = 1 # Storing frequencies in a list freq = [] for key,value in hm.items(): freq.append(value) # Sorting the list freq.sort() size = n idx = len(freq) - 1 count = 0 # Counting number of elements to be deleted while (size > n// 2): size -= freq[idx] idx -= 1 count += 1 return count# Driver Codeif __name__ == '__main__': arr = [2, 2, 2, 2, 4, 4,6, 7, 7, 3, 3, 3] n = len(arr) count = reduceArrSize(arr, n) print(count) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG { // Function to calculate the minimum // elements removed public static int reduceArrSize(int[] arr) { Dictionary<int, int> hm = new Dictionary<int, int>(); // Making frequency map of elements for (int i = 0; i < arr.Length; i++) { if(hm.ContainsKey(arr[i])){ hm[arr[i]] = hm[arr[i]]+1; } else{ hm.Add(arr[i], 1); } } // Storing frequencies in a list List<int> freq = new List<int>(hm.Values); // Sorting the list freq.Sort(); int size = arr.Length; int idx = freq.Count - 1; int count = 0; // Counting number of elements to be deleted while (size > arr.Length / 2) { size -= freq[idx--]; count++; } return count; } // Driver Code public static void Main(String[] args) { int[] arr = { 2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3 }; int count = reduceArrSize(arr); Console.WriteLine(count); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach// Function to calculate the minimum// elements removedfunction reduceArrSize(arr){ var hm = new Map(); // Making frequency map of elements for (var i = 0; i < arr.length; i++) { if(hm.has(arr[i])){ hm.set(arr[i], hm.get(arr[i])+1); } else{ hm.set(arr[i], 1); } } // Storing frequencies in a list var freq =[ ...hm.values()]; // Sorting the list freq.sort(); var size = arr.length; var idx = freq.length - 1; var count = 0; // Counting number of elements to be deleted while (size > arr.length / 2) { size -= freq[idx--]; count++; } return count;}// Driver Codevar arr = [2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3];var count = reduceArrSize(arr);document.write(count);// This code is contributed by rutvik_56.</script> |
Output:
2
Time Complexity: O(N*logN), sorting the frequency list
Auxiliary Space: O(N), hashmap to store the frequencies
Another efficient approach(Using a Priority Queue):
We can solve the problem by using a priority queue instead of sorting the frequency of elements. So we initialize a priority queue and store the frequency of every unique element which we get from an unordered map. Then we run a while loop and start deleting the elements from the priority queue which has the highest frequency to the lowest frequencies in decreasing order and hence we get the count of elements to be deleted. The time complexity of this approach is O(n*logn) but it will take very less time if elements are more frequent.
Below is the implementation of the above approach:
C++
// C++ Program to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the minimum// elements removedint reduceArrSize(int arr[],int n) { unordered_map<int,int> hm; // Making frequency map of elements for (int i = 0; i < n; i++) { hm[arr[i]]++; } priority_queue<int> pq; for(auto it = hm.begin(); it != hm.end(); it++) { pq.push(it->second); } int size = n; int count = 0; // Counting number of elements to be deleted while (size > n/ 2) { size -= pq.top(); pq.pop(); count++; } return count; }// Driver Codeint main() { int arr[] = { 2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3 }; int n = sizeof(arr)/sizeof(arr[0]); int count = reduceArrSize(arr, n); cout<<(count); return 0; }// This code is contributed by Pushpesh raj |
Java
// Java Program to implement the above approachimport java.util.*;public class Main { // Function to calculate the minimum // elements removed public static int reduceArrSize(int[] arr) { Map<Integer, Integer> hm = new HashMap<>(); // Making frequency map of elements for (int i = 0; i < arr.length; i++) { hm.put(arr[i], hm.getOrDefault(arr[i], 0) + 1); } PriorityQueue<Integer> pq = new PriorityQueue<>(new Comparator<Integer>() { @Override public int compare(Integer o1, Integer o2) { return o2 - o1; } }); for (Map.Entry<Integer, Integer> entry : hm.entrySet()) { pq.add(entry.getValue()); } int size = arr.length; int count = 0; // Counting number of elements to be deleted while (size > arr.length / 2) { size -= pq.poll(); count++; } return count; } public static void main(String[] args) { int[] arr = { 2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3 }; int count = reduceArrSize(arr); System.out.println(count); }} |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;class GFG { // Function to calculate the minimum // elements removed public static int ReduceArrSize(int[] arr) { Dictionary<int, int> hm = new Dictionary<int, int>(); // Making frequency map of elements for (int i = 0; i < arr.Length; i++) { if (hm.ContainsKey(arr[i])) { hm[arr[i]]++; } else { hm.Add(arr[i], 1); } } // Declaring a sorted set SortedSet<int> pq = new SortedSet<int>( Comparer<int>.Create((o1, o2) => o2 - o1)); // Building the sorted set foreach(KeyValuePair<int, int> entry in hm) { pq.Add(entry.Value); } int size = arr.Length; int count = 0; // Counting number of elements to be deleted while (size > arr.Length / 2) { size -= pq.Min; pq.Remove(pq.Min); count++; } return count; } // Driver code public static void Main(string[] args) { int[] arr = { 2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3 }; // Function call int count = ReduceArrSize(arr); Console.WriteLine(count); }}// This code is contributed by phasing17 |
Javascript
// JavaScript Program to implement the above approach// Function to calculate the minimum// elements removedfunction reduceArrSize(arr, n) { let hm = new Map(); // Making frequency map of elements for (let i = 0; i < n; i++) { if (hm.has(arr[i])) { hm.set(arr[i], hm.get(arr[i]) + 1); } else { hm.set(arr[i], 1); } } let pq = []; for (let [key, value] of hm) { pq.push(value); } pq.sort((a, b) => b - a); let size = n; let count = 0; // Counting number of elements to be deleted while (size > n / 2) { size -= pq.shift(); count++; } return count;}// Driver Codelet arr = [2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3];let n = arr.length;let count = reduceArrSize(arr, n);console.log(count);// This code is contributed by phasing17. |
Python3
import heapqimport collections #Function to calculate the minimum elements removed def reduceArrSize(arr): # Creating frequency map of elements hm = collections.Counter(arr) pq = [] # Adding frequency of each element to priority queue for frequency in hm.values(): heapq.heappush(pq, frequency) size = len(arr) count = 0 # Counting number of elements to be deleted while size > len(arr)/2: size -= heapq.heappop(pq) count += 1 return count//2# Driver codeif __name__ == "__main__": arr = [ 2, 2, 2, 2, 4, 4, 6, 7, 7, 3, 3, 3 ] count = reduceArrSize(arr) print(count) |
Output
2
Time Complexity: O(N*logN)
Auxiliary Space: O(N)
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