Maximum distinct elements after removing k elements
Given an array arr[] containing n elements. The problem is to find maximum number of distinct elements (non-repeating) after removing k elements from the array.
Note: 1 <= k <= n.
Examples:
Input : arr[] = {5, 7, 5, 5, 1, 2, 2}, k = 3 Output : 4 Remove 2 occurrences of element 5 and 1 occurrence of element 2. Input : arr[] = {1, 2, 3, 4, 5, 6, 7}, k = 5 Output : 2 Input : arr[] = {1, 2, 2, 2}, k = 1 Output : 1
Approach: Following are the steps:
- Create a hash table to store the frequency of each element.
- Insert frequency of each element in a max heap.
- Now, perform the following operation k times. Remove an element from the max heap. Decrement its value by 1. After this if element is not equal to 0, then again push the element in the max heap.
- After the completion of step 3, the number of elements in the max heap is the required answer.
C++
// C++ implementation to find maximum distinct // elements after removing k elements #include <bits/stdc++.h> using namespace std; // function to find maximum distinct elements // after removing k elements int maxDistinctNum( int arr[], int n, int k) { // 'um' implemented as hash table to store // frequency of each element unordered_map< int , int > um; // priority_queue 'pq' implemented as // max heap priority_queue< int > pq; // storing frequency of each element in 'um' for ( int i = 0; i < n; i++) um[arr[i]]++; // inserting frequency of each element in 'pq' for ( auto it = um.begin(); it != um.end(); it++) pq.push(it->second); while (k--) { // get the top element of 'pq' int temp = pq.top(); // remove top element from 'pq' pq.pop(); // decrement the popped element by 1 temp--; // if true, then push the element in 'pq' if (temp != 0) pq.push(temp); } // Count all those elements that appear // once after above operations. int res = 0; while (pq.empty() == false ) { int x = pq.top(); pq.pop(); if (x == 1) res++; } return res; } // Driver program to test above int main() { int arr[] = { 5, 7, 5, 5, 1, 2, 2 }; int n = sizeof (arr) / sizeof (arr[0]); int k = 3; cout << "Maximum distinct elements = " << maxDistinctNum(arr, n, k); return 0; } |
chevron_right
filter_none
Java
// Java implementation to find maximum distinct // elements after removing k elements import java.util.*; class GFG { // function to find maximum distinct elements // after removing k elements static int maxDistinctNum( int [] arr, int n, int k) { // hash map to store // frequency of each element HashMap<Integer, Integer> map = new HashMap<>(); // priority_queue 'pq' implemented as // max heap PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder()); // storing frequency of each element in map for ( int i = 0 ; i < n; i++) { if (map.containsKey(arr[i])) { int val = map.get(arr[i]); val++; map.remove(arr[i]); map.put(arr[i], val); } else map.put(arr[i], 1 ); } // inserting frequency of each element in 'pq' for (Map.Entry<Integer, Integer> entry : map.entrySet()) { pq.add(entry.getValue()); } while (k > 0 ) { // get the top element of 'pq' int temp = pq.poll(); // decrement the popped element by 1 temp--; // if true, then push the element in 'pq' if (temp > 0 ) pq.add(temp); k--; } // Count all those elements that appear // once after above operations. int res = 0 ; while (pq.size() != 0 ) { pq.poll(); res++; } return res; } // Driver code public static void main(String args[]) { int [] arr = { 5 , 7 , 5 , 5 , 1 , 2 , 2 }; int n = arr.length; int k = 3 ; System.out.println( "Maximum distinct elements = " + maxDistinctNum(arr, n, k)); } } // This code is contributed by rachana soma |
chevron_right
filter_none
Output:
Maximum distinct elements = 4
Time Complexity: O(k*logd), where d is the number of distinct elements in the given array.
Recommended Posts:
- Minimum number of distinct elements after removing m items
- Average of remaining elements after removing K largest and K smallest elements from array
- Count of subsequences having maximum distinct elements
- Equally divide into two sets such that one set has maximum distinct elements
- Sum of distinct elements when elements are in range 1 to n
- Make two sets disjoint by removing minimum elements
- Minimize the sum calculated by repeatedly removing any two elements and inserting their sum to the Array
- Subarrays with distinct elements
- Longest subarray not having more than K distinct elements
- Distinct adjacent elements in an array
- Distinct elements in subarray using Mo's Algorithm
- Count distinct elements in an array
- Check if all array elements are distinct
- Arrange N elements in circular fashion such that all elements are strictly less than sum of adjacent elements
- Print All Distinct Elements of a given integer array
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Improved By : rachana soma