# 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
```

## Recommended: Please solve it on PRACTICE first, before moving on to the solution.

Approach: Following are the steps:

1. Create a hash table to store the frequency of each element.
2. Insert frequency of each element in a max heap.
3. 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.
4. 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 ` `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); ` `    ``int` `k = 3; ` `    ``cout << ``"Maximum distinct elements = "` `         ``<< maxDistinctNum(arr, n, k); ` `    ``return` `0; ` `} `

## 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 map = ``new` `HashMap<>(); ` ` `  `             ``// priority_queue 'pq' implemented as  ` `             ``// max heap  ` `             ``PriorityQueue 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 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 `

Output:

```Maximum distinct elements = 4
```

Time Complexity: O(k*logd), where d is the number of distinct elements in the given array.

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Improved By : rachana soma