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# Minimum number of distinct elements after removing m items

• Difficulty Level : Medium
• Last Updated : 28 Mar, 2023

Given an array of items, an i-th index element denotes the item id’s, and given a number m, the task is to remove m elements such that there should be minimum distinct id’s left. Print the number of distinct id’s.

Examples:

```Input : arr[] = { 2, 2, 1, 3, 3, 3}
m = 3
Output : 1
Remove 1 and both 2's.So, only 3 will be
left that's why distinct id is 1.

Input : arr[] = { 2, 4, 1, 5, 3, 5, 1, 3}
m = 2
Output : 3
Remove 2 and 4 completely. So, remaining ids
are 1, 3 and 5 i.e. 3```

1. Count the occurrence of elements and store them in the hash.
2. Sort the hash.
3. Start removing elements from the hash whose frequency count is less than m.
4. Return the number of values left in the hash.

Implementation:

## C++

 `// C++ program for above implementation` `#include ` `using` `namespace` `std;`   `// Function to find distinct id's` `int` `distinctIds(``int` `arr[], ``int` `n, ``int` `mi)` `{` `    ``unordered_map<``int``, ``int``> m;` `    ``vector > v;` `    ``int` `count = 0;`   `    ``// Store the occurrence of ids` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``m[arr[i]]++;`   `    ``// Store into the vector second as first and vice-versa` `    ``for` `(``auto` `it = m.begin(); it != m.end(); it++)` `        ``v.push_back(make_pair(it->second, it->first));`   `    ``// Sort the vector` `    ``sort(v.begin(), v.end());`   `    ``int` `size = v.size();`   `    ``// Start removing elements from the beginning` `    ``for` `(``int` `i = 0; i < size; i++) {`   `        ``// Remove if current value is less than` `        ``// or equal to mi` `        ``if` `(v[i].first <= mi) {` `            ``mi -= v[i].first;` `            ``count++;` `        ``}`   `        ``// Return the remaining size` `        ``else` `            ``return` `size - count;` `    ``}` `    ``return` `size - count;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 2, 3, 1, 2, 3, 3 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``int` `m = 3;`   `    ``cout << distinctIds(arr, n, m);` `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `class` `Solution {` `    ``static` `int` `distinctIds(``int` `arr[], ``int` `n, ``int` `m)` `    ``{` `        ``// Creating HashMap to store frequency count` `        ``HashMap h = ``new` `HashMap<>();`   `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(h.containsKey(arr[i])) {` `                ``h.put(arr[i], h.get(arr[i]) + ``1``);` `            ``}` `            ``else` `{` `                ``h.put(arr[i], ``1``);` `            ``}` `        ``}`   `        ``// Creating a list to sort HashMap according to` `        ``// values` `        ``List > l` `            ``= ``new` `LinkedList >(` `                ``h.entrySet());`   `        ``// sorting using Comparator` `        ``Collections.sort(` `            ``l,` `            ``new` `Comparator >() {` `                ``public` `int` `compare(` `                    ``Map.Entry o1,` `                    ``Map.Entry o2)` `                ``{` `                    ``return` `o1.getValue() - o2.getValue();` `                ``}` `            ``});`   `        ``// Creating new map after sorting and also` `        ``// maintaining insertion order` `        ``LinkedHashMap lh` `            ``= ``new` `LinkedHashMap<>();` `        ``for` `(Map.Entry e : l) {` `            ``lh.put(e.getKey(), e.getValue());` `        ``}`   `        ``for` `(Integer i : lh.keySet()) {` `            ``// removing element from whose frequency count is` `            ``// less than m ,Sorted manner to get minimum` `            ``// distinct ids` `            ``if` `(h.get(i) <= m) {` `                ``m -= h.get(i);` `                ``h.remove(i);` `            ``}` `        ``}`   `        ``return` `h.size();` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// TODO Auto-generated method stub` `        ``int` `arr[] = { ``2``, ``4``, ``1``, ``5``, ``3``, ``5``, ``1``, ``3` `};` `        ``int` `m = ``2``;`   `        ``System.out.println(distinctIds(arr, arr.length, m));` `    ``}` `}`

## Python3

 `# Python program for above implementation`   `# Function to find distinct id's` `def` `distinctIds(arr, n, mi):` `  ``m ``=` `{}` `  ``v ``=` `[]` `  ``count ``=` `0`   `  ``# Store the occurrence of ids` `  ``for` `i ``in` `range``(n):` `    ``if` `arr[i] ``in` `m:` `      ``m[arr[i]] ``+``=` `1` `    ``else``:` `      ``m[arr[i]] ``=` `1`   `  ``# Store into the list value as key and vice-versa` `  ``for` `i ``in` `m:` `    ``v.append([m[i],i])`   `  ``v.sort()` `  ``size ``=` `len``(v)`   `  ``# Start removing elements from the beginning` `  ``for` `i ``in` `range``(size):` `    `  `    ``# Remove if current value is less than ` `    ``# or equal to mi` `    ``if` `(v[i][``0``] <``=` `mi):` `      ``mi ``-``=` `v[i][``0``]` `      ``count ``+``=` `1` `        `  `    ``else``:   ``# Return the remaining size` `      ``return` `size ``-` `count` `  ``return` `size ``-` `count`   `# Driver code` `arr ``=` `[ ``2``, ``3``, ``1``, ``2``, ``3``, ``3` `]` `n ``=` `len``(arr)`   `m ``=` `3`   `# To display the result` `print``(distinctIds(arr, n, m))`   `# This code is contributed by rohitsingh07052`

## C#

 `// C# program for Minimum number of` `// distinct elements after removing m items` `using` `System;` `using` `System.Collections.Generic;  ` `class` `GFG` `{`   `  ``// Function to find distinct id's` `  ``static` `int` `distinctIds(``int``[] arr, ``int` `n, ``int` `mi)` `  ``{`   `    ``Dictionary<``int``, ``int``> m = ``new` `Dictionary<``int``, ``int``>();  ` `    ``int` `count = 0;` `    ``int` `size = 0;`   `    ``// Store the occurrence of ids` `    ``for` `(``int` `i = 0; i < n; i++)` `    ``{`   `      ``// If the key is not add it to map` `      ``if` `(m.ContainsKey(arr[i]) == ``false``)` `      ``{` `        ``m[arr[i]] = 1;` `        ``size++;` `      ``}`   `      ``// If it is present then increase the value by 1` `      ``else` `      ``{` `        ``m[arr[i]]++;` `      ``}` `    ``}`   `    ``// Start removing elements from the beginning` `    ``foreach``(KeyValuePair<``int``, ``int``> mp ``in` `m) ` `    ``{`   `      ``// Remove if current value is less than` `      ``// or equal to mi` `      ``if` `(mp.Key <= mi)` `      ``{` `        ``mi -= mp.Key;` `        ``count++;` `      ``}` `    ``}` `    ``return` `size - count;` `  ``}`   `  ``// Driver code` `  ``static` `void` `Main() ` `  ``{`   `    ``// TODO Auto-generated method stub` `    ``int``[] arr = {2, 3, 1, 2, 3, 3};` `    ``int` `m = 3;`   `    ``Console.WriteLine(distinctIds(arr, arr.Length, m));` `  ``}` `}`   `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output

`1`

Time Complexity: O(n log n)

Space Complexity: O(n), As we are using map to store elements

## Approach 2: Using Priority Queue and Map

1. Create a frequency map:
An unordered map is created to store the frequency of each element in the given array. The key of the map is the element itself and the value is its frequency.
2. Create a priority queue:
A priority queue is created to store the frequency of the elements in ascending order. The priority queue is a min heap, which means that the smallest element will be at the top of the queue.
3. Insert the frequency of each element into the priority queue:
Iterate through the frequency map and insert the frequency of each element into the priority queue.
4. Remove elements:
Now, remove elements from the priority queue until k becomes less than or equal to zero. While removing an element from the queue, subtract its frequency from k. If k becomes negative, break the loop and return the size of the priority queue.
5. Return the size of the priority queue:
After removing k elements, return the size of the priority queue.

## C++

 `// C++ program for above implementation` `#include ` `using` `namespace` `std;`   `// Function to find distinct id's` `int` `distinctIds(``int` `arr[], ``int` `n ,``int` `k) {` `    ``// Step 1: Create a frequency map of the elements in the array` `    ``unordered_map<``int` `, ``int``> mp;` `    ``for``(``int` `i = 0; i,greater<``int``>>pq;`   `    ``// Step 3: Insert the frequencies of the elements into the priority queue` `    ``for``(``auto` `it : mp){` `        ``pq.push(it.second);` `    ``}`   `    ``// Step 4: Remove elements from the priority queue until k becomes less than or equal to 0` `    ``while``(k > 0){` `        ``k -= pq.top();  ``// Subtract the frequency of the smallest element from k` `        ``if``(k >= 0) pq.pop();  ``// Remove the smallest element from the queue if k is still positive` `    ``}`   `    ``// Step 5: Return the size of the priority queue` `    ``return` `pq.size();` `}` `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 2, 3, 1, 2, 3, 3 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``int` `k = 3;`   `    ``cout << distinctIds(arr, n, k);` `    ``return` `0;` `}` `//this code is contributed by Ravi Singh`

## Java

 `import` `java.util.*;`   `class` `Solution {` `   ``static` `int` `distinctIds(``int``[] arr , ``int` `n, ``int` `k) {` `    ``// Step 1: Create a frequency map of the elements in the array` `    ``HashMap map = ``new` `HashMap();` `    ``for` `(``int` `num : arr) {` `        ``map.put(num, map.getOrDefault(num, ``0``) + ``1``);` `    ``}`   `    ``// Step 2: Create a priority queue to store the frequencies in ascending order` `    ``PriorityQueue pq = ``new` `PriorityQueue();`   `    ``// Step 3: Insert the frequencies of the elements into the priority queue` `    ``for` `(``int` `freq : map.values()) {` `        ``pq.offer(freq);` `    ``}`   `    ``// Step 4: Remove elements from the priority queue until k becomes less than or equal to 0` `    ``while` `(k > ``0` `&& !pq.isEmpty()) {` `        ``k -= pq.poll();  ``// Subtract the frequency of the smallest element from k` `    ``}`   `    ``// Step 5: Return the size of the priority queue` `    ``return` `pq.size();` `}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// TODO Auto-generated method stub` `        ``int` `arr[] = { ``2``, ``4``, ``1``, ``5``, ``3``, ``5``, ``1``, ``3` `};` `        ``int` `m = ``2``;`   `        ``System.out.println(distinctIds(arr, arr.length, m));` `    ``}` `}` `//this code is contributed by Ravi Singh`

## Python3

 `# Python program for above implementation` `import` `heapq`   `# Function to find distinct id's` `def` `distinctIds(arr, n, k):` `    ``# Step 1: Create a frequency map of the elements in the array` `    ``mp ``=` `{}` `    ``for` `i ``in` `range``(n):` `        ``mp[arr[i]] ``=` `mp.get(arr[i], ``0``) ``+` `1`   `    ``# Step 2: Create a priority queue to store the frequencies in ascending order` `    ``pq ``=` `[]`   `    ``# Step 3: Insert the frequencies of the elements into the priority queue` `    ``for` `_, freq ``in` `mp.items():` `        ``pq.append(freq)`   `    ``heapq.heapify(pq)  ``# heapify to get a min-heap`   `    ``# Step 4: Remove elements from the priority queue until k becomes less than or equal to 0` `    ``while` `k > ``0``:` `        ``k ``-``=` `heapq.heappop(pq)  ``# Subtract the frequency of the smallest element from k` `        ``if` `k >``=` `0` `and` `not` `pq:  ``# if k is still positive and the priority queue is empty, return -1` `            ``return` `-``1`   `    ``# Step 5: Return the size of the priority queue` `    ``return` `len``(pq)`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``2``, ``3``, ``1``, ``2``, ``3``, ``3``]` `    ``n ``=` `len``(arr)` `    ``k ``=` `3` `    ``print``(distinctIds(arr, n, k))`   `# This code is contributed by rishabmalhdijo`

Output

`1`

Time Complexity: O(n log n)

Space Complexity: O(n)

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