Here, we are going to find the minimum number of distinct elements after removing M number of items from array.

**Given: **An array **arr[]** of items, an **i**‘th index element denotes the item id’s and given a number **m**.

**Problem: **Remove **m** elements such that there should be minimum distinct id’s left and then print the number of distinct id’s.

**Example:**

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

**Solution:**

- Count the occurrence of elements and store in the hash.
- Sort the hash.
- Start removing elements from hash.
- Return the number of values left in the hash.

### Program to Find minimum number of distinct elements after removing M items in C++:

// C++ program for above implementation #include <bits/stdc++.h> using namespace std; // Function to find distinct id's int distinctIds(int arr[], int n, int mi) { unordered_map<int, int> m; vector<pair<int, int> > 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; } // Main int main() { int arr[] = { 2, 3, 1, 2, 3, 3 }; int n = sizeof(arr) / sizeof(arr[0]); int m = 3; cout << distinctIds(arr, n, m); return 0; }

**OUTPUT:**

1

**Time Complexity:** O(n log n)

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