Reduce Array Size to The Half

Description

You are given an integer array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.

Return the minimum size of the set so that at least half of the integers of the array are removed.

Example 1:

Input: arr = [3,3,3,3,5,5,5,2,2,7] Output: 2 Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array). Possible sets of size 2 are {3,5},{3,2},{5,2}. Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has a size greater than half of the size of the old array.

Example 2:

Input: arr = [7,7,7,7,7,7] Output: 1 Explanation: The only possible set you can choose is {7}. This will make the new array empty.

Constraints:

• 2 <= arr.length <= 105
• arr.length is even.
• 1 <= arr[i] <= 105

Code

Max Heap

Time Complexity: $O(n\log n)$, Space Complexity: $O(n)$

class Solution {
public:
    int minSetSize(vector<int>& arr) {
        unordered_map<int, int> mp;
        for(int i = 0; i < arr.size(); i++) {
            mp[arr[i]]++;
        }
 
        priority_queue<int, vector<int>> max_heap;
        for(auto it = mp.begin(); it != mp.end(); it++) {
            max_heap.push(it->second);
        }
        int count = 0;
        int res = 0;
        int n = arr.size();
        while(!max_heap.empty() && count < n / 2) {
            count += max_heap.top();
            max_heap.pop();
            res++;
        }
 
        return res;
 
    }
};

Counting Sort

Time Complexity: $O(n)$, Space Complexity: $O(n)$

class Solution {
public:
    int minSetSize(vector<int>& arr) {
        unordered_map<int, int> mp;
        for(int i = 0; i < arr.size(); i++) {
            mp[arr[i]]++;
        }
 
        int n = arr.size();
        vector<int> fre(n + 1, 0);
        for(auto it = mp.begin(); it != mp.end(); it++) {
            fre[it->second]++;
        }
 
        int count = 0;
        int res = 0;
        int idx = n;
        while(count < n / 2) {
            while(fre[idx] == 0) idx--;
            res += 1;
            count += idx;
            --fre[idx];
        }
 
        return res;
 
    }
};

Source

• Reduce Array Size to The Half - LeetCode