# [LeetCode 1338. Reduce Array Size to The Half](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge-2021/608/week-1-july-1st-july-7th/3804/)
### Daily challenge Jul 6, 2021 (MEDIAN)

>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.

:::info
**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 size greater than half of the size of the old array.
:::

:::info
**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.
:::

:::info
**Example 3:**
**Input:** arr = [1,9]
**Output:** 1
:::

:::info
**Example 4:**
**Input:** arr = [1000,1000,3,7]
**Output:** 1
:::

:::info
**Example 5:**
**Input:** arr = [1,2,3,4,5,6,7,8,9,10]
**Output:** 5
:::

:::warning
**Constraints:**
* 1 <= arr.length <= 105
* arr.length is even.
* 1 <= arr[i] <= 105
:::

---

### Approach 1 : HashMap & Sort :bulb:
**`156 ms ( 83% )`** **`O(nlogn)`**

>1. Using **HashMap** to count frequency of each number.
>2. Get the array of frequency and **sort** it.
>3. Removing the elements from the largest frequency until we have removed at least half of **`arr`**.

class Solution {
public:
    int minSetSize(vector<int>& arr) {
        int size = arr.size();
        unordered_map<int, int > freq;
        
        for(int i=0; i<size; i++){
            freq[arr[i]]++;
        }
        
        vector<int > list;
        for(auto it : freq){
            list.push_back(it.second);
        }
        sort(list.begin(), list.end());
        
        int ans = 0;
        int count = 0;
        for(int i=list.size()-1; i>=0; i--){
            ans++;
            count += list[i];
            
            if(count >= size / 2)   return ans;
        }
        
        return size;
    }
};