README_EN.md

1338. Reduce Array Size to The Half

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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

Solutions

Python3

class Solution:
    def minSetSize(self, arr: List[int]) -> int:
        cnt = Counter(arr)
        ans = m = 0
        for _, v in cnt.most_common():
            m += v
            ans += 1
            if m * 2 >= len(arr):
                break
        return ans

Java

class Solution {
    public int minSetSize(int[] arr) {
        int mx = 0;
        for (int x : arr) {
            mx = Math.max(mx, x);
        }
        int[] cnt = new int[mx + 1];
        for (int x : arr) {
            ++cnt[x];
        }
        Arrays.sort(cnt);
        int ans = 0;
        int m = 0;
        for (int i = mx;; --i) {
            if (cnt[i] > 0) {
                m += cnt[i];
                ++ans;
                if (m * 2 >= arr.length) {
                    return ans;
                }
            }
        }
    }
}

C++

class Solution {
public:
    int minSetSize(vector<int>& arr) {
        int mx = *max_element(arr.begin(), arr.end());
        int cnt[mx + 1];
        memset(cnt, 0, sizeof(cnt));
        for (int& x : arr) {
            ++cnt[x];
        }
        sort(cnt, cnt + mx + 1, greater<int>());
        int ans = 0;
        int m = 0;
        for (int& x : cnt) {
            if (x) {
                m += x;
                ++ans;
                if (m * 2 >= arr.size()) {
                    break;
                }
            }
        }
        return ans;
    }
};

Go

func minSetSize(arr []int) (ans int) {
	mx := 0
	for _, x := range arr {
		mx = max(mx, x)
	}
	cnt := make([]int, mx+1)
	for _, x := range arr {
		cnt[x]++
	}
	sort.Ints(cnt)
	for i, m := mx, 0; ; i-- {
		if cnt[i] > 0 {
			m += cnt[i]
			ans++
			if m >= len(arr)/2 {
				return
			}
		}
	}
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

TypeScript

function minSetSize(arr: number[]): number {
    const counter = new Map<number, number>();
    for (const v of arr) {
        counter.set(v, (counter.get(v) ?? 0) + 1);
    }
    const t = Array.from(counter.values());
    t.sort((a, b) => b - a);
    let ans = 0;
    let n = 0;
    for (const cnt of t) {
        n += cnt;
        ++ans;
        if (n * 2 >= arr.length) {
            break;
        }
    }
    return ans;
}

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