LeetCode每日一题(Reduce Array Size to The Half)

Given an 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 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.

Example 3:

Input: arr = [1,9]
Output: 1

Example 4:

Input: arr = [1000,1000,3,7]
Output: 1

Example 5:

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

Constraints:

• 1 <= arr.length <= 10^5
• arr.length is even.
• 1 <= arr[i] <= 10^5

思路:

直觉告诉我们应该每次找arr里出现最多的整数进行删除，直到剩下的整数的数量小于等于原有数量的一半。

代码(Rust):

use std::collections::{BinaryHeap, HashMap};

impl Solution {
    pub fn min_set_size(arr: Vec<i32>) -> i32 {
        let len = arr.len() as i32;
        let half = if len % 2 == 0 { len / 2 } else { len / 2 + 1 };
        let map: HashMap<i32, i32> = arr.into_iter().fold(HashMap::new(), |mut m, v| {
            *m.entry(v).or_insert(0) += 1;
            m
        });
        let mut heap: BinaryHeap<i32> = map.values().map(|v| *v).collect();
        let mut ans = 0;
        let mut count = 0;
        while let Some(v) = heap.pop() {
            if count >= half {
                break;
            }
            count += v;
            ans += 1;
        }
        ans
    }
}