1338. Reduce Array Size to The Half

2021/7/7 6:06:37

本文主要是介绍1338. Reduce Array Size to The Half,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

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
class Solution {
    public int minSetSize(int[] arr) {
        Map<Integer, Integer> map = new HashMap();
        for(int i : arr) map.put(i, map.getOrDefault(i, 0) + 1);
        List<Integer>[] bucket = new ArrayList[arr.length + 1];
        for(int i = 0; i <= arr.length; i++) bucket[i] = new ArrayList();
        for(int k : map.keySet()) {
            int f = map.get(k);
            bucket[f].add(k);
        }
        int n = arr.length;
        int res = 0, tmp = 0;
        for(int i = n; i >= 1; i--) {
            int cursize = bucket[i].size();
            if(cursize == 0) continue;
            while(cursize > 0) {
                tmp += i;
                res++;
                if(tmp >= n / 2) return res;
                cursize--;
            }
            
        }
        return res;
    }
}

数组频率有关的就hashmap + bucket sort



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