leetcode 1481. Least Number of Unique Integers after K Removals（python）

描述

Given an array of integers arr and an integer k. Find the least number of unique integers after removing exactly k elements.

Example 1:

Input: arr = [5,5,4], k = 1
Output: 1
Explanation: Remove the single 4, only 5 is left.	

Example 2:

Input: arr = [4,3,1,1,3,3,2], k = 3
Output: 2
Explanation: Remove 4, 2 and either one of the two 1s or three 3s. 1 and 3 will be left.

Note:

1 <= arr.length <= 10^5
1 <= arr[i] <= 10^9
0 <= k <= arr.length

解析

根据题意，在 arr 中去掉 k 个元素之后，最后至少剩下多少独一无二的元素。其实很好理解只要对 arr 中的元素进行计数形成字典 c ，然后按照 c 中的值进行生序排序，只要从个数最少的元素开始剔除，最后肯定越能多剔除几类元素，自然剩下的元素种类会越少。

解答

class Solution(object):
    def findLeastNumOfUniqueInts(self, arr, k):
        """
        :type arr: List[int]
        :type k: int
        :rtype: int
        """
        c = {}
        for i in arr:
            if i not in c:
                c[i]=1
            else:
                c[i]+=1
        c = sorted(c.items(), key=lambda x:x[1])
        n = len(c)
        for key, value in c:
            if value<=k:
                k -= value
                n -= 1
            else:
                break
        return n

运行结果

Runtime: 432 ms, faster than 86.90% of Python online submissions for Least Number of Unique Integers after K Removals.
Memory Usage: 38.5 MB, less than 33.73% of Python online submissions for Least Number of Unique Integers after K Removals.

原题链接：https://leetcode.com/problems/least-number-of-unique-integers-after-k-removals/

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