【leetcode】1338. Reduce Array Size to The Half

标签：str   this   constrain   lis   append   ota   length   代码   put   

题目如下：

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

解题思路：贪心算法，从出现次数最多的元素的删起，然后删除次多的元素，直到满足条件为止。

代码如下：

class Solution(object):
    def minSetSize(self, arr):
        """
        :type arr: List[int]
        :rtype: int
        """
        dic = {}
        val = []
        for i in arr:
            dic[i] = dic.setdefault(i,0) + 1
        
        
        for i in dic:
            val.append(dic[i])
        val.sort()
        
        count = 0
        total = 0
        for i in val[::-1]:
            total += i
            count += 1
            if total >= len(arr)/2:
                break
        return count 
        

【leetcode】1338. Reduce Array Size to The Half

标签：str   this   constrain   lis   append   ota   length   代码   put   

原文地址：https://www.cnblogs.com/seyjs/p/12283376.html