How many pairs can an array of leetcode simple questions form

subject

I'll give you a subscript from 0 The starting array of integers nums . In one step , You can do the following ：

from nums elect Two equal Integers
from nums Remove these two integers , To form a Number pair
Please come in nums Perform this operation several times on until it cannot be continued .

Returns a subscript from 0 Start 、 The length is 2 Array of integers for answer As the answer , among answer[0] Is the number of pairs formed ,answer[1] It's right nums Try to count the number of integers left after the above operation .

Example 1：

Input ：nums = [1,3,2,1,3,2,2]
Output ：[3,1]
explain ：
nums[0] and nums[3] Form a number pair , And from nums Remove ,nums = [3,2,3,2,2] .
nums[0] and nums[2] Form a number pair , And from nums Remove ,nums = [2,2,2] .
nums[0] and nums[1] Form a number pair , And from nums Remove ,nums = [2] .
Cannot form more pairs . A total of 3 Pairs of numbers ,nums The rest of the world is 1 A digital .
Example 2：

Input ：nums = [1,1]
Output ：[1,0]
explain ：nums[0] and nums[1] Form a number pair , And from nums Remove ,nums = [] .
Cannot form more pairs . A total of 1 Pairs of numbers ,nums The rest of the world is 0 A digital .
Example 3：

Input ：nums = [0]
Output ：[0,1]
explain ： Cannot form a number of pairs ,nums The rest of the world is 1 A digital .

Tips ：

1 <= nums.length <= 100
0 <= nums[i] <= 100

source ： Power button （LeetCode）

Their thinking

   To form a number pair, you must first find the character frequency of the array , Where the frequency is 2 Multiples of will not produce the remaining unpaired characters , If not 2 Multiples of can be generated 1 position single.

class Solution:
    def numberOfPairs(self, nums: List[int]) -> List[int]:
        couple,remain=0,0
        for i in Counter(nums).values():
            a,b=divmod(i,2)
            couple+=a
            remain+=b
        return [couple,remain]