9/6 每日一題(返回最短的連續子數組)

 

Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

 

Example 1:

Input: nums = [1,2,2,3,1]
Output: 2
Explanation: 
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.

Example 2:

Input: nums = [1,2,2,3,1,4,2]
Output: 6
Explanation: 
The degree is 3 because the element 2 is repeated 3 times.
So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.

 

Constraints:

• nums.length will be between 1 and 50,000.
• nums[i] will be an integer between 0 and 49,999.

class Solution:

    def findShortestSubArray(self, nums: List[int]) -> int:

       

        ele_times=Counter(nums)

        print(ele_times)

        max_freq=max(ele_times.values())

        #檢查是否存在最大重複次數是否有相同的其他元素

        my_dict={}

        for k,v in ele_times.items():

            if v == max_freq:

                my_dict[k]=v

        print(my_dict)

        #ex: my_dict={1: 6, 2: 6}

        res={} #打造一個key是重複最高,valu是他對應的索引

        for n,i in enumerate(nums):

            if i in my_dict :

                if i not in res:

                    res[i]=[n]

                else:

                    res[i].append(n)

       

        #ex, res={1: [0, 3, 5, 6, 7, 8], 2: [1, 2, 4, 9, 10, 11]}

        min_len=float('inf')

        for i in res.values():

            min_len=min(min_len,i[-1]-i[0]+1)

        return min_len

   

     #思考方向

1.找出重複次數, 將其中重複平率最高的元素拉出來, 所以用字典 或 List

2.利用找到的最高重複元素, 其記錄該元素的第一次出現索引值,以及最後一次出現的索引值,

找出之間的最短長度