【leetcode】1343. Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold

题目如下：

Given an array of integers arr and two integers k and threshold.

Return the number of sub-arrays of size k and average greater than or equal to threshold.

Example 1:

Input: arr = [2,2,2,2,5,5,5,8], k = 3, threshold = 4
Output: 3
Explanation: Sub-arrays [2,5,5],[5,5,5] and [5,5,8] have averages 4, 5 and 6 respectively. All other sub-arrays of size 3 have averages less than 4 (the threshold).

Example 2:

Input: arr = [1,1,1,1,1], k = 1, threshold = 0
Output: 5

Example 3:

Input: arr = [11,13,17,23,29,31,7,5,2,3], k = 3, threshold = 5
Output: 6
Explanation: The first 6 sub-arrays of size 3 have averages greater than 5. Note that averages are not integers.

Example 4:

Input: arr = [7,7,7,7,7,7,7], k = 7, threshold = 7
Output: 1

Example 5:

Input: arr = [4,4,4,4], k = 4, threshold = 1
Output: 1

Constraints:

• 1 <= arr.length <= 10^5
• 1 <= arr[i] <= 10^4
• 1 <= k <= arr.length
• 0 <= threshold <= 10^4

解题思路：滑动窗口问题，把每一个区间的和都算出来吧。

代码如下：

class Solution(object):
    def numOfSubarrays(self, arr, k, threshold):
        """
        :type arr: List[int]
        :type k: int
        :type threshold: int
        :rtype: int
        """
        res = 0
        count = 0
        for i in range(len(arr)):
            if i < k - 1 :
                count += arr[i]
            elif i == k - 1:
                count += arr[i]
                if count / k >= threshold:
                    res += 1
            else:
                count += arr[i]
                count -= arr[i-k]
                if count / k >= threshold:
                    res += 1
        return res

原文地址：https://www.cnblogs.com/seyjs/p/12287791.html