Maximum average subarray i

class Solution:
    def findMaxAverage(self, nums: List[int], k: int) -> float:
        best = now = sum(nums[:k])
        for i in range(k, len(nums)):
            now += nums[i]
            now -= nums[i - k]
            best = max(best, now)

        return best / k

Maximum Average Subarray I

Difficulty: Easy

You are given an integer array nums consisting of n elements, and an integer k.

Find a contiguous subarray whose length is equal to k that has the maximum average value and return this value. Any answer with a calculation error less than 10-5 will be accepted.

 

Example 1:

Input: nums = [1,12,-5,-6,50,3], k = 4
Output: 12.75000
Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75

Example 2:

Input: nums = [5], k = 1
Output: 5.00000

 

Constraints:

  • n == nums.length
  • 1 <= k <= n <= 105
  • -104 <= nums[i] <= 104