# 643.maximum-average-subarray-i

## Statement

``````输入：nums = [1,12,-5,-6,50,3], k = 4

``````

``````输入：nums = , k = 1

``````

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

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 = , k = 1
Output: 5.00000
``````

Constraints:

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

## Solution

``````from typing import List

class Solution:
def findMaxAverage(self, nums: List[int], k: int) -> float:
i = 0
sum = 0
res = -10000
for j in range(len(nums)):
sum += nums[j]
while j - i + 1 > k:
sum -= nums[i]
i += 1
if j - i + 1 == k:
res = max(res, sum / k)
return res
``````