[link](https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/)

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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 = [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.

#### Constraints:

- 1 <= arr.length <= 105
- 1 <= arr[i] <= 104
- 1 <= k <= arr.length
- 0 <= threshold <= 104

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The function works by maintaining a running sum of the elements in arr. The running sum is initialized to the sum of the first k-1 elements in arr. Then, the function iterates over the elements of arr starting from index k-1. For each element, the function adds the element to the running sum and then checks if the average of the running sum is greater than or equal to threshold. If it is, the function increments a counter. Finally, the function returns the value of the counter.

#### Solution 1

class Solution:
    def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
        res = 0
        curSum = sum(arr[0:k-1])

        for L in range(len(arr) - k + 1):
            curSum += arr[L + k - 1]
            if (curSum / k) >= threshold:
                res += 1
            curSum -= arr[L]

        return res
        

O(T): O(n)
O(S): O(1)