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

## Description

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

## Solutions

• class Solution {
public int numOfSubarrays(int[] arr, int k, int threshold) {
int s = 0;
for (int i = 0; i < k; ++i) {
s += arr[i];
}
int ans = s / k >= threshold ? 1 : 0;
for (int i = k; i < arr.length; ++i) {
s += arr[i] - arr[i - k];
ans += s / k >= threshold ? 1 : 0;
}
return ans;
}
}

• class Solution {
public:
int numOfSubarrays(vector<int>& arr, int k, int threshold) {
int s = accumulate(arr.begin(), arr.begin() + k, 0);
int ans = s >= k * threshold;
for (int i = k; i < arr.size(); ++i) {
s += arr[i] - arr[i - k];
ans += s >= k * threshold;
}
return ans;
}
};

• class Solution:
def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
s = sum(arr[:k])
ans = int(s / k >= threshold)
for i in range(k, len(arr)):
s += arr[i]
s -= arr[i - k]
ans += int(s / k >= threshold)
return ans

• func numOfSubarrays(arr []int, k int, threshold int) (ans int) {
s := 0
for _, x := range arr[:k] {
s += x
}
if s/k >= threshold {
ans++
}
for i := k; i < len(arr); i++ {
s += arr[i] - arr[i-k]
if s/k >= threshold {
ans++
}
}
return
}

• function numOfSubarrays(arr: number[], k: number, threshold: number): number {
let s = arr.slice(0, k).reduce((acc, cur) => acc + cur, 0);
let ans = s >= k * threshold ? 1 : 0;
for (let i = k; i < arr.length; ++i) {
s += arr[i] - arr[i - k];
ans += s >= k * threshold ? 1 : 0;
}
return ans;
}