LeetCode 1343. Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Sliding Window

We can multiply threshold by k, so that we can directly compare the sum within the window with threshold.

We maintain a sliding window of length k, and for each window, we calculate the sum s. If s is greater than or equal to threshold, we increment the answer.

The time complexity is O(n), where n is the length of the array arr. The space complexity is O(1).

PythonJavaC++GoTypeScript

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

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

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

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

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