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

SourceBiweekly Contest 19 Q2DifficultyMediumRating1317

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

Thinking

Count windows of length \(k\) whose average is at least \(\textit{threshold}\). \(n \le 10^5\) forbids rebuilding each window. The average test is the sum versus \(k \times \textit{threshold}\). A sliding sum of width \(k\) updates in \(O(1)\) and counts qualifying windows in one pass.

We can multiply \(\textit{threshold}\) by \(k\), so that we can directly compare the sum within the window with \(\textit{threshold}\).

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 \(\textit{threshold}\), we increment the answer.

The time complexity is \(O(n)\), and the space complexity is \(O(1)\), where \(n\) is the length of the array \(\textit{arr}\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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