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

CNoodle發表於2024-11-10

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

大小為 K 且平均值大於等於閾值的子陣列數目。

給你一個整數陣列 arr 和兩個整數 k 和 threshold 。

請你返回長度為 k 且平均值大於等於 threshold 的子陣列數目。

思路

這道題是一道視窗尺寸固定的滑動視窗題。注意儘量不要在中間過程求平均值因為會涉及到精度問題。在過程中我們可以只求數字的 sum,不求平均值,到最後再計算平均值。

複雜度

時間O(n)
空間O(1)

程式碼

Java實現

class Solution {
    public int numOfSubarrays(int[] arr, int k, int threshold) {
        int n = arr.length;
        int sum = 0;
        for (int i = 0; i < k; i++) {
            sum += arr[i];
        }

        int res = 0;
        if (sum >= threshold * k) {
            res++;
        }
        for (int i = k; i < n; i++) {
            sum += arr[i];
            sum -= arr[i - k];
            if (sum >= threshold * k) {
                res++;
            }
        }
        return res;
    }
}

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