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Formatted question description: https://leetcode.ca/all/1343.html
1343. Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold (Medium)
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 = [1,1,1,1,1], k = 1, threshold = 0 Output: 5
Example 3:
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.
Example 4:
Input: arr = [7,7,7,7,7,7,7], k = 7, threshold = 7 Output: 1
Example 5:
Input: arr = [4,4,4,4], k = 4, threshold = 1 Output: 1
Constraints:
1 <= arr.length <= 10^5
1 <= arr[i] <= 10^4
1 <= k <= arr.length
0 <= threshold <= 10^4
Related Topics:
Array
Solution 1.
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class Solution { public int numOfSubarrays(int[] arr, int k, int threshold) { int count = 0; int sumThreshold = threshold * k; int sum = 0; for (int i = 0; i < k; i++) sum += arr[i]; if (sum >= sumThreshold) count++; int length = arr.length; for (int i = k; i < length; i++) { sum -= arr[i - k]; sum += arr[i]; if (sum >= sumThreshold) count++; } return count; } } ############ 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; } }
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// OJ: https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/ // Time: O(N) // Space: O(1) class Solution { public: int numOfSubarrays(vector<int>& arr, int k, int threshold) { int sum = 0, cnt = 0; for (int i = 0; i < arr.size(); ++i) { sum += arr[i]; if (i >= k) sum -= arr[i - k]; if (i + 1 >= k && sum / k >= threshold) ++cnt; } return cnt; } };
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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
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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 }
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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; }