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
class Solution { public int numOfSubarrays(int[] arr, int k, int threshold) { int res = 0; for(int i = 0; i <= arr.length - k; i++){ if(helper(arr, i, k) >= threshold) res++; } return res; } public int helper(int[] arr, int l, int k){ int res = 0; for(int i = 0; i < k; i++){ res += arr[l + i]; } return res / k; } }
class Solution { public int numOfSubarrays(int[] arr, int k, int threshold) { int n = arr.length, ans = 0, s = 0; for(int i = 0; i < k - 1; i++) s += arr[i]; for(int i = k - 1; i < n; i++) { s += arr[i]; if(s / k >= threshold) ans++; s -= arr[i - k + 1]; } return ans; } }
下面的方法聪明点
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/12302081.html
时间: 2024-11-03 17:25:21