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


    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

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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/12302081.html
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