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

问题：

给定一个数组，求其中连续K个元素组成的子数组中，平均值>=Threshold的子数组个数。

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

　　

解法：

⚠️ 注意：

关于 Sub-arrays 的定义：

Subarrays are arrays within another array.Subarrays contains contiguous elements whereas subsequences are not.

An Example would make it clear

Consider an array {1,2,3,4}

 

List of all its subarrays are {},{1},{2},{3},{4},{1,2},{2,3},{3,4},{1,2,3,4}

List of all its subsequences are {},{1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}

解法一：

Sliding windows 滑动窗口

遍历累加元素，到第k个元素后，没向后移动一个元素，同时减去前面的一个元素。

按照窗口大小为k，向右移动。

同时判断，当前窗口的元素和sum是否>=threshold*k

满足：res++

代码参考：

class Solution {
public:
    int numOfSubarrays(vector<int>& arr, int k, int threshold) {
        int sum=0, thresum=threshold*k;
        int res=0;
        for(int i=0; i<arr.size(); i++){
            sum+=arr[i];
            if(i>=k) sum-=arr[i-k];
            if(i>=k-1 && sum>=thresum) res++;
        }
        return res;
    }
};

解法二：

Prefix Sum 前缀和

计算到当前元素为止，所有元素之和，记录到presum列表中。

初始化presum[0]=0

到第k个元素开始，判断presum[i+1]-presum[i-k+1]=当前元素开始向前推k个元素和 是否>=threshold*k

满足：res++

代码参考：

class Solution {
public:
    int numOfSubarrays(vector<int>& arr, int k, int threshold) {
        vector<int> presum(arr.size()+1, 0);
        int thresum=threshold*k;
        int res=0;
        for(int i=0; i<arr.size(); i++){
            presum[i+1]=presum[i]+arr[i];
            if(i>=k-1 && presum[i+1]-presum[i-k+1]>=thresum) res++;
        }
        return res;
    }
};