713. Subarray Product Less Than K-优快云博客

本文介绍了一种高效算法来计算一个整数数组中乘积小于特定阈值K的所有连续子数组的数量。通过维护一个滑动窗口，确保了算法的时间效率。举例说明了如何在添加新元素时更新子数组计数。

Your are given an array of positive integers nums.

Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than k.

Example 1:

Input: nums = [10, 5, 2, 6], k = 100
Output: 8
Explanation: The 8 subarrays that have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6].
Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.

Note:

0 < nums.length <= 50000.
0 < nums[i] < 1000.
0 <= k < 10^6.

Approach #1. Math. [Java]

class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        int pro = 1;
        int cnt = 0;
        
        for (int i = 0, j = 0; j < nums.length; ++j) {
            pro *= nums[j];
            while (i <= j && pro >= k) {
                pro /= nums[i++];
            }
            cnt += j - i + 1;
        }
        
        return cnt;
    }
}

Analysis:

The idea is always keep an max-product-window less than K;

Every time shift window by adding a new number on the right(j), if the product is greater than K, then try to reduce numbers on the left(i), untill the subarray product fit less than K again, (subarray could be empty);

Each step introduces x new subarrays, where x is the size of the current window (j - i + 1);

Example:

for window (5, 2), when 6 is ntroduced, it add 3 new subarray: (5, (2, (6)))