Formatted question description: https://leetcode.ca/all/713.html

Medium

## Description

You 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], [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.

## Solution

Initialize product = 1, count = 0, start = 0 and end = 0. Loop over nums using end as the index. Each time let product *= nums[end]. If product >= k, then do product /= nums[start++] until product < k. Then calculate the length of the current subarray, which is end - start + 1. The length of the current subarray is also the number of subarrays with product less than k and ends at index end. Add the length of the current subarray to count. Finally, return count.

Java Code

Java

class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (nums == null || nums.length == 0 || k <= 1)
return 0;
int product = 1;
int count = 0;
int start = 0, end = 0;
int length = nums.length;
while (end < length) {
product *= nums[end];
while (product >= k) {
product /= nums[start];
start++;
}
count += end - start + 1;
end++;
}
return count;
}
}