713. Subarray Product Less Than K

Description

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.

Problem URL

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Solution

给一个数组和最大的成绩单，要求找到数组中有多少组连续的子串，使得它们的成绩单小于k。

At first glance, I may think about a brute force approach by count how many subarray with window size 1, then 2 then 3… have the product less than k. But there definityly has much better solution.

So we could keep a sliding window, each time the window move, we could get window.size subarrays which have product less than k. If their product is more than k, pop out left element and move I forward.

The calculate of count is tricky, every time j move forward, it could bring j - I - 1 subarrays. Because we get a nums[j] it self to nums[j] * … * nums[I]. So the total number of subarrays to be add is j - I + 1.

Code

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

Time Complexity: O(n)
Space Complexity: O(1)

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