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.

Solution

給一個數組和最大的成績單，要求找到陣列中有多少組連續的子串，使得它們的成績單小於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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