713. Subarray Product Less Than K
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)))
Reference:
https://leetcode.com/problems/subarray-product-less-than-k/discuss/108861/JavaC%2B%2B-Clean-Code-with-Explanation
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