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713. Subarray Product Less Than K (Medium)

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.

Related Topics:
Array, Two Pointers

Similar Questions:

• Maximum Product Subarray (Medium)
• Maximum Size Subarray Sum Equals k (Medium)
• Subarray Sum Equals K (Medium)
• Two Sum Less Than K (Easy)

Solution 1. Shrinkable Sliding Window

Check out "C++ Maximum Sliding Window Cheatsheet Template!".

• state: prod is the product of the numbers in window
• invalid: prod >= k is invalid.

Note that since we want to make sure the window [i, j] is valid at the end of the for loop, we need i <= j check for the inner for loop. i == j + 1 means this window is empty.

Each maximum window [i, j] can generate j - i + 1 valid subarrays, so we need to add j - i + 1 to the answer.

// OJ: https://leetcode.com/problems/subarray-product-less-than-k/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& A, int k) {
        if (k == 0) return 0;
        long i = 0, j = 0, N = A.size(), prod = 1, ans = 0;
        for (; j < N; ++j) {
            prod *= A[j];
            while (i <= j && prod >= k) prod /= A[i++];
            ans += j - i + 1;
        }
        return ans;
    }
};

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https://leetcode.com/problems/subarray-product-less-than-k/solutions/1501834/c-sliding-window-cheat-sheet/