713. Subarray Product Less Than K
1. 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.
2. Example
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.
3. Note
- 0 < nums.length <= 50000.
- 0 < nums[i] < 1000.
- 0 <= k < $10^6$.
4. Solutions
My Accepted Solution
n = i_nums.size()
Time complexity: O(n)
Space complexity: O(1)
// sliding windows
class Solution {
public:
// int numSubarrayProductLessThanK(vector<int>& nums, int k)
int numSubarrayProductLessThanK(const vector<int>& i_nums, int k) {
int count = 0;
for (int product = 1, left = 0, right = 0; right < i_nums.size(); ++right) {
product *= i_nums[right];
while (product >= k && left <= right) {
product /= i_nums[left];
++left;
}
// if the subarrray is [left, left + 1, left + 2, ..., right - 2, right - 1, right]
// when we add a new right, we add combinations:
// [right]
// [right - 1, right]
// [right - 2, right - 1, right]
// [..., right - 2, right - 1, right]
// [left, ..., right - 2, right - 1, right]
// so, its count is (right - left + 1)
count += right - left + 1;
}
return count;
}
};