Description
Given an array of integers nums
and an integer k
, return the number of contiguous subarrays where the product of all the elements in the subarray is strictly 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.
Example 2:
Input: nums = [1,2,3], k = 0 Output: 0
Constraints:
1 <= nums.length <= 3 * 104
1 <= nums[i] <= 1000
0 <= k <= 106
Solution
Python3
class Solution:
def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
N = len(nums)
res = 0
curr = 1
i = 0
for j, x in enumerate(nums):
curr *= x
while i <= j and curr >= k:
curr //= nums[i]
i += 1
res += j - i + 1
return res
C++
class Solution {
public:
int numSubarrayProductLessThanK(vector<int>& nums, int k) {
int c = 1, res = 0;
for (int i = 0,j = 0; j < nums.size(); j++){
c *= nums[j];
while (i <= j && c >= k)
c /= nums[i++];
res += (j - i + 1);
}
return res;
}
};