# 713. Subarray Product Less Than K

## 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

## Solutions

• class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
int ans = 0;
for (int i = 0, j = 0, s = 1; i < nums.length; ++i) {
s *= nums[i];
while (j <= i && s >= k) {
s /= nums[j++];
}
ans += i - j + 1;
}
return ans;
}
}

• class Solution {
public:
int numSubarrayProductLessThanK(vector<int>& nums, int k) {
int ans = 0;
for (int i = 0, j = 0, s = 1; i < nums.size(); ++i) {
s *= nums[i];
while (j <= i && s >= k) s /= nums[j++];
ans += i - j + 1;
}
return ans;
}
};

• class Solution:
def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
ans, s, j = 0, 1, 0
for i, v in enumerate(nums):
s *= v
while j <= i and s >= k:
s //= nums[j]
j += 1
ans += i - j + 1
return ans


• func numSubarrayProductLessThanK(nums []int, k int) int {
ans := 0
for i, j, s := 0, 0, 1; i < len(nums); i++ {
s *= nums[i]
for ; j <= i && s >= k; j++ {
s /= nums[j]
}
ans += i - j + 1
}
return ans
}

• function numSubarrayProductLessThanK(nums: number[], k: number): number {
let ans = 0;
for (let i = 0, j = 0, s = 1; i < nums.length; ++i) {
s *= nums[i];
while (j <= i && s >= k) {
s /= nums[j++];
}
ans += i - j + 1;
}
return ans;
}


• /**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var numSubarrayProductLessThanK = function (nums, k) {
const n = nums.length;
let ans = 0;
let s = 1;
for (let i = 0, j = 0; i < n; ++i) {
s *= nums[i];
while (j <= i && s >= k) {
s = Math.floor(s / nums[j++]);
}
ans += i - j + 1;
}
return ans;
};


• impl Solution {
pub fn num_subarray_product_less_than_k(nums: Vec<i32>, k: i32) -> i32 {
if k <= 1 {
return 0;
}

let mut res = 0;
let mut product = 1;
let mut i = 0;
nums.iter()
.enumerate()
.for_each(|(j, v)| {
product *= v;
while product >= k {
product /= nums[i];
i += 1;
}
res += j - i + 1;
});
res as i32
}
}