Algorithm
Problem Name: 713. Subarray Product Less Than K
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 * 1041 <= nums[i] <= 10000 <= k <= 106
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int numSubarrayProductLessThanK(vector<int>& nums, int k) {
int count = 0, product = 1;
for(int i = 0, j = 0, pre = 0; j < nums.size(); i++, pre = j, j = max(i, j)){
while(j < nums.size() && product * nums[j] < k) product *= nums[j++];
count += (long)(j - pre) * (1 + (j - pre)) / 2 + (j - pre) * (pre - i);
product = max(product / nums[i], 1);
}
return count;
}
};
Input
nums = [10,5,2,6], k = 100
Output
8
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (k <= 1) {
return 0;
}
int prod = 1;
int ans = 0;
int left = 0;
for (int right = 0; right < nums.length; right++) {
prod *= nums[right];
while (prod >= k) {
prod /= nums[left++];
}
ans += right - left + 1;
}
return ans;
}
}
Input
nums = [10,5,2,6], k = 100
Output
8
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const numSubarrayProductLessThanK = function(nums, k) {
if (k == 0) return 0
let cnt = 0
let pro = 1
for (let i = 0, j = 0, len = nums.length; j < len; j++) {
pro *= nums[j]
while (i <= j && pro >= k) {
pro /= nums[i++]
}
cnt += j - i + 1
}
return cnt
}
Input
nums = [1,2,3], k = 0
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def numSubarrayProductLessThanK(self, nums, k):
l, res, cur = 0, 0, 1
for i in range(len(nums)):
cur *= nums[i]
while cur >= k and l < i: l, cur = l + 1, cur // nums[l]
if cur < k: res += i - l + 1
return res
Input
nums = [1,2,3], k = 0
#5 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0713_SubarrayProductLessThanK
{
public int NumSubarrayProductLessThanK(int[] nums, int k)
{
if (k <= 1) return 0;
int prod = 1, result = 0, left = 0;
for (int right = 0; right < nums.Length; right++)
{
prod *= nums[right];
while (prod >= k)
prod /= nums[left++];
result += right - left + 1;
}
return result;
}
}
}
Input
nums = [10,5,2,6], k = 100
Output
8