Problem Statement
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
Try here before watching the video.
Similar Problem
Video Solution
Java Code
class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if(k <= 1) return 0;
int start = 0, subarrayCount = 0;
int product = 1;
for(int end = 0; end < nums.length; end++) {
product *= nums[end];
while(product >= k) {
product /= nums[start];
start += 1;
}
subarrayCount += (end - start + 1);
}
return subarrayCount;
}
}
C++ Code
class Solution {
public:
int numSubarrayProductLessThanK(vector<int>& nums, int k) {
if(k <= 1) return 0;
int start = 0, subarrayCount = 0;
int product = 1;
for(int end = 0; end < nums.size(); end++) {
product *= nums[end];
while(product >= k) {
product /= nums[start];
start += 1;
}
subarrayCount += (end - start + 1);
}
return subarrayCount;
}
};
Python Code
class Solution:
def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
if k <= 1:
return 0
start = 0
subarray_count = 0
product = 1
for end in range(len(nums)):
product *= nums[end]
while product >= k:
product /= nums[start]
start += 1
subarray_count += (end - start + 1)
return subarray_count
Javascript Code
var numSubarrayProductLessThanK = function(nums, k) {
if (k <= 1) return 0;
let start = 0;
let subarrayCount = 0;
let product = 1;
for (let end = 0; end < nums.length; end++) {
product *= nums[end];
while (product >= k) {
product /= nums[start];
start += 1;
}
subarrayCount += (end - start + 1);
}
return subarrayCount;
};
Go Code
func numSubarrayProductLessThanK(nums []int, k int) int {
if k <= 1 {
return 0
}
start := 0
subarrayCount := 0
product := 1
for end := 0; end < len(nums); end++ {
product *= nums[end]
for product >= k {
product /= nums[start]
start += 1
}
subarrayCount += (end - start + 1)
}
return subarrayCount
}
Complexity Analysis
Time Complexity: O(N)
Space Complexity: O(1)