#### Approach #1: Binary Search on Logarithms [Accepted]

Intuition

Because , we can reduce the problem to subarray sums instead of subarray products. The motivation for this is that the product of some arbitrary subarray may be way too large (potentially 1000^50000), and also dealing with sums gives us some more familiarity as it becomes similar to other problems we may have solved before.

Algorithm

After this transformation where every value x becomes log(x), let us take prefix sums prefix[i+1] = nums + nums + ... + nums[i]. Now we are left with the problem of finding, for each i, the largest j so that nums[i] + ... + nums[j] = prefix[j] - prefix[i] < k.

Because prefix is a monotone increasing array, this can be solved with binary search. We add the width of the interval [i, j] to our answer, which counts all subarrays [i, k] with k <= j.

Python

class Solution(object):
def numSubarrayProductLessThanK(self, nums, k):
if k == 0: return 0
k = math.log(k)

prefix = 
for x in nums:
prefix.append(prefix[-1] + math.log(x))

ans = 0
for i, x in enumerate(prefix):
j = bisect.bisect(prefix, x + k - 1e-9, i+1)
ans += j - i - 1
return ans


Java

class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (k == 0) return 0;
double logk = Math.log(k);
double[] prefix = new double[nums.length + 1];
for (int i = 0; i < nums.length; i++) {
prefix[i+1] = prefix[i] + Math.log(nums[i]);
}

int ans = 0;
for (int i = 0; i < prefix.length; i++) {
int lo = i + 1, hi = prefix.length;
while (lo < hi) {
int mi = lo + (hi - lo) / 2;
if (prefix[mi] < prefix[i] + logk - 1e-9) lo = mi + 1;
else hi = mi;
}
ans += lo - i - 1;
}
return ans;
}
}


Complexity Analysis

• Time Complexity: , where is the length of nums. Inside our for loop, each binary search operation takes time.

• Space Complexity: , the space used by prefix.

#### Approach #2: Sliding Window [Accepted]

Intuition

For each right, call opt(right) the smallest left so that the product of the subarray nums[left] * nums[left + 1] * ... * nums[right] is less than k. opt is a monotone increasing function, so we can use a sliding window.

Algorithm

Our loop invariant is that left is the smallest value so that the product in the window prod = nums[left] * nums[left + 1] * ... * nums[right] is less than k.

For every right, we update left and prod to maintain this invariant. Then, the number of intervals with subarray product less than k and with right-most coordinate right, is right - left + 1. We'll count all of these for each value of right.

Python

class Solution(object):
def numSubarrayProductLessThanK(self, nums, k):
if k <= 1: return 0
prod = 1
ans = left = 0
for right, val in enumerate(nums):
prod *= val
while prod >= k:
prod /= nums[left]
left += 1
ans += right - left + 1
return ans


Java

class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (k <= 1) return 0;
int prod = 1, ans = 0, 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;
}
}


Complexity Analysis

• Time Complexity: , where is the length of nums. left can only be incremented at most N times.

• Space Complexity: , the space used by prod, left, and ans.

Analysis written by: @awice.