How do you solve Kth Smallest Product of Two Sorted Arrays on LeetCode?

Kth Smallest Product of Two Sorted Arrays (LeetCode #2040) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max_product)) time complexity and O(1) space complexity. Key insight: Understanding how to efficiently count products is crucial for optimizing the solution.

What is the brute force solution for Kth Smallest Product of Two Sorted Arrays?

The brute force approach for Kth Smallest Product of Two Sorted Arrays is Brute Force, with O(n²) time complexity and O(n²) space complexity. The brute force approach generates all possible products of elements from both arrays and sorts them to find the k-th smallest. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n log(max_product)).

What algorithm or data structure is used to solve Kth Smallest Product of Two Sorted Arrays?

Kth Smallest Product of Two Sorted Arrays uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Two Pointers.

What are the interview tips for Kth Smallest Product of Two Sorted Arrays?

Always clarify the constraints and edge cases with the interviewer. Explain your thought process clearly while transitioning from brute force to optimal solutions. Practice binary search problems to become comfortable with the technique.

What are common mistakes when solving Kth Smallest Product of Two Sorted Arrays?

Not considering the edge cases where products can be negative. Failing to optimize the counting of products, leading to unnecessary computations.

#2040

Kth Smallest Product of Two Sorted Arrays

Hard

ArrayBinary SearchBinary SearchTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(max_product))
Space	O(n²)	O(1)

💡

Intuition

Time O(n log(max_product))Space O(1)

Instead of generating all products, we can use binary search to efficiently find the k-th smallest product. By determining how many products are less than or equal to a mid value, we can narrow down our search range.

⚙️

Algorithm

3 steps

1Step 1: Define the search range for products from min(nums1) * min(nums2) to max(nums1) * max(nums2).
2Step 2: Use binary search to find the k-th smallest product by checking how many products are less than or equal to a mid value.
3Step 3: Adjust the search range based on the count of products found.

solution.py21 lines

1# Full working Python code
2from bisect import bisect_right
3
4def count_less_equal(mid, nums1, nums2):
5    count = 0
6    for x in nums1:
7        count += bisect_right(nums2, mid // x)
8    return count
9
10def kthSmallestProduct(nums1, nums2, k):
11    left, right = min(nums1) * min(nums2), max(nums1) * max(nums2)
12    while left < right:
13        mid = (left + right) // 2
14        if count_less_equal(mid, nums1, nums2) < k:
15            left = mid + 1
16        else:
17            right = mid
18    return left
19
20result = kthSmallestProduct([2, 5], [3, 4], 2)
21print(result)

ℹ

Complexity note: The time complexity is O(n log(max_product)) because we perform binary search on the product range, and for each mid value, we count the number of products in O(n) time. Space complexity is O(1) as we use only a few variables.

1Understanding how to efficiently count products is crucial for optimizing the solution.
2Binary search can significantly reduce the time complexity by narrowing down the search space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.