What is the time complexity of Kth Smallest Product of Two Sorted Arrays?

The optimal solution has a time complexity of O(m * log(n) * log(10000000000)).

What data structures are used to solve Kth Smallest Product of Two Sorted Arrays?

This problem uses Array, Binary Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Kth Smallest Product of Two Sorted Arrays?

Insight 1: Binary Search is the key. The problem can be efficiently solved by using binary search on the range of possible product values. We don't need to generate all products. Insight 2: Counting products less than or equal to a given value. For a given value `mid`, efficiently determining the number of products less than or equal to `mid` is crucial for the binary search process.

Which companies ask LeetCode 2040. Kth Smallest Product of Two Sorted Arrays in interviews?

This problem is frequently asked at LinkedIn.

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Kth Smallest Product of Two Sorted Arrays - LeetCode 2040 Solution

Q: How to solve Kth Smallest Product of Two Sorted Arrays?

The solution uses binary search on the range of possible product values to find the k-th smallest product. The `count_le` function is a helper function that, given a value `val`, counts the number of products `nums1[i] * nums2[j]` that are less than or equal to `val`. The binary search iteratively narrows down the search space until it finds the smallest value `ans` such that there are at least `k` products less than or equal to `ans`.

Kth Smallest Product of Two Sorted Arrays - Complete Solution Guide

Kth Smallest Product of Two Sorted Arrays is LeetCode problem 2040, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two sorted 0-indexed integer arrays nums1 and nums2 as well as an integer k , return the k th ( 1-based ) smallest product of nums1[i] * nums2[j] where 0 <= i < nums1.length and 0 <= j < nums2.length . Example 1: Input: nums1 = [2,5], nums2 = [3,4], k = 2 Output: 8 Explanation: The 2 smallest products are: - nums1[0] * nums2[0] = 2 * 3 = 6 - nums1[0] * nums2[1] = 2 * 4 = 8 The 2 nd smallest product is 8. Example 2: Input: nums1 = [-4,-2,0,3], nums2 = [2,4], k = 6 Output: 0 Explanation: The

Detailed Explanation

The problem requires finding the k-th smallest product formed by multiplying elements from two sorted integer arrays, `nums1` and `nums2`. We need to consider all possible pairs of elements, one from each array, calculate their product, and then identify the k-th smallest among these products. Crucially, we don't need to actually calculate all the products and sort them; the goal is to find an efficient method to determine the k-th smallest product without explicitly generating all possible products.

Solution Approach

The solution uses binary search on the range of possible product values to find the k-th smallest product. The `count_le` function is a helper function that, given a value `val`, counts the number of products `nums1[i] * nums2[j]` that are less than or equal to `val`. The binary search iteratively narrows down the search space until it finds the smallest value `ans` such that there are at least `k` products less than or equal to `ans`.

Step-by-Step Algorithm

Step 1: Initialize `low` and `high` for binary search. `low` is the smallest possible product (-10^10 - 1) and `high` is the largest possible product (10^10 + 1).
Step 2: Implement the `count_le` function. This function takes a value `val` and counts the number of pairs `(nums1[i], nums2[j])` such that their product is less than or equal to `val`. It iterates through `nums1` and, for each element `n1`, counts how many elements `n2` in `nums2` satisfy the condition `n1 * n2 <= val`. Binary search is used within `count_le` to efficiently find this count for each `n1`.
Step 3: Perform binary search on the range [low, high]. In each iteration, calculate the `mid` value and call `count_le(mid)`. If `count_le(mid)` is greater than or equal to `k`, it means that the k-th smallest product is less than or equal to `mid`. Update `ans` to `mid` and reduce the search space to `[low, mid - 1]`. Otherwise, increase the search space to `[mid + 1, high]`.
Step 4: Return the final `ans` value, which represents the k-th smallest product.

Key Insights

Insight 1: Binary Search is the key. The problem can be efficiently solved by using binary search on the range of possible product values. We don't need to generate all products.
Insight 2: Counting products less than or equal to a given value. For a given value `mid`, efficiently determining the number of products less than or equal to `mid` is crucial for the binary search process.
Insight 3: Handling negative and zero values. The arrays can contain negative, zero, and positive numbers. The count_le function must handle these cases correctly, as the order changes with the negative numbers.
Insight 4: Potential Integer Overflow. The product of two integers from the input arrays might exceed the range of `int`. It is necessary to use `long` to store products to avoid overflow.

Complexity Analysis

Time Complexity: O(m * log(n) * log(10000000000))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: LinkedIn.

Kth Smallest Product of Two Sorted Arrays | LeetCode 2040

Hard

Array Binary Search

Time: O(m * log(n) * log(10000000000))Space: O(1)

Solution Code

Available in:

class Solution {
    public long kthSmallestProduct(int[] nums1, int[] nums2, long k) {
        
        
        long low = (long) -1e10 - 1;
        long high = (long) 1e10 + 1;
        long ans = high;

        while (low <= high) {
            long mid = low + (high - low) / 2;
            if (count_le(nums1, nums2, mid) >= k) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        
        return ans;
    }

    private long count_le(int[] nums1, int[] nums2, long val) {
        long count = 0;
        int m = nums2.length;

        for (int n1 : nums1) {
            if (n1 > 0) {
                int l = 0, r = m - 1;
                int p = -1;
                while (l <= r) {
                    int mid_idx = l + (r - l) / 2;
                    if ((long) n1 * nums2[mid_idx] <= val) {
                        p = mid_idx;
                        l = mid_idx + 1;
                    } else {
                        r = mid_idx - 1;
                    }
                }
                count += (p + 1);
            } else if (n1 < 0) {
                int l = 0, r = m - 1;
                int p = m;
                while (l <= r) {
                    int mid_idx = l + (r - l) / 2;
                    if ((long) n1 * nums2[mid_idx] <= val) {
                        p = mid_idx;
                        r = mid_idx - 1;
                    } else {
                        l = mid_idx + 1;
                    }
                }
                count += (m - p);
            } else {  // n1 == 0
                if (val >= 0) {
                    count += m;
                }
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 2040: Kth Smallest Product of Two Sorted Arrays - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2040

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: Initialize `low` and `high` for binary search. `low` is the smallest possible product (-10^10 - 1) and `high` is the largest possible product (10^10 + 1).
2
Step 2: Implement the `count_le` function. This function takes a value `val` and counts the number of pairs `(nums1[i], nums2[j])` such that their product is less than or equal to `val`. It iterates through `nums1` and, for each element `n1`, counts how many elements `n2` in `nums2` satisfy the condition `n1 * n2 <= val`. Binary search is used within `count_le` to efficiently find this count for each `n1`.
3
Step 3: Perform binary search on the range [low, high]. In each iteration, calculate the `mid` value and call `count_le(mid)`. If `count_le(mid)` is greater than or equal to `k`, it means that the k-th smallest product is less than or equal to `mid`. Update `ans` to `mid` and reduce the search space to `[low, mid - 1]`. Otherwise, increase the search space to `[mid + 1, high]`.
4
Step 4: Return the final `ans` value, which represents the k-th smallest product.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m * log(n) * log(10000000000))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search on the potential range of product values gives O(log(range)) which is O(log(2 * 10^10)). Within each step of this binary search, we use another binary search to find the count. Hence the final complexity is O(m * log(n) * log(10^10)).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Heap-based approach

Maintain a min-heap of size k. Iterate through all product pairs, and for each product, insert it into the heap. If the heap size exceeds k, remove the largest element from the heap. After processing all product pairs, the k-th smallest element can be found at the top of the heap. However, it is less efficient than the implemented solution.

Time: O(m * n * log(k)), where m and n are the lengths of nums1 and nums2 respectively.Space: O(k)

Trade-offs: This approach can be simpler to implement but has a higher time complexity compared to the binary search approach, especially when k is relatively large. Use it when k is small and implementation simplicity is prioritized.

✨

Key Insights

1Insight 1: Binary Search is the key. The problem can be efficiently solved by using binary search on the range of possible product values. We don't need to generate all products.
2Insight 2: Counting products less than or equal to a given value. For a given value `mid`, efficiently determining the number of products less than or equal to `mid` is crucial for the binary search process.
3Insight 3: Handling negative and zero values. The arrays can contain negative, zero, and positive numbers. The count_le function must handle these cases correctly, as the order changes with the negative numbers.
4Insight 4: Potential Integer Overflow. The product of two integers from the input arrays might exceed the range of `int`. It is necessary to use `long` to store products to avoid overflow.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Zeroes in the input arrays. If either `nums1` or `nums2` contains zeros, then all products involving zero will be zero. The algorithm correctly handles these cases by appropriately incrementing the `count` in the `count_le` function.
Edge case 2: Negative numbers in the input arrays. The algorithm must properly count the numbers when either or both numbers are negative, reversing the sort order. The solution counts correctly because it considers cases where n1 is positive, negative or zero.
Edge case 3: k = 1. The solution will return the smallest value, which could be a negative value
Edge case 4: Integer Overflow. Because `nums1[i]` and `nums2[j]` can be as large as 10^5, their product can be up to 10^10. Using `int` to store the product will cause overflow. Therefore, the code uses `long` to store the product to avoid overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating products. Forgetting to use `long` can lead to incorrect results due to overflow.

✗Common mistake 2: Incorrectly counting the products less than or equal to a given value. Care must be taken when handling negative numbers and zeros to ensure that the products are counted correctly in the `count_le` function.

✗Common mistake 3: Not handling the case where k is equal to 1. It is important to test this edge case to ensure that the code works for the smallest possible value of k.

✗Common mistake 4: Forgetting that the arrays are already sorted. Not taking advantage of the sorted property would increase time complexity unnecessarily.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Recommendation systems. Finding the k-th smallest cost of purchasing different bundles of goods, where each bundle contains one item from each of several product categories.

🏦Real-world scenario 2: Resource allocation. Optimizing resource allocation in a cloud computing environment by finding the k-th smallest cost combination of using different virtual machines and storage options.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Kth Smallest Product of Two Sorted Arrays"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?