What is the time complexity of Kth Smallest Number in Multiplication Table?

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

What data structures are used to solve Kth Smallest Number in Multiplication Table?

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

What are the key insights for solving Kth Smallest Number in Multiplication Table?

Insight 1: The multiplication table is inherently sorted. However, a complete sort is computationally expensive. We can leverage the sorted nature implicitly using binary search. Insight 2: Binary search can be applied on the possible values in the multiplication table. The lower bound is 1, and the upper bound is m * n. For any 'mid' value in the search space, we need to efficiently count how many elements are less than or equal to 'mid' in the multiplication table.

Which companies ask LeetCode 668. Kth Smallest Number in Multiplication Table in interviews?

This problem is frequently asked at DE Shaw.

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Kth Smallest Number in Multiplication Table - LeetCode 668 Solution

Q: How to solve Kth Smallest Number in Multiplication Table?

The solution uses binary search to find the k-th smallest element. It searches through the possible values from 1 to m*n. For each mid-value during the binary search, the `count_le` function counts the number of elements in the multiplication table that are less than or equal to the `mid` value. If the count is greater than or equal to k, then we know the k-th smallest element is less than or equal to mid, so we narrow down the search space by setting high = mid. Otherwise, if the count is less than k, we narrow the search space by setting low = mid + 1. Eventually, low and high converge to the k-th smallest element.

Kth Smallest Number in Multiplication Table - Complete Solution Guide

Kth Smallest Number in Multiplication Table is LeetCode problem 668, 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

Nearly everyone has used the Multiplication Table . The multiplication table of size m x n is an integer matrix mat where mat[i][j] == i * j ( 1-indexed ). Given three integers m , n , and k , return the k th smallest element in the m x n multiplication table . Example 1: Input: m = 3, n = 3, k = 5 Output: 3 Explanation: The 5 th smallest number is 3. Example 2: Input: m = 2, n = 3, k = 6 Output: 6 Explanation: The 6 th smallest number is 6. Constraints: 1 <= m, n <= 3 * 10 4 1 <= k <= m * n

Detailed Explanation

The problem asks us to find the k-th smallest number in an m x n multiplication table. The multiplication table is a matrix where the element at row i and column j is equal to i * j (1-indexed). We are given m, n, and k, and need to return the k-th smallest element.

Solution Approach

The solution uses binary search to find the k-th smallest element. It searches through the possible values from 1 to m*n. For each mid-value during the binary search, the `count_le` function counts the number of elements in the multiplication table that are less than or equal to the `mid` value. If the count is greater than or equal to k, then we know the k-th smallest element is less than or equal to mid, so we narrow down the search space by setting high = mid. Otherwise, if the count is less than k, we narrow the search space by setting low = mid + 1. Eventually, low and high converge to the k-th smallest element.

Step-by-Step Algorithm

Step 1: Initialize low to 1 and high to m * n. These represent the range of possible values for the k-th smallest element.
Step 2: Perform binary search while low < high.
Step 3: Calculate mid = low + (high - low) // 2. This avoids potential integer overflow issues.
Step 4: Call the `count_le` function to count the number of elements in the multiplication table that are less than or equal to mid.
Step 5: If `count_le(mid)` is greater than or equal to k, it means the k-th smallest number is less than or equal to mid, so set high = mid.
Step 6: Otherwise, if `count_le(mid)` is less than k, it means the k-th smallest number is greater than mid, so set low = mid + 1.
Step 7: After the while loop finishes, low will be equal to the k-th smallest number. Return low.

Key Insights

Insight 1: The multiplication table is inherently sorted. However, a complete sort is computationally expensive. We can leverage the sorted nature implicitly using binary search.
Insight 2: Binary search can be applied on the possible values in the multiplication table. The lower bound is 1, and the upper bound is m * n. For any 'mid' value in the search space, we need to efficiently count how many elements are less than or equal to 'mid' in the multiplication table.
Insight 3: Counting elements <= mid can be optimized. For each row 'i', the number of elements <= mid is min(n, mid // i). If mid // i is 0, it means there are no elements in the row that are <= mid, and we can break early.

Complexity Analysis

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

Space Complexity: O(1)

Topics

This problem involves: Math, Binary Search.

Companies

Asked at: DE Shaw.

Kth Smallest Number in Multiplication Table | LeetCode 668

Hard

Math Binary Search

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

Solution Code

Available in:

class Solution {
    public int findKthNumber(int m, int n, int k) {
        if (m > n) {
            int temp = m;
            m = n;
            n = temp;
        }

        int low = 1;
        int high = m * n;

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (count_le(mid, m, n) >= k) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }

        return low;
    }

    private int count_le(int x, int m, int n) {
        int count = 0;
        for (int i = 1; i <= m; i++) {
            int add = Math.min(n, x / i);
            if (add == 0) {
                break;
            }
            count += add;
        }
        return count;
    }
}

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Problem Info

Description

Problem 668: Kth Smallest Number in Multiplication Table - Detailed explanation and solution approaches.

Topics

Math Binary Search

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:668

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

7 steps

1
Step 1: Initialize low to 1 and high to m * n. These represent the range of possible values for the k-th smallest element.
2
Step 2: Perform binary search while low < high.
3
Step 3: Calculate mid = low + (high - low) // 2. This avoids potential integer overflow issues.
4
Step 4: Call the `count_le` function to count the number of elements in the multiplication table that are less than or equal to mid.
5
Step 5: If `count_le(mid)` is greater than or equal to k, it means the k-th smallest number is less than or equal to mid, so set high = mid.
6
Step 6: Otherwise, if `count_le(mid)` is less than k, it means the k-th smallest number is greater than mid, so set low = mid + 1.
7
Step 7: After the while loop finishes, low will be equal to the k-th smallest number. Return low.

💡

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(m*n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search runs in O(log(m * n)) time because the search space is from 1 to m * n. The `count_le` function iterates through the rows of the multiplication table, which takes O(m) time. Therefore, the overall time complexity is O(m * log(m * n)). The algorithm uses a constant amount of extra space, so the space complexity is O(1).

📖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

Using a min-heap to store the elements of the multiplication table, initially containing the first element of each row. Then, repeatedly pop the smallest element from the heap and add the next element from the same row until the k-th element is popped.

Time: O(k * log(m))Space: O(m)

Trade-offs: This approach can be faster than the binary search approach if k is significantly smaller than m * n, but it uses more space (O(m) vs O(1)). If k is closer to m * n, the binary search is better due to lower space complexity.

✨

Key Insights

1Insight 1: The multiplication table is inherently sorted. However, a complete sort is computationally expensive. We can leverage the sorted nature implicitly using binary search.
2Insight 2: Binary search can be applied on the possible values in the multiplication table. The lower bound is 1, and the upper bound is m * n. For any 'mid' value in the search space, we need to efficiently count how many elements are less than or equal to 'mid' in the multiplication table.
3Insight 3: Counting elements <= mid can be optimized. For each row 'i', the number of elements <= mid is min(n, mid // i). If mid // i is 0, it means there are no elements in the row that are <= mid, and we can break early.

📚

Key Terms Explained

For Students

Binary Search

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

Edge Cases

Edge case 1: m = 1 or n = 1. The algorithm correctly handles this case as the count_le will still function correctly, summing from i=1 to m (which could be just 1)
Edge case 2: k = 1. The binary search will correctly converge to the smallest element, which is 1.
Edge case 3: k = m * n. The binary search will correctly converge to the largest element, which is m * n.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using integer division without considering the order of operations in `count_le`. Ensure the division `x / i` is performed before comparing with n or using in min.

✗Common mistake 2: Integer overflow when calculating `mid = (low + high) / 2`. Use `mid = low + (high - low) // 2` to prevent this.

✗Common mistake 3: Incorrectly implementing the `count_le` function leading to an incorrect count of elements less than or equal to mid. Specifically, failing to break the inner loop when `add == 0` resulting in unnecessary iterations.

✅

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: Optimizing resource allocation where tasks can be represented in a matrix. Determining the 'k-th most efficient' allocation combination.

🏦Real-world scenario 2: Ranking systems that involve multiplicative factors. Finding the k-th ranked item based on the product of multiple attributes.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: October 7, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

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 Number in Multiplication Table"?

For this hard-level Math 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?