What is the time complexity of K-th Smallest in Lexicographical Order?

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

What data structures are used to solve K-th Smallest in Lexicographical Order?

This problem uses Trie. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving K-th Smallest in Lexicographical Order?

Insight 1: The key is to realize that we don't need to generate all numbers from 1 to n and sort them. Instead, we can traverse the lexicographical tree and count how many numbers fall under each prefix. Insight 2: The count_steps function is crucial. It determines how many numbers are lexicographically between 'curr' and 'curr + 1' while being less than or equal to 'n'. This helps determine if we should move deeper into the tree (multiply by 10) or move to the next sibling (increment by 1).

Which companies ask LeetCode 440. K-th Smallest in Lexicographical Order in interviews?

This problem is frequently asked at DE Shaw, Hulu, Samsung.

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K-th Smallest in Lexicographical Order - LeetCode 440 Solution

Q: How to solve K-th Smallest in Lexicographical Order?

The solution uses a depth-first traversal approach on a conceptual lexicographical tree. The root of the tree is '1', and each node's children are formed by appending digits 0-9 to it. The algorithm starts with the current number 'curr' initialized to 1 and reduces k until it reaches 0. In each iteration, it calculates the number of integers between 'curr' and 'curr + 1' which are also less than or equal to 'n' using the `count_steps` function. If the number of integers 'steps' is less than or equal to the current value of 'k', it means the k-th smallest integer is not in the subtree rooted at 'curr', so we increment 'curr' by 1 and reduce 'k' by 'steps'. Otherwise, it means the k-th smallest integer is in the subtree rooted at 'curr', so we move deeper into the tree by multiplying 'curr' by 10 and reduce 'k' by 1.

K-th Smallest in Lexicographical Order - Complete Solution Guide

K-th Smallest in Lexicographical Order is LeetCode problem 440, 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 integers n and k , return the k th lexicographically smallest integer in the range [1, n] . Example 1: Input: n = 13, k = 2 Output: 10 Explanation: The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10. Example 2: Input: n = 1, k = 1 Output: 1 Constraints: 1 <= k <= n <= 10 9

Detailed Explanation

The problem asks us to find the k-th smallest integer in lexicographical order among the integers from 1 to n. Lexicographical order is similar to alphabetical order but for numbers. For example, if n = 13 and k = 2, the lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], and the 2nd smallest number is 10. The constraints specify that 1 <= k <= n <= 10^9.

Solution Approach

The solution uses a depth-first traversal approach on a conceptual lexicographical tree. The root of the tree is '1', and each node's children are formed by appending digits 0-9 to it. The algorithm starts with the current number 'curr' initialized to 1 and reduces k until it reaches 0. In each iteration, it calculates the number of integers between 'curr' and 'curr + 1' which are also less than or equal to 'n' using the `count_steps` function. If the number of integers 'steps' is less than or equal to the current value of 'k', it means the k-th smallest integer is not in the subtree rooted at 'curr', so we increment 'curr' by 1 and reduce 'k' by 'steps'. Otherwise, it means the k-th smallest integer is in the subtree rooted at 'curr', so we move deeper into the tree by multiplying 'curr' by 10 and reduce 'k' by 1.

Step-by-Step Algorithm

Step 1: Initialize 'curr' to 1 and decrement 'k' by 1. The reason for decrementing k by one initially is to convert it to a zero-based index, making calculations easier during traversal.
Step 2: While 'k' is greater than 0, calculate the number of steps between 'curr' and 'curr + 1' which are less than or equal to 'n' using the `count_steps` function.
Step 3: If 'steps' is less than or equal to 'k', subtract 'steps' from 'k' and increment 'curr' by 1. This means the k-th smallest number is not in the subtree of 'curr' and we advance to the next sibling.
Step 4: Otherwise, if 'steps' is greater than 'k', subtract 1 from 'k' and multiply 'curr' by 10. This means the k-th smallest number is in the subtree of 'curr' and we go down one level in the lexicographical tree.
Step 5: Repeat steps 2-4 until 'k' is equal to 0. At this point, 'curr' will be the k-th smallest number.

Key Insights

Insight 1: The key is to realize that we don't need to generate all numbers from 1 to n and sort them. Instead, we can traverse the lexicographical tree and count how many numbers fall under each prefix.
Insight 2: The count_steps function is crucial. It determines how many numbers are lexicographically between 'curr' and 'curr + 1' while being less than or equal to 'n'. This helps determine if we should move deeper into the tree (multiply by 10) or move to the next sibling (increment by 1).
Insight 3: It's important to use long data types (in C++ and C) or be mindful of potential integer overflow when calculating steps, especially with the multiplication and addition operations.

Complexity Analysis

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

Space Complexity: O(1)

Topics

This problem involves: Trie.

Companies

Asked at: DE Shaw, Hulu, Samsung.

K-th Smallest in Lexicographical Order | LeetCode 440

Hard

Trie

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

Solution Code

Available in:

class Solution {
    public int findKthNumber(int n, int k) {
        
        
        k -= 1;
        int curr = 1;
        
        while (k > 0) {
            int steps = count_steps(curr, curr + 1, n);
            
            if (steps <= k) {
                k -= steps;
                curr += 1;
            } else {
                k -= 1;
                curr *= 10;
            }
        }
        
        return curr;
    }
    
    private int count_steps(int n1, int n2, int n) {
        int steps = 0;
        while (n1 <= n) {
            steps += Math.min(n + 1, n2) - n1;
            n1 *= 10;
            n2 *= 10;
        }
        return steps;
    }
}

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Problem Info

Description

Problem 440: K-th Smallest in Lexicographical Order - Detailed explanation and solution approaches.

Topics

Trie

Companies

DE ShawHuluSamsung

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:440

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

5 steps

1
Step 1: Initialize 'curr' to 1 and decrement 'k' by 1. The reason for decrementing k by one initially is to convert it to a zero-based index, making calculations easier during traversal.
2
Step 2: While 'k' is greater than 0, calculate the number of steps between 'curr' and 'curr + 1' which are less than or equal to 'n' using the `count_steps` function.
3
Step 3: If 'steps' is less than or equal to 'k', subtract 'steps' from 'k' and increment 'curr' by 1. This means the k-th smallest number is not in the subtree of 'curr' and we advance to the next sibling.
4
Step 4: Otherwise, if 'steps' is greater than 'k', subtract 1 from 'k' and multiply 'curr' by 10. This means the k-th smallest number is in the subtree of 'curr' and we go down one level in the lexicographical tree.
5
Step 5: Repeat steps 2-4 until 'k' is equal to 0. At this point, 'curr' will be the k-th smallest number.

💡

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(log(n)*log(n))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates roughly log(n) times in the worst case as 'curr' increases. The count_steps function also iterates a maximum of log(n) times because the numbers are multiplied by 10 each time. Therefore, the overall time complexity is O(log(n) * log(n)). The algorithm uses only a few variables and doesn't use any extra data structures that scale with the input size, resulting in O(1) space complexity.

📖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

Brute Force

Generate all numbers from 1 to n, store them in a list, sort the list lexicographically, and return the element at index k-1.

Time: O(n log n)Space: O(n)

Trade-offs: This is a simple approach but inefficient for large values of n due to the sorting step. It has a higher time and space complexity compared to the optimized approach.

✨

Key Insights

1Insight 1: The key is to realize that we don't need to generate all numbers from 1 to n and sort them. Instead, we can traverse the lexicographical tree and count how many numbers fall under each prefix.
2Insight 2: The count_steps function is crucial. It determines how many numbers are lexicographically between 'curr' and 'curr + 1' while being less than or equal to 'n'. This helps determine if we should move deeper into the tree (multiply by 10) or move to the next sibling (increment by 1).
3Insight 3: It's important to use long data types (in C++ and C) or be mindful of potential integer overflow when calculating steps, especially with the multiplication and addition operations.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: k = 1. The algorithm correctly handles this case by initializing curr=1, and then decrementing k, thus returning 1 if it meets the exit condition right away.
Edge case 2: n is a small number (e.g., n=1). The while loop condition (n1 <= n) in count_steps and the main while loop are correctly evaluated to handle these cases, and count_steps returns valid numbers.
Edge case 3: Large n and k values approaching n. The use of long data types where applicable (C, C++) and/or use of Math.min in Java helps to prevent overflow issues that might occur in step calculations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. When calculating steps in the count_steps function, intermediate results (especially with multiplication) can exceed the maximum value of an integer, leading to incorrect results. Using long types in languages like C++ and C is critical.

✗Common mistake 2: Incorrectly calculating the number of steps. The count_steps function must carefully calculate the number of integers between curr and curr+1 while respecting the upper bound n. Off-by-one errors are common here.

✗Common mistake 3: Not handling k=1 properly. The initial k-=1 adjustment must be considered in the exit condition for early termination.

✗Common mistake 4: Confusing lexicographical order with numerical order.

✅

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: Autocompletion feature in search engines or IDEs. The lexicographical order is used to suggest possible completions based on the user's input.

🏦Real-world scenario 2: Database indexing. Lexicographical order is often used for creating indexes in databases to enable efficient searching and retrieval of data.

💡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.

👨‍💻 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: August 20, 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 13

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

@AlgoMasterJul 6

❓

Frequently Asked Questions

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 "K-th Smallest in Lexicographical Order"?

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

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