What is the time complexity of Lexicographical Numbers?

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

What data structures are used to solve Lexicographical Numbers?

This problem uses Depth-First Search, Trie. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Lexicographical Numbers?

Insight 1: Lexicographical order can be thought of as a modified depth-first search (DFS) on a tree where each node's children are its multiples by 10 up to n. Insight 2: The core idea is to start with 1, then try multiplying by 10. If the result is within the bound `n`, we continue. Otherwise, we increment by 1. If the last digit is 9, or incrementing causes us to exceed the bound, we backtrack to a higher level of the implicit DFS tree.

Which companies ask LeetCode 386. Lexicographical Numbers in interviews?

This problem is frequently asked at Barclays.

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Lexicographical Numbers - LeetCode 386 Solution

Q: How to solve Lexicographical Numbers?

The solution uses an iterative approach to simulate a depth-first search. It starts with the number 1 and iteratively generates the next lexicographically ordered number. The main logic involves three cases: 1. If multiplying the current number by 10 results in a number less than or equal to `n`, we multiply it by 10 (go deeper in the lexicographical tree). 2. Otherwise, if incrementing the current number by 1 doesn't result in a number ending in 0 (meaning the last digit wasn't 9) and remains within the bound of `n`, we increment it by 1 (go to the next sibling). 3. If neither of the above two conditions is met, we need to backtrack to a higher-level node in the lexicographical tree. This is done by repeatedly dividing by 10 until we reach a number whose last digit is not 9. Then, we increment that number by 1.

Lexicographical Numbers - Complete Solution Guide

Lexicographical Numbers is LeetCode problem 386, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer n , return all the numbers in the range [1, n] sorted in lexicographical order. You must write an algorithm that runs in O(n) time and uses O(1) extra space. Example 1: Input: n = 13 Output: [1,10,11,12,13,2,3,4,5,6,7,8,9] Example 2: Input: n = 2 Output: [1,2] Constraints: 1 <= n <= 5 * 10 4

Detailed Explanation

The problem asks us to generate a list of numbers from 1 to `n` (inclusive) and sort them in lexicographical order (dictionary order). For example, with `n = 13`, the lexicographical order is `[1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9]`. Essentially, we want to treat the numbers as strings and sort them as strings, but the output should be a list of integers. The constraint is that we should aim for O(n) time complexity and O(1) extra space (excluding the output list).

Solution Approach

The solution uses an iterative approach to simulate a depth-first search. It starts with the number 1 and iteratively generates the next lexicographically ordered number. The main logic involves three cases: 1. If multiplying the current number by 10 results in a number less than or equal to `n`, we multiply it by 10 (go deeper in the lexicographical tree). 2. Otherwise, if incrementing the current number by 1 doesn't result in a number ending in 0 (meaning the last digit wasn't 9) and remains within the bound of `n`, we increment it by 1 (go to the next sibling). 3. If neither of the above two conditions is met, we need to backtrack to a higher-level node in the lexicographical tree. This is done by repeatedly dividing by 10 until we reach a number whose last digit is not 9. Then, we increment that number by 1.

Step-by-Step Algorithm

Step 1: Initialize an empty result array of size `n` and initialize the current number `curr` to 1.
Step 2: Iterate `n` times. In each iteration, add the current number `curr` to the result array.
Step 3: Check if `curr * 10` is less than or equal to `n`. If it is, update `curr` to `curr * 10` (move to the next level in the lexicographical tree).
Step 4: If `curr * 10 > n`, check if incrementing `curr` by 1 is possible: check if `curr % 10 != 9` and `curr + 1 <= n`. If it is, update `curr` to `curr + 1` (move to the next sibling in lexicographical order).
Step 5: If neither step 3 nor step 4 is possible, it means we need to backtrack. Keep dividing `curr` by 10 until `curr % 10 != 9`. After finding such `curr`, divide it by 10 and increment by 1. `curr = curr / 10 + 1`.
Step 6: Repeat steps 2-5 until the result array is filled.

Key Insights

Insight 1: Lexicographical order can be thought of as a modified depth-first search (DFS) on a tree where each node's children are its multiples by 10 up to n.
Insight 2: The core idea is to start with 1, then try multiplying by 10. If the result is within the bound `n`, we continue. Otherwise, we increment by 1. If the last digit is 9, or incrementing causes us to exceed the bound, we backtrack to a higher level of the implicit DFS tree.
Insight 3: The provided solution implicitly simulates a depth-first traversal of a 'lexicographical tree' to construct the lexicographically sorted sequence.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Depth-First Search, Trie.

Companies

Asked at: Barclays.

Lexicographical Numbers | LeetCode 386

Medium

Depth-First Search Trie

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<Integer> lexicalOrder(int n) {
        List<Integer> result = new ArrayList<>(n);
        int curr = 1;
        for (int i = 0; i < n; i++) {
            result.add(curr);
            if (curr * 10 <= n) {
                curr *= 10;
            } else if (curr % 10 != 9 && curr + 1 <= n) {
                curr += 1;
            } else {
                while ((curr / 10) % 10 == 9) {
                    curr /= 10;
                }
                curr = curr / 10 + 1;
            }
        }
        return result;
    }
}

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Problem Info

Description

Problem 386: Lexicographical Numbers - Detailed explanation and solution approaches.

Topics

Depth-First Search Trie

Companies

Barclays

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:386

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

6 steps

1
Step 1: Initialize an empty result array of size `n` and initialize the current number `curr` to 1.
2
Step 2: Iterate `n` times. In each iteration, add the current number `curr` to the result array.
3
Step 3: Check if `curr * 10` is less than or equal to `n`. If it is, update `curr` to `curr * 10` (move to the next level in the lexicographical tree).
4
Step 4: If `curr * 10 > n`, check if incrementing `curr` by 1 is possible: check if `curr % 10 != 9` and `curr + 1 <= n`. If it is, update `curr` to `curr + 1` (move to the next sibling in lexicographical order).
5
Step 5: If neither step 3 nor step 4 is possible, it means we need to backtrack. Keep dividing `curr` by 10 until `curr % 10 != 9`. After finding such `curr`, divide it by 10 and increment by 1. `curr = curr / 10 + 1`.
6
Step 6: Repeat steps 2-5 until the result array is filled.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates exactly `n` times to populate the result array. The operations inside the loop (multiplication, increment, division) take constant time. Therefore, the overall time complexity is O(n). The space complexity is O(1) because we are only using a few integer variables (`curr`, `i`) and not using any extra data structures that scale with the input `n`. The output list does not count towards extra space as defined by the problem.

📖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

Sorting Strings

Convert numbers from 1 to `n` to strings, sort the strings lexicographically, and then convert the sorted strings back to integers.

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

Trade-offs: Simpler to implement, but less efficient than the O(n) solution due to the sorting step. This approach is generally less optimal given problem constraints.

Depth-First Search (Recursive)

Use recursion to perform a depth-first traversal of the lexicographical tree. This approach will maintain a global list to store sorted numbers.

Time: O(n)Space: O(n) for the recursion stack in worst case

Trade-offs: Can be more intuitive to understand conceptually, but has O(n) space complexity for recursion stack. The iterative approach uses constant extra space.

✨

Key Insights

1Insight 1: Lexicographical order can be thought of as a modified depth-first search (DFS) on a tree where each node's children are its multiples by 10 up to n.
2Insight 2: The core idea is to start with 1, then try multiplying by 10. If the result is within the bound `n`, we continue. Otherwise, we increment by 1. If the last digit is 9, or incrementing causes us to exceed the bound, we backtrack to a higher level of the implicit DFS tree.
3Insight 3: The provided solution implicitly simulates a depth-first traversal of a 'lexicographical tree' to construct the lexicographically sorted sequence.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: When `n` is a single digit number (e.g., `n = 5`), the algorithm correctly generates `[1, 2, 3, 4, 5]` because it simply increments the current number in each iteration.
Edge case 2: When `n` ends with a string of 9s (e.g., `n = 199`), the algorithm correctly backtracks to find the next lexicographical number. For instance, if `curr` is 199, it will be divided by 10 twice to become 1, then incremented to become 2.
Edge case 3: When n is 10, 11,12,..,19, the program will correctly generate [1,10,11,...,19,2,3,4,5,6,7,8,9]

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle backtracking correctly when the current number is near the upper bound `n` or ends with 9.

✗Common mistake 2: Using sorting based solutions that have O(n log n) time complexity, which do not meet the requirement of O(n) time complexity.

✗Common mistake 3: Incorrectly implementing the DFS traversal, leading to incorrect lexicographical ordering or missing numbers.

✅

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: File system navigation, where files and directories are often displayed in lexicographical order.

🏦Real-world scenario 2: Autocomplete and search suggestions, where suggestions are typically presented in lexicographical order to improve user experience.

🎮Real-world scenario 3: Database indexing, to ensure quick retrieval based on lexicographical ordering.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: June 27, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Lexicographical Numbers"?

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

🤖 Need help with this problem?