How do you solve Lexicographical Numbers on LeetCode?

Lexicographical Numbers (LeetCode #386) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Lexicographical order is similar to dictionary order.

What is the brute force solution for Lexicographical Numbers?

The brute force approach for Lexicographical Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all numbers from 1 to n, converting them to strings, and then sorting them lexicographically. This is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Lexicographical Numbers?

Lexicographical Numbers uses the following concepts: Depth-First Search, Trie. The recommended patterns to study are: Depth-First Search, Backtracking.

What are the interview tips for Lexicographical Numbers?

Explain your thought process clearly while coding. Consider edge cases, such as when n is less than 10. Practice writing recursive functions, as they are common in DFS problems.

What are common mistakes when solving Lexicographical Numbers?

Not considering leading zeros when generating numbers. Forgetting to check if the current number exceeds n.

#386

Lexicographical Numbers

Medium

Depth-First SearchTrieDepth-First SearchBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses a depth-first search (DFS) strategy to generate numbers in lexicographical order without needing to sort them afterwards. This is efficient and meets the space requirements.

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Algorithm

5 steps

1Step 1: Start with an empty list to hold the result.
2Step 2: Define a recursive function that takes the current number and builds the next number in lexicographical order.
3Step 3: For each number, append it to the result if it is less than or equal to n.
4Step 4: Multiply the current number by 10 to go deeper (e.g., from 1 to 10) and repeat.
5Step 5: If the current number is greater than n, return.

solution.py14 lines

1def lexicalOrder(n):
2    result = []
3    def dfs(current):
4        if current > n:
5            return
6        result.append(current)
7        for i in range(10):
8            dfs(current * 10 + i)
9    for i in range(1, 10):
10        dfs(i)
11    return result
12
13# Example usage:
14print(lexicalOrder(13))

ℹ

Complexity note: The time complexity is O(n) since we are generating each number once, and the space complexity is O(1) because we are not using any additional data structures that grow with n.

1Lexicographical order is similar to dictionary order.
2Using DFS allows us to explore numbers in the correct order without sorting.

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