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Given an integer n, return 1 through n in lexicographical order.

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]

Clarifying Questions

  1. Q: What is the input range limitations for n?
    • A: The problem does not specify, but typical competitive programming constraints would mean assuming 1 <= n <= 5000 is reasonable unless specified otherwise.
  2. Q: Should the solution handle large inputs efficiently?
    • A: Yes, it should be efficient enough for larger values within typical constraints.
  3. Q: Do we need to handle invalid inputs for n?
    • A: We can assume n is always a positive integer according to the problem.

Strategy

  1. Brute-force Approach:
    • Generate the numbers from 1 to n and then sort them lexicographically.
    • This solution works but is not optimal given its (O(n \log n)) complexity due to sorting.
  2. Optimal Approach:
    • Utilize a depth-first search (DFS) strategy to generate numbers in lexicographical order directly.
    • This approach avoids the requirement to sort, thereby optimizing the solution.
    • We will initiate DFS from every digit from 1 to 9.

Detailed Steps:

  1. Initialize an empty result list.
  2. Perform DFS starting from each digit 1 to 9.
    • In each step of DFS, append the current number to the result list.
    • Recursively append each digit from 0 to 9 as the next digit unless it exceeds n.

Time Complexity

Code

Here is the Python implementation of the described approach:

def lexicalOrder(n: int) -> list:
    def dfs(current):
        if current > n:
            return
        result.append(current)
        for i in range(10):
            if 10 * current + i <= n:
                dfs(10 * current + i)
            else:
                break

    result = []
    for i in range(1, 10):
        dfs(i)
    return result

# Example Usage
n = 13
print(lexicalOrder(n))  # Output: [1,10,11,12,13,2,3,4,5,6,7,8,9]

This code will generate the lexicographical order of numbers from 1 to n by using DFS starting from each number 1 through 9 and recursively building upon those.

This approach ensures that the numbers are naturally constructed in lexicographical order without needing an extra sort step, providing an efficient solution.

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