LeetCode #386 — MEDIUM

Lexicographical Numbers

Move from brute-force thinking to an efficient approach using trie strategy.

The Problem

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 * 10⁴

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Trie

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #386: Lexicographical Numbers
class Solution {
    public List<Integer> lexicalOrder(int n) {
        List<Integer> ans = new ArrayList<>(n);
        int v = 1;
        for (int i = 0; i < n; ++i) {
            ans.add(v);
            if (v * 10 <= n) {
                v *= 10;
            } else {
                while (v % 10 == 9 || v + 1 > n) {
                    v /= 10;
                }
                ++v;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #386: Lexicographical Numbers
func lexicalOrder(n int) (ans []int) {
	v := 1
	for i := 0; i < n; i++ {
		ans = append(ans, v)
		if v*10 <= n {
			v *= 10
		} else {
			for v%10 == 9 || v+1 > n {
				v /= 10
			}
			v++
		}
	}
	return
}

# Accepted solution for LeetCode #386: Lexicographical Numbers
class Solution:
    def lexicalOrder(self, n: int) -> List[int]:
        ans = []
        v = 1
        for _ in range(n):
            ans.append(v)
            if v * 10 <= n:
                v *= 10
            else:
                while v % 10 == 9 or v + 1 > n:
                    v //= 10
                v += 1
        return ans

// Accepted solution for LeetCode #386: Lexicographical Numbers
impl Solution {
    pub fn lexical_order(n: i32) -> Vec<i32> {
        let mut ans = Vec::with_capacity(n as usize);
        let mut v = 1;
        for _ in 0..n {
            ans.push(v);
            if v * 10 <= n {
                v *= 10;
            } else {
                while v % 10 == 9 || v + 1 > n {
                    v /= 10;
                }
                v += 1;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #386: Lexicographical Numbers
function lexicalOrder(n: number): number[] {
    const ans: number[] = [];
    let v = 1;
    for (let i = 0; i < n; ++i) {
        ans.push(v);
        if (v * 10 <= n) {
            v *= 10;
        } else {
            while (v % 10 === 9 || v === n) {
                v = Math.floor(v / 10);
            }
            ++v;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(L)

Space

O(N × L)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.