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The idea is to perform a depth-first search (DFS) to generate numbers in lexicographical order. Starting from the number 1, we explore all numbers that can be formed by appending digits from 0 to 9 as long as the generated number is less than or equal to n. If we reach a number greater than n, backtrack and explore the next number at the same level and repeat the process.
Time Complexity: O(n), Space Complexity: O(1) (excluding the output array).
1def dfs(current, n, result):
2 if current > n:
3 return
4 result.append(current)
5 for i in range(10):
6 if current * 10 + i > n:
7 return
8 dfs(current * 10 + i, n, result)
9
10def lexicalOrder(n):
11 result = []
12 for i in range(1, 10):
13 dfs(i, n, result)
14 return result
15
16n = 13
17print(lexicalOrder(n))This Python implementation leverages a result list and recursively calls dfs to append lexicographically ordered numbers until the bounds are met.
This iterative approach attempts to simulate the depth-first search behavior using an iterative procedure. It begins by using a number and incrementally moving to the next lexicographical number by clever digit manipulations. If a number reaches an immediate boundary, it backs off accordingly to stay within valid digits.
Time Complexity: O(n), Space Complexity: O(1) (ignoring return array).
1using System.Collections.Generic;
public class IterativeLexicalOrder {
public IList<int> LexicalOrder(int n) {
IList<int> result = new List<int>();
int num = 1;
for (int i = 0; i < n; ++i) {
result.Add(num);
if (num * 10 <= n) {
num *= 10;
} else {
while (num % 10 == 9 || num + 1 > n) {
num /= 10;
}
num++;
}
}
return result;
}
public static void Main(string[] args) {
IterativeLexicalOrder obj = new IterativeLexicalOrder();
int n = 13;
IList<int> result = obj.LexicalOrder(n);
foreach (int num in result) {
Console.Write(num + " ");
}
}
}This C# solution efficiently and iteratively computes the lexicographical order by leveraging systematic incrementation and resetting techniques to generate valid sequential numbers.