[
backtracking
trie
]
BinarySearch 0380 N Lexicographic Integers
Problem statement
https://binarysearch.com/problems/N-Lexicographic-Integers/
Solution
Equal to Leetcode 0386 Lexicographical Numbers.
Complexity
It is O(n)
for time and space.
Code
class Solution:
def solve(self, n):
def dfs(num):
self.result.append(num)
for i in range(1 if num == 0 else 0, 10):
if num*10 + i > n: break
dfs(num*10 + i)
self.result = []
dfs(0)
return self.result[1:]