Problem
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.
Examples
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]
Solution
Method 1 - DFS
Using DFS in a preorder manner(i.e. add the result as soon as we find it and then process the children) to generate numbers in lexicographical order:
- Start from the number
1
. - Recursively try to go deeper (multiplying by 10) to maintain the smallest lexical order.
- If multiplying is not possible, i.e. number exceed n, we backtrack , and consider the next number (increment), and go to siblings
- Continue this process until all numbers are processed.
For eg.. number 133, here is how the dfs tree:
graph TD A["Start"] A --> B1["1"] A --> B2["2"] A --> B3["3"] A --> B4["..."] B1 --> C1["10"] B1 --> C2["11"] B1 --> C3["12"] B1 --> C4["13"] B1 --> C5["..."] B1 --> C6["19"] C1 --> D1["100"] C1 --> D2["101"] C1 --> D3["..."] C1 --> D4["109"] D1 ---> E1["1000"] C2 --> F1["110"] C2 --> F2["111"] C2 --> F3["..."] C2 --> F4["119"] C3 --> G1["120"] C3 --> G2["..."] C4 --> G3["130"] C4 --> G4["131"] C4 --> G5["132"] C4 --> G6["133"] E1:::red classDef red fill:#f96,stroke:#333,stroke-width:2px;
Code
Java
public class Solution {
public List<Integer> lexicalOrder(int n) {
List<Integer> result = new ArrayList<>();
for (int i = 1; i < 10; i++) {
dfs(i, n, ans);
}
return result;
}
private void dfs(int current, int n, List<Integer> ans) {
if (current > n) {
return;
}
ans.add(current);
for (int i = 0; i < 10; i++) {
if (10 * current + i > n) {
return;
}
dfs(10 * current + i, n, ans);
}
}
}
Python
class Solution:
def lexicalOrder(self, n: int):
result = []
for i in range(1, 10):
self.dfs(i, n, result)
return result
def dfs(self, current, n, result):
if current > n:
return
result.append(current)
for i in range(10):
next_num = 10 * current + i
if next_num > n:
return
self.dfs(next_num, n, result)
Complexity
- ⏰ Time complexity:
O(n)
, because we generate and process each number exactly once. - 🧺 Space complexity:
O(1)
- Typical recursion depth in the worst case can be (O(\log_{10} n)), but this is not considered extra space complexity based on the problem constraints.
Method 2 - Iterative
To generate numbers in lexicographical order from 1
to n
, we simulate the process of numbering in dictionary order by:
- Initializing the current number as
1
. - Attempting to multiply the
current
number by10
if possible to maintain the smallest lexical order. - Incrementing the number when multiplying results in a value greater than
n
. - Handling cases where the incremented number results need adjusting to valid lexicographical numbers.
Code
Java
public class Solution {
public List<Integer> lexicalOrder(int n) {
List<Integer> ans = new ArrayList<>(n);
int current = 1;
for (int i = 0; i < n; i++) {
ans.add(current);
if (current * 10 <= n) {
current *= 10;
} else {
if (current >= n) {
current /= 10;
}
current++;
while (current % 10 == 0) {
current /= 10;
}
}
}
return result;
}
}
Python
class Solution:
def lexicalOrder(self, n: int):
result = []
current = 1
for _ in range(n):
result.append(current)
if current * 10 <= n:
current *= 10
else:
if current >= n:
current //= 10
current += 1
while current % 10 == 0:
current //= 10
return result
Complexity
- ⏰ Time complexity:
O(n)
- 🧺 Space complexity:
O(1)