Leetcode #2375: Construct Smallest Number From DI String
In this guide, we solve Leetcode #2375 Construct Smallest Number From DI String in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed string pattern of length n consisting of the characters 'I' meaning increasing and 'D' meaning decreasing. A 0-indexed string num of length n + 1 is created using the following conditions: num consists of the digits '1' to '9', where each digit is used at most once.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Greedy, String, Backtracking
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: pattern = "IIIDIDDD"
Output: "123549876"
Explanation:
At indices 0, 1, 2, and 4 we must have that num[i] < num[i+1].
At indices 3, 5, 6, and 7 we must have that num[i] > num[i+1].
Some possible values of num are "245639871", "135749862", and "123849765".
It can be proven that "123549876" is the smallest possible num that meets the conditions.
Note that "123414321" is not possible because the digit '1' is used more than once.
Python Solution
class Solution:
def smallestNumber(self, pattern: str) -> str:
def dfs(u):
nonlocal ans
if ans:
return
if u == len(pattern) + 1:
ans = ''.join(t)
return
for i in range(1, 10):
if not vis[i]:
if u and pattern[u - 1] == 'I' and int(t[-1]) >= i:
continue
if u and pattern[u - 1] == 'D' and int(t[-1]) <= i:
continue
vis[i] = True
t.append(str(i))
dfs(u + 1)
vis[i] = False
t.pop()
vis = [False] * 10
t = []
ans = None
dfs(0)
return ans
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.