Leetcode #1625: Lexicographically Smallest String After Applying Operations
In this guide, we solve Leetcode #1625 Lexicographically Smallest String After Applying Operations 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 string s of even length consisting of digits from 0 to 9, and two integers a and b. You can apply either of the following two operations any number of times and in any order on s: Add a to all odd indices of s (0-indexed).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, String, Enumeration
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: s = "5525", a = 9, b = 2
Output: "2050"
Explanation: We can apply the following operations:
Start: "5525"
Rotate: "2555"
Add: "2454"
Add: "2353"
Rotate: "5323"
Add: "5222"
Add: "5121"
Rotate: "2151"
Add: "2050"
There is no way to obtain a string that is lexicographically smaller than "2050".
Python Solution
class Solution:
def findLexSmallestString(self, s: str, a: int, b: int) -> str:
q = deque([s])
vis = {s}
ans = s
while q:
s = q.popleft()
if ans > s:
ans = s
t1 = ''.join(
[str((int(c) + a) % 10) if i & 1 else c for i, c in enumerate(s)]
)
t2 = s[-b:] + s[:-b]
for t in (t1, t2):
if t not in vis:
vis.add(t)
q.append(t)
return ans
Complexity
The time complexity is and the space complexity is , where is the length of string . The space complexity is , where is the length of string .
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.