Leetcode #2663: Lexicographically Smallest Beautiful String
In this guide, we solve Leetcode #2663 Lexicographically Smallest Beautiful 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
A string is beautiful if: It consists of the first k letters of the English lowercase alphabet. It does not contain any substring of length 2 or more which is a palindrome.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, String
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: s = "abcz", k = 26
Output: "abda"
Explanation: The string "abda" is beautiful and lexicographically larger than the string "abcz".
It can be proven that there is no string that is lexicographically larger than the string "abcz", beautiful, and lexicographically smaller than the string "abda".
Python Solution
class Solution:
def smallestBeautifulString(self, s: str, k: int) -> str:
n = len(s)
cs = list(s)
for i in range(n - 1, -1, -1):
p = ord(cs[i]) - ord('a') + 1
for j in range(p, k):
c = chr(ord('a') + j)
if (i > 0 and cs[i - 1] == c) or (i > 1 and cs[i - 2] == c):
continue
cs[i] = c
for l in range(i + 1, n):
for m in range(k):
c = chr(ord('a') + m)
if (l > 0 and cs[l - 1] == c) or (l > 1 and cs[l - 2] == c):
continue
cs[l] = c
break
return ''.join(cs)
return ''
Complexity
The time complexity is , where is the length of the string. The space complexity is .
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.