How do you solve Break a Palindrome on LeetCode?

Break a Palindrome (LeetCode #1328) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Changing the first non-'a' character to 'a' gives the smallest lexicographical result.

What is the brute force solution for Break a Palindrome?

The brute force approach for Break a Palindrome is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying to replace each character in the palindrome with every possible lowercase letter and checking if the result is not a palindrome. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Break a Palindrome?

Break a Palindrome uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, Two Pointers.

What are the interview tips for Break a Palindrome?

Always consider edge cases, such as single-character strings. Think about the properties of palindromes and how they can be manipulated. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Break a Palindrome?

Not considering the case where the string length is 1, which should return an empty string. Failing to check for palindromes correctly after making a change.

#1328

Break a Palindrome

Medium

StringGreedyGreedyTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach focuses on the fact that we can make the smallest lexicographical change by changing the first non-'a' character to 'a', or the last character to 'b' if all are 'a'. This avoids unnecessary checks and is efficient.

⚙️

Algorithm

4 steps

1Step 1: If the length of the string is 1, return an empty string.
2Step 2: Iterate through the string until the middle (or the end for even lengths).
3Step 3: If a character is found that is not 'a', change it to 'a' and return the modified string.
4Step 4: If all characters are 'a', change the last character to 'b' and return the modified string.

solution.py10 lines

1# Full working Python code
2
3def breakPalindrome(palindrome):
4    if len(palindrome) == 1:
5        return ""
6    n = len(palindrome)
7    for i in range(n // 2):
8        if palindrome[i] != 'a':
9            return palindrome[:i] + 'a' + palindrome[i + 1:]
10    return palindrome[:-1] + 'b'  # Change last character to 'b'

ℹ

Complexity note: This complexity is linear because we only traverse half of the string to find the first non-'a' character, and we make a single modification, which is efficient.

1Changing the first non-'a' character to 'a' gives the smallest lexicographical result.
2If all characters are 'a', changing the last character to 'b' is the only option.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.