Thoughts and Talk
Palindromes have stood the test of time as a classic interview question due to their fundamental nature. Although there are various approaches to solving them, the traditional two-pointer method remains the simplest. However, in this current referenced leetcode problem there's a twist to this problem❗ —it's not just about determining if a string is a palindrome. The real challenge lies in identifying whether removing one character from the string would transform it into a palindrome.
This added complexity makes the question intriguing, even if it's categorized as easy. It requires a keen eye for detail and a thorough understanding of palindrome properties to solve effectively.
The problem of determining if a string can be transformed into a palindrome by removing at most one character introduces a fascinating twist on a classic problem. This task isn't just about palindrome detection; it also tests our ability to handle slight modifications in string properties efficiently. It's a problem that requires not just understanding basic string manipulation but also a deeper insight into decision-making processes in algorithms.
Problem Statement
We need to determine whether a given string can be rendered a palindrome by deleting at most one character. This problem extends the classic palindrome check by adding the possibility of minor adjustments, presenting an interesting challenge in algorithm design.
Hints
- Utilize two pointers to compare characters from the beginning and end of the string.
- Upon finding non-matching characters, decide whether skipping the current character from the beginning or the end might help.
- Only one such skip is allowed. If skipping doesn't help, the answer should be false.
Approaches Explored
Approach 1: Two-pointer with single skip check
Pros:
- Efficient, since it typically requires scanning the string at most twice.
- Simple to implement using conditional checks to handle the single allowed skip.
Cons:
- Deciding which character to skip (left or right) can require checking both possibilities in some cases, adding complexity.
Approach 2: Modified two-pointer with state tracking (not so different!)
This is the approach reflected in our original code. ouIt uses a
breaker variable to handle complex decision points when both characters to the left and right could potentially be skipped to maintain a palindrome.Pros:
- Handles ambiguities effectively by revisiting the decision point if the initial choice does not resolve the palindrome.
- Keeps track of the state to avoid unnecessary rechecks.
Cons:
- More complex due to additional variables and conditionals.
- Slightly harder to understand and maintain due to its stateful design.
Finalized Approach
Based on the analysis, the modified two-pointer approach seems the most suitable. It addresses the challenge of potentially needing to skip either of two characters by tracking the state and allowing for a revisit of critical decisions. Let's dissect the given code to understand this approach better.
Executable Code
TheTechCruise.com Pyodide Terminalclass Solution:
def validPalindrome(self, s: str) -> bool:
ct, p, q, breaker = 0, 0, len(s)-1, False
p1, q1 = 0, 0
while p<q:
if s[p]!=s[q]:
# print(p,q,s[p],s[q])
if ct==1:
if breaker: break
else: return False
ct += 1
choose_p = (p+1<q and s[p+1]==s[q])
choose_q = (p<q-1 and s[p]==s[q-1])
if choose_p and choose_q:
breaker = True
p1, q1 = p, q-1
if choose_p: p += 1
elif choose_q: q -= 1
else: return p+1==q
p += 1
q -= 1
else: return ct<=1
for i in range(p1+q1//2):
if s[p1+i]!=s[q1-i]: return False
return True
# List of strings to test
test_strings = [
"ebcbbececabbacecbbcbe",
"abca",
"aba",
"abc",
"cbbcc",
"aguokepatgbnvfqmgmlcupuufxoohdfpgjdmysgvhmvffcnqxjjxqncffvmhvgsymdjgpfdhooxfuupuculmgmqfvnbgtapekouga"
]
# Create an instance of Solution
solution = Solution()
# Check each string for the palindrome property
results = [solution.validPalindrome(s) for s in test_strings]
# Output results
for num_string, result in zip(test_strings, results):
print(f"String: {num_string} -> {'Palindrome possible' if result else 'Palindrome not possible'}")Code Analysis
This code initiates a two-pointer approach but integrates a complex decision-making process (
breaker). If both skipping left and right characters could potentially solve the problem, it saves the state and tries one path first. If the loop completes without a conclusive palindrome check due to breaker, it revisits the decision by checking the alternate path saved in p1 and q1.Logic visualization
Discussion on Complexity
- Time Complexity: O(n), as the string is scanned a maximum of two times.
- Space Complexity: O(1), as it only uses a few auxiliary variables for indices and flags, independent of input size.
This approach is robust, handling edge cases gracefully with its backup strategy to revisit critical decisions, ensuring that all possibilities are adequately explored.
