Which statement best describes the algorithm to check if a string is a palindrome and its time complexity?

Which statement best describes the algorithm to check if a string is a palindrome and its time complexity?

• A. Check if the string contains no repeating characters.
• B. Compare characters from both ends moving toward the center; time complexity O(n).
• C. Convert the string to uppercase and perform character comparisons.
• D. Check only the first and last characters.

Answer: (B)

Palindromicity hinges on symmetry: you compare characters from the two ends moving toward the center. If every mirrored pair matches, the string reads the same forward and backward; if any pair differs, it isn't a palindrome. The best algorithm uses two pointers: left at the start and right at the end, compare the characters at those positions. If they differ, you can stop early and declare not a palindrome. If they match, move the pointers inward and continue until they meet or cross. This approach directly checks the defining property and requires only a single pass through the string, so the time grows linearly with length: about n/2 comparisons in the worst case, giving O(n) time. Space usage is constant because you only store two indices. Why this is preferred over other ideas: simply checking that there are no repeating characters doesn’t relate to palindromicity—palindromes can have repeated characters, and non-palindromes can also share repeats. Normalizing case (like converting to uppercase) without performing the mirrored comparisons won’t determine palindromicity. Checking only the first and last characters misses all the inner positions and can miss mismatches deep inside. The two-pointer symmetric comparison captures the actual property and yields the expected linear time bound.

Palindromicity hinges on symmetry: you compare characters from the two ends moving toward the center. If every mirrored pair matches, the string reads the same forward and backward; if any pair differs, it isn't a palindrome. The best algorithm uses two pointers: left at the start and right at the end, compare the characters at those positions. If they differ, you can stop early and declare not a palindrome. If they match, move the pointers inward and continue until they meet or cross. This approach directly checks the defining property and requires only a single pass through the string, so the time grows linearly with length: about n/2 comparisons in the worst case, giving O(n) time. Space usage is constant because you only store two indices.

Why this is preferred over other ideas: simply checking that there are no repeating characters doesn’t relate to palindromicity—palindromes can have repeated characters, and non-palindromes can also share repeats. Normalizing case (like converting to uppercase) without performing the mirrored comparisons won’t determine palindromicity. Checking only the first and last characters misses all the inner positions and can miss mismatches deep inside. The two-pointer symmetric comparison captures the actual property and yields the expected linear time bound.