ITPEC FE Subject B October 2024 Question 12

Palindrome check with two pointers — fill in the blanks for a function that checks whether a character array is a palindrome using left/right index convergence.

The function uses two pointers: left starts at 1, right starts at the array length. Each iteration compares s[left] and s[right]:

• •If they match, both pointers move inward: left ← left + 1, right ← right - 1
• •If they don't match, ok ← false and the loop breaks

Blank A must be s[left] = s[right] (equality check — matching means we continue).
Blank B must be right ← right - 1 (move the right pointer inward symmetrically).

Why not others:
- (a) right ← right + 1 moves right outward, causing an out-of-bounds access
- (c) right ← right + left produces meaningless index values
- (d) right ← right - left does not mirror the symmetric convergence pattern
- (e–h) use s[left] ≠ s[right] as condition A, which inverts the logic: matching characters would trigger ok ← false

Key rule: Two-pointer palindrome check — when characters match, move both pointers inward (left + 1, right - 1); when they don't, the string is not a palindrome.

AI-generated — may contain errors