Given a string s, return true if it can become a palindrome by deleting at most one character. Walking two pointers inward from both ends handles the matching characters for free. The only real decision point is the first mismatch: since only one deletion is allowed, there are exactly two ways to fix it — drop the character at the left pointer, or drop the one at the right pointer. Whichever choice leaves a palindrome in the remaining range makes the answer true. Checking both candidate ranges still keeps the whole algorithm at O(n), since every character is visited only a small, constant number of times.