A palindrome reads the same in both directions. It is fixed by its center: the letters on the left of the center mirror the letters on the right. If you remove the first and the last letter of a palindrome, what is left is still a palindrome with the same center.

So expand(lo, hi) starts on a center and grows it. While lo >= 0, hi < n and s[lo] == s[hi], it moves lo one letter left and hi one letter right. The palindromes around this center appear from the shortest to the longest. The first mismatch ends the loop. A longer palindrome around the same center would need that mismatched pair to match, so none exists.

The loop stops one step past each end of the palindrome. So the palindrome is s[lo + 1:hi], and its length is hi - lo - 1.

A string of n letters has 2n - 1 centers: n letters and n - 1 gaps between two letters. Every palindromic substring has exactly one center. So trying every center finds the longest palindrome. The board does this for s = "cbbd", one expand call at a time.

A palindrome with an odd length, such as "aba", has a middle letter. A palindrome with an even length, such as "abba", has no middle letter. Its center is the gap between the two middle letters.

expand(i, i) starts on the letter s[i], so it only finds odd palindromes. expand(i, i + 1) starts on the gap after s[i], so it only finds even palindromes. Each i needs both calls.

In the board, expand(1, 2) is the call that finds "bb": the letters at 1 and 2 match, then lo = 0 holds "c" and hi = 3 holds "d", which differ. It returns 2. Every expand(i, i) call returns 1 for "cbbd". A loop with only the odd call keeps "c" and returns the wrong answer.

The other example of Longest Palindromic Substring (LC 5) has s = "babad". The call expand(1, 1) grows "b" to "bab", so the function returns "bab". "aba" is also accepted, because the problem allows either.

Only start and length are stored. The slice s[start:start + length] is made once, in the return.

For an odd palindrome of length 3 centered on i = 1, there is one letter on each side of i, so start = 0. The formula gives 1 - (3 - 1) // 2 = 0. For an even palindrome of length 2 whose center is the gap after i = 1, the left half has one letter and it is s[1]. The formula gives 1 - (2 - 1) // 2 = 1. In both cases the left half has (length - 1) // 2 letters before i.

A shorter formula, i - length // 2, gives the same start for odd lengths. For the even "bb" it gives 1 - 1 = 0, and the answer is "cb".

The bound lo >= 0 is also needed. In Python, s[-1] is the last letter, not an error. For s = "aa" without the bound, the loop compares the first a with the last one, keeps growing, and the function returns "a" instead of "aa".

There are 2n - 1 centers, and one expand call can grow up to about n / 2 times. So the time is O(N^2). For s made of ten letters a, the loops grow 55 times in total. The space is O(1), because only start, length and a few indices are stored.

A table of all pairs (lo, hi) takes O(N^2) time too, but it also takes O(N^2) space. Manacher's algorithm finds the answer in O(N) time. It is rarely needed in an interview, and expansion is much shorter to write.

Palindromic Substrings (LC 647) counts all palindromic substrings instead of finding the longest. The same expansion works. Each growth step of a call is one palindrome. expand(i, i) with result L adds (L + 1) // 2 palindromes, and expand(i, i + 1) adds L // 2. For "aaa" the total is 6.

This card needs letters next to each other. If the palindrome may skip letters, as in Longest Palindromic Subsequence (LC 516) or Valid Palindrome III (LC 1216), the center idea does not apply and a table over (lo, hi) is the standard approach.