Valid palindrome II (one delete) Explained: Tracing "abca"

Imagine you are handed the string s = "abca". Your first instinct is likely to check if it reads the same forwards and backwards. But wait: a-b-c-a backwards is a-c-b-a. They do not match. However, the problem asks a more forgiving question: can we make it a palindrome by deleting at most one character?

If you inspect "abca", dropping the b leaves "aca", which is a clean palindrome. Dropping the c leaves "aba", which is also a palindrome. The answer for "abca" should be true.

Contrast this with s = "abc". No single character deletion turns it into a mirror image ("bc" vs "ac" vs "ab"), so the answer is false. Meanwhile, s = "aba" requires zero deletions, so it should also return true.

Standard strict palindrome checks run two pointers inward from the ends, immediately flagging a mismatch and failing. But with one deletion allowed, a mismatch is not a game over—it is a fork in the road. You have to decide whether to skip the left character or skip the right character and see if either path resolves into a valid palindrome.

When we check if s = "abca" is a palindrome, we start with a left pointer at index 0 (a) and a right pointer at index 3 (a). They match! We shrink inward: left moves to 1 (b), right moves to 2 (c). At this point, s[1] is b and s[2] is c. A mismatch! Normally, a standard palindrome check would immediately return false. But our rules grant us one free deletion.

Because we hit a mismatch between index 1 and index 2, we face a critical fork in the road: should we delete the character at the left pointer, or the character at the right pointer? If we delete the left character (b), we must verify whether the remaining substring from index 2 to index 3 ("ca"? Wait, from left+1 to right) forms a valid palindrome. If we delete the right character (c), we must verify the substring from index 1 to index 2 ("bc"? From left to right-1).

This gives rise to our core mental model: a standard inward-moving two-pointer scan that, upon encountering its first and only allowed mismatch, spawns two independent sub-scans. The overall string is valid if at least one of these sub-scans succeeds completely without any further mismatches. Let us examine how this branching logic behaves on our working examples "abca", "abc", and "aba" to see why trying both paths is mandatory rather than optional.

To see how the two-pointer branching model handles a single deletion, let us trace our locked example: s = "abca". We initialize our left pointer L = 0 at character a and our right pointer R = 3 at character a. Our goal is to determine if s can become a palindrome by deleting at most one character, verifying the expected result of true.

Given

- Input string s = "abca"
- Left pointer L = 0, Right pointer R = 3

Steps

- Compare s[L] and s[R]: s[0] is a, and s[3] is a. They match!
- Increment L to 1 and decrement R to 2.
- Current pointers: L = 1, R = 2, inspecting substring "bc".

- Compare s[1] and s[2]: s[1] is b, and s[2] is c. They do not match!
- Since we have encountered our first mismatch, we must branch and test two sub-problems using a helper function isPalindrome(string, left, right):
- Branch A (Delete left character): Test the substring from L + 1 to R, which is isPalindrome("abca", 2, 2). Here, L = 2 and R = 2, inspecting substring "c".
- Branch B (Delete right character): Test the substring from L to R - 1, which is isPalindrome("abca", 1, 1). Here, L = 1 and R = 1, inspecting substring "b".

- In Branch A (L = 2, R = 2), the pointers meet at the same index, trivially forming a valid palindrome (true).
- In Branch B (L = 1, R = 1), the pointers also meet at the same index, trivially forming a valid palindrome (true).

Result

Since at least one of the branches (true or true) successfully validates as a palindrome, the overall algorithm returns true for "abca". Deleting the character at index 1 (b) leaves "aca", which is a valid palindrome.

Now that you have seen how abca splits into two helper checks upon hitting a mismatch, it is time to test your mental model on a slightly different input. Take the string s = "deeee" and trace how the two-pointer scan behaves from the outer edges inward. Remember the core rule from the branching model: when left and right characters differ for the first time, you do not immediately fail—you must test both the string with left removed and the string with right removed.

Work through the pointers step by step on s = "deeee". Note which indices match, where the first mismatch occurs, and what the two helper checks look like before deciding whether the overall function returns true or false.

The two-pointer branching strategy we used to solve abca—where a single mismatch forks into two independent palindrome sub-problems (isPal(L+1, R) or isPal(L, R-1))—is a specific instance of a broader algorithmic pattern. Whenever a problem allows a fixed number of localized modifications (like one deletion, one character replacement, or skipping one mismatch), you can generalize this logic without escalating to expensive $O(n^2)$ dynamic programming matrices. When facing a variation where you can delete up to $k$ characters, the recursive call stack simply increments a deletion counter at each branch until $k$ is exhausted.

Consider how this applies to a slightly different task: determining if two strings are one edit away, or if a string can become a palindrome after deleting up to two characters. By keeping the core two-pointer contract intact—advancing inward on matches and branching on mismatches—you maintain linear time complexity $O(n)$ while handling combinatorial flexibility. The key takeaway from our journey with abca, abc, and aba is that you do not need to generate all possible deleted strings; you only need to explore the specific branch created by the first failure of symmetry.

def can_form_palindrome_with_two_deletes(s: str) -> bool:
    # How would you adapt the one-delete two-pointer branching
    # model to handle up to two deletions?
    pass