Problem Description

You are given a string s. Your task is to determine if this string can become a palindrome by deleting at most one character from it.

A palindrome is a string that reads the same forwards and backwards. For example, "racecar" and "noon" are palindromes.

The key constraint is that you can delete either:

• Zero characters (the string is already a palindrome), or
• Exactly one character from any position in the string

You need to return true if the string can be made into a palindrome under these conditions, and false otherwise.

Examples:

• If s = "aba", return true (already a palindrome, no deletion needed)
• If s = "abca", return true (delete 'c' to get "aba" which is a palindrome)
• If s = "abc", return false (no single deletion can make it a palindrome)

The solution uses a two-pointer technique. Starting from both ends of the string, we compare characters. When we find a mismatch, we have two options: skip the left character or skip the right character. We check if either option results in a palindrome for the remaining substring. If we never find a mismatch, the string is already a palindrome.

Intuition

The natural way to check if a string is a palindrome is to compare characters from both ends moving towards the center. If all corresponding pairs match, it's a palindrome.

Now, what happens when we're allowed to delete at most one character? Let's think through this step by step.

When we're comparing characters from both ends and find a mismatch, we have a decision to make. For example, if we have s = "abca" and we're comparing s[0] = 'a' with s[3] = 'a' - they match. Next, we compare s[1] = 'b' with s[2] = 'c' - they don't match.

At this mismatch point, we know one of these two characters needs to be "deleted" (or skipped) for the string to potentially become a palindrome. But which one? We don't know immediately, so we need to try both possibilities:

1. Skip the left character ('b') and check if the remaining part is a palindrome
2. Skip the right character ('c') and check if the remaining part is a palindrome

If either option works, then we can make the string a palindrome with one deletion.

The key insight is that we only get one chance to handle a mismatch. Once we encounter the first mismatch and decide to skip a character, the remaining substring must be a perfect palindrome - no more deletions allowed.

This leads us to a two-pointer approach where:

• We start with pointers at both ends
• Move them towards each other as long as characters match
• When we hit a mismatch, we branch into two scenarios (skip left or skip right)
• For each scenario, we check if the remaining part forms a valid palindrome
• If we never hit a mismatch, the string is already a palindrome

Learn more about Greedy and Two Pointers patterns.

Solution Approach

The implementation uses the two-pointer technique to efficiently solve this problem.

Main Algorithm Structure:

1. Helper Function check(i, j):

   • This function verifies if the substring from index i to j is a palindrome
   • It uses two pointers moving towards each other
   • Returns False immediately if any mismatch is found
   • Returns True if all characters match

2. Main Logic:

   • Initialize two pointers: i = 0 (left) and j = len(s) - 1 (right)
   • Move pointers towards each other comparing characters at each step

Step-by-step Implementation:

i, j = 0, len(s) - 1
while i < j:
    if s[i] != s[j]:
        # Found first mismatch - try both options
        return check(i, j - 1) or check(i + 1, j)
    i, j = i + 1, j - 1
return True

Key Decision Point: When s[i] != s[j], we have two choices:

• check(i, j - 1): Skip the character at position j (right pointer)
• check(i + 1, j): Skip the character at position i (left pointer)

We use the logical or operator because we only need one of these options to work.

Example Walkthrough with s = "abca":

1. Compare s[0]='a' with s[3]='a' → match, move pointers
2. Compare s[1]='b' with s[2]='c' → mismatch
3. Try check(1, 1) (skip right 'c'): checks if "b" is palindrome → True
4. Return True

Time Complexity: O(n) where n is the length of the string. In the worst case, we traverse the string once in the main loop and once more in the helper function.

Space Complexity: O(1) as we only use a constant amount of extra space for pointers.

Example Walkthrough

Let's walk through the solution with s = "raceacar".

Initial Setup:

• String: "raceacar" (indices 0-7)
• Left pointer i = 0, Right pointer j = 7

Step-by-step Execution:

1. First comparison: s[0]='r' vs s[7]='r'

   • They match! Move pointers inward
   • Update: i = 1, j = 6

2. Second comparison: s[1]='a' vs s[6]='a'

   • They match! Move pointers inward
   • Update: i = 2, j = 5

3. Third comparison: s[2]='c' vs s[5]='c'

   • They match! Move pointers inward
   • Update: i = 3, j = 4

4. Fourth comparison: s[3]='e' vs s[4]='a'

   • Mismatch found! Now we try both deletion options:

   Option 1: Skip left character (delete 'e' at index 3)

   • Call check(4, 4) to verify if substring from index 4 to 4 is palindrome
   • This checks just "a", which is trivially a palindrome
   • Returns True

   Option 2: Skip right character (delete 'a' at index 4)

   • Call check(3, 3) to verify if substring from index 3 to 3 is palindrome
   • This checks just "e", which is trivially a palindrome
   • Returns True

5. Final Result: Since check(4, 4) returns True, the function returns True

The string "raceacar" can become a palindrome "racecar" by deleting the 'a' at index 4.

Visual representation of the process:

r a c e a c a r
^             ^  (match: r=r)
  ^         ^    (match: a=a)
    ^     ^      (match: c=c)
      ^ ^        (mismatch: e≠a)
        ↓
   Try deleting one:
   - Delete 'e': rac_acar → "racacar" (check remaining)
   - Delete 'a': race_car → "racecar" ✓ (palindrome!)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the string s.

The algorithm uses a two-pointer approach starting from both ends of the string. In the main while loop, we traverse the string at most once with pointers i and j moving towards each other. When a mismatch is found at positions i and j, we call the helper function check() twice at most - once for check(i, j-1) and potentially once for check(i+1, j).

Each call to check() also performs at most O(n) comparisons in the worst case. However, the key insight is that:

• The main loop processes at most n/2 characters before finding a mismatch (or completing if no mismatch)
• When a mismatch is found, check() validates the remaining substring, which combined with the already processed part, totals to at most n character comparisons
• The two check() calls are connected by an or operator, so in the worst case we check at most 2n characters total

Therefore, the overall time complexity remains O(n).

Space Complexity: O(1).

The algorithm only uses a constant amount of extra space:

• Two pointers i and j in the main function
• Two local pointers i and j in the helper function check()
• No additional data structures are created
• The recursion depth is at most 1 (the helper function doesn't call itself recursively)

All variables use constant space regardless of the input size, resulting in O(1) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting Multiple Deletions

A common mistake is trying to continue checking after the first mismatch, potentially allowing multiple deletions.

Incorrect approach:

def validPalindrome(self, s: str) -> bool:
    left, right = 0, len(s) - 1
    deletion_used = False
  
    while left < right:
        if s[left] != s[right]:
            if deletion_used:  # Wrong: trying to track deletions manually
                return False
            # This approach fails because we don't know which character to skip
            left += 1  # or right -= 1?
            deletion_used = True
        else:
            left += 1
            right -= 1
    return True

Why it fails: When a mismatch occurs, we don't know whether to skip the left or right character without checking both possibilities.

2. Checking Only One Deletion Option

Some implementations only check one side when a mismatch occurs, missing valid palindromes.

Incorrect approach:

def validPalindrome(self, s: str) -> bool:
    left, right = 0, len(s) - 1
  
    while left < right:
        if s[left] != s[right]:
            # Only checking by skipping left character
            return is_palindrome(left + 1, right)
        left += 1
        right -= 1
    return True

Example failure: For s = "cbbcc", this would return False when it should return True (delete the first 'c').

3. Incorrect Helper Function Bounds

Off-by-one errors in the palindrome checking helper function.

Incorrect approach:

def is_palindrome(left: int, right: int) -> bool:
    while left <= right:  # Wrong: should be left < right
        if s[left] != s[right]:
            return False
        left += 1
        right -= 1
    return True

Why it matters: Using <= instead of < causes unnecessary comparison when pointers meet at the same index (which is always valid for odd-length palindromes).

4. Not Handling Edge Cases

Forgetting to handle strings of length 0, 1, or 2.

Solution: The correct implementation naturally handles these:

• Empty string or single character: The main while loop never executes, returns True
• Two characters: Checked once, and if different, both deletion options lead to single characters (always palindromes)

5. Recursion Without Deletion Tracking

Using recursion without properly tracking whether a deletion has been used.

Incorrect approach:

def validPalindrome(self, s: str, deleted=False) -> bool:
    left, right = 0, len(s) - 1
  
    while left < right:
        if s[left] != s[right]:
            if not deleted:
                # Recursive calls with string slicing (inefficient)
                return (self.validPalindrome(s[left+1:right+1], True) or 
                       self.validPalindrome(s[left:right], True))
            return False
        left += 1
        right -= 1
    return True

Problems:

• String slicing creates new strings (space inefficient)
• Recursive approach adds unnecessary complexity
• Function signature doesn't match the expected interface

Correct Solution Reminder: The provided solution avoids all these pitfalls by:

• Using a helper function that checks palindromes on index ranges (no string copying)
• Checking both deletion options when a mismatch occurs
• Returning immediately after finding the first mismatch (ensuring at most one deletion)
• Using simple iteration instead of recursion