680. Valid Palindrome II

Problem Description

The problem requires determining if a given string s can be made into a palindrome by removing at most one character. A palindrome is a word, phrase, number, or other sequences of characters which reads the same forward and backward, ignoring spaces, punctuation, and capitalization. For example, 'radar' is a palindrome, while 'radio' is not. However, 'radio' can become a palindrome by removing the 'i', which becomes 'rado', and 'raod' is a palindrome because it reads the same backward as forward. The challenge here is to decide whether such a removal is possible to achieve a palindrome with at most one deletion.

Intuition

To understand the given solution, we should recognize two things:

1. If a string does not need any character removal to be a palindrome, it means that the characters at the start and end of the string (and so on moving inward) match.
2. If one mismatch is found, we get a choice - to remove a character from either the left or the right at the point of mismatch and check if the resulting substrings could form a palindrome.

The solution uses a two-pointer approach:

• Start with two pointers, one at the beginning (i) and one at the end (j) of the string.
• Move both pointers towards the center, checking if the characters are the same.
• If a mismatch is discovered, there are two substrings to check: one without the character at i and one without the character at j. We use the helper function check() to verify if either substrings can form a palindrome.
• If we can successfully verify one substring as a palindrome, we conclude that the original string can be a palindrome after removing at most one character.
• We continue checking until we either find a proper substring that is a palindrome or exit because we have checked all characters without violating the palindrome property.

The key is the helper function check(i, j) which checks if the substring from i to j is a palindrome by iterating through the substring and comparing characters at the start and end, moving inwards.

By carefully applying the check() function only when a mismatch is found, the given algorithm efficiently decides whether one can obtain a palindrome with at most one deletion.

Learn more about Greedy and Two Pointers patterns.

Solution Approach

The solution's core relies on a two-pointer approach while also incorporating a helper method to reduce redundancy. Here's the process step-by-step:

1. Initial Two-pointer Setup: Initialize two pointers, i and j, representing the start and the end of the string s. These two pointers will progressively move towards the center.

2. First Pass - Checking Palindrome: Move both pointers towards the center, implied by i < j. If the characters at i and j match (s[i] == s[j]), we can safely continue. This loop continues until a mismatch is found (when s[i] != s[j]) or until the entire string is checked.

3. Handling Mismatches - Utilizing the Helper Function: Upon encountering a mismatch, the solution must determine whether omitting one of the characters can lead to a palindrome. This is where the helper method, check(i, j), comes into play. When s[i] != s[j], two checks are made:

   • One check leaves out the character at the end (check(i, j - 1)), assuming this might be the extra character causing a non-palindrome.
   • The other check omits the character at the start (check(i + 1, j)), assuming this might be the non-matching odd one out.

4. The Helper Method - check(i, j): The method takes two indices and checks if the substring between them (s[i] to s[j]) is a palindrome. It uses the same two-pointer technique, now applying it within the narrower range. It returns True if a palindrome is detected, False if not.

5. Single Character Removal Decision: After calling the check() method for both possible single removals, we use logical OR (or) to combine the results. If either case returns True, the original string can be converted to a palindrome by removing one character. The function then returns True.

6. Completing the Loop: If no mismatch is found, the loop ends, and we can assume the string is already a palindrome or can be made into one with a single removal (might be a character at the start or the end which doesn't interfere with the palindrome property).

7. Final Return: If we reach outside the loop without returning False, the string must be a palindrome, and the function returns True.

This approach provides an optimal solution as it only scans the string once and performs the minimum necessary comparisons, obeying the constraints set by the problem (at most one character removal).

Example Walkthrough

Let's use the string s = "abca" to illustrate the solution approach. We need to determine if it's possible to make s into a palindrome by removing at most one character.

1. Initial Two-pointer Setup: We start with two pointers, i = 0 (pointing to 'a') and j = 3 (pointing to 'a'). The string looks like this: abca, with i at the first character and j at the last.

2. First Pass - Checking Palindrome: We compare s[i] and s[j]. Since s[0] is 'a' and s[3] is also 'a', there's a match, so we move both i and j towards each other (now i = 1, j = 2).

3. Finding a Mismatch: Now i = 1 is pointing to 'b' and j = 2 is pointing to 'c'. They don't match (s[i] != s[j]). We need to check if removing one character makes it a palindrome. We call the check() function twice as follows:

   • check(1, 1): This omits the character at j (the 'c') and checks if ab is a palindrome.
   • check(2, 3): This omits the character at i (the 'b') and checks if ca is a palindrome.

4. The Helper Method - check(i, j): When we run check(1, 1), it instantly returns True as it's effectively checking a single character 'b', which is trivially a palindrome. We don't need to perform check(2, 3) as we've already found a potential palindrome by removing 'c'.

5. Single Character Removal Decision: Because check(1, 1) returned True, we have confirmed that by removing one character ('c'), the string s could be turned into a palindrome. Therefore, for the input abca, the function would return True.

6. Completing the Loop: In this example, the mismatch was found, and the check() function indicated a palindrome is possible, so the loop is completed successfully.

7. Final Return: Since we found that a single removal can lead to a palindrome, the function would return True for the string s = "abca".

This walkthrough demonstrates how we can efficiently determine that the string "abca" can become a palindrome by removing at most one character, 'c' in this case, leading to the palindrome "aba".

Solution Implementation

Time and Space Complexity

The given Python code defines a method validPalindrome to determine if a given string can be considered a palindrome by removing at most one character.

Time Complexity:

The time complexity of the main function is generally O(n), where n is the length of the input string s. This is because the function includes a while loop that iterates over each character of the string at most twice - once in the main while loop and once in the check function, which is called at most twice if a non-matching pair is found.

To break it down:

• The main while loop iterates over the string s, comparing characters from i to j. Each loop iteration takes O(1) time.
• If a mismatch is found, there are two calls to the helper function check, each with a worst-case time complexity of O(n/2), which simplifies to O(n).

In the worst case, you compare up to n - 1 characters twice (once for each call of check), so the upper bound is 2 * (n - 1) operations, which still results in a linear time complexity: O(n).

Space Complexity:

The space complexity of the code is O(1). No additional space is proportional to the input size is being used, aside from a constant number of integer variables to keep track of indices. The check function is called with different indices but does not use any extra space apart from the input string s, which is passed by reference and not copied.

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