Valid Palindrome II

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

A palindrome is a word that reads the same backward as forward.

Algorithm

  1. Initialize Pointers: Initialize two pointers, left and right, pointing to the start and the end of the string.

  2. Check Palindrome: Iterate through the string, comparing the characters at the left and right pointers.

    • If the characters match, move both pointers closer to the center of the string.
    • If the characters don’t match, check if the remaining part of the string (ignoring one of the mismatched characters) is a palindrome.

  3. Sub-Palindrome Check: If there’s a mismatch, you’ll need to check two possibilities: ignoring the character at the left pointer or ignoring the character at the right pointer. If either of these results in a palindrome, return true.

  4. Result: If the loop completes without finding any mismatches (or only one that can be ignored), return true. Otherwise, return false.

Code

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class Solution:
    def validPalindrome(self, s: str) -> bool:
        # Function to check if a substring is a palindrome
        def is_palindrome(sub_s):
            return sub_s == sub_s[::-1]

        left, right = 0, len(s) - 1

        # Iterate through the string
        while left < right:
            if s[left] != s[right]:
                # Check if ignoring one of the mismatched characters results in a palindrome
                return is_palindrome(s[left:right]) or is_palindrome(s[left + 1:right + 1])
            left += 1
            right -= 1

        return True

Explanation

  • The code checks if the string is a palindrome by iterating through it with two pointers.
  • If a mismatch is found, it checks if ignoring either of the mismatched characters results in a palindrome.

Insights

  • Time Complexity: The main loop runs in (O(n)), and checking a sub-palindrome is also (O(n)), resulting in a total time complexity of (O(n)).
  • Space Complexity: (O(1)), as we are only using constant extra space.

This solution efficiently determines if a given string can be turned into a palindrome by deleting at most one character, fulfilling the given constraints.

Deleting

Implement palindrome method. Create a new string by deleting one character Check if the new string is a palindrome.

This will create many string objects.

Skipping

Mark which index can be skipped. Let the palindrome method skip this character when comparing characters

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def valid_palindrome(s)
  palindrome?(s, 0, s.size - 1)
end

def palindrome?(s, l, r)
  while l < r
    if s[l] == s[r]
      l += 1
      r -= 1
    else
      skip_left = s[l+1..r]
      skip_right = s[l..r-1]

      return skip_left.reverse == skip_left || skip_right.reverse == skip_right
    end
  end

  true
end

Identifying Problem Isomorphism

“Valid Palindrome II” can be approximately mapped to “Longest Palindromic Subsequence”.

Reasoning:

Both problems are related to palindromes and string manipulation.

  1. In “Valid Palindrome II”, you are given a non-empty string and you can delete at most one character to determine if the string can become a palindrome.

  2. “Longest Palindromic Subsequence” asks you to find the length of the longest palindromic subsequence in a string. If you view the input string of “Valid Palindrome II” as a subsequence of itself, the problem essentially becomes finding if there exists a subsequence (with at most one character deleted) that is a palindrome.

The mapping is approximate. While both problems deal with palindromes and subsequences, “Valid Palindrome II” specifically focuses on the ability to delete one character to form a palindrome, while “Longest Palindromic Subsequence” is about finding the longest possible palindromic subsequence in a string.

“Valid Palindrome II” is simpler since it only requires checking the string from both ends and comparing the characters. If there is a mismatch, we can try removing either the left character or the right character and check if the rest of the string forms a palindrome. “Longest Palindromic Subsequence” is more complex since it involves dynamic programming to solve.

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def validPalindrome(self, s: str) -> bool:
    def verify(s, left, right, deleted):
        while left < right:
            if s[left] != s[right]:
                if deleted:
                    return False
                else:
                    return verify(s, left+1, right, True) or verify(s, left, right-1, True)
            else:
                left += 1
                right -= 1
        return True
    return verify(s, 0, len(s)-1, False)

Problem Classification

The problem can be classified as a String Manipulation problem.

Here are the ‘What’ components of the problem:

  1. A String: The primary input to the problem is a string s.

  2. Palindromes: A palindrome is a string that reads the same forward and backward. In this problem, the concept of a palindrome is a key part of the problem statement.

  3. Deletion of Characters: The problem involves potentially deleting at most one character from the input string to possibly make it a palindrome.

  4. Boolean Output: The output of the problem is a boolean value (true or false). The function should return true if the string can be made into a palindrome by deleting at most one character, and false otherwise.

The problem can further be classified as a Decision Problem, since the solution involves deciding whether or not a certain condition (the ability to form a palindrome with at most one deletion) can be met. The problem also has elements of String Processing and Palindrome Checking, which are common subcategories in the domain of string manipulation problems. The solution will likely involve iterative or recursive traversal of the string, and could potentially involve dynamic programming depending on the implementation.

Language Agnostic Coding Drills

  1. Dissect the Code

This piece of code involves several distinct programming concepts:

  • Function definition: The code defines a main function validPalindrome with a single parameter (s), and a nested function verify with four parameters (s, left, right, deleted).

  • String/Data Structure Manipulation: The function processes a string (s). It accesses elements from both ends of the string, moving towards the center.

  • Conditional Statements: The code includes an if statement to check a condition (if s[left] is not equal to s[right]), and nested if-else statements to determine the next action based on the deleted flag.

  • Loops: The code uses a while loop to iterate until left is no longer less than right.

  • Recursion: The verify function calls itself within the else statement, signifying recursion.

  • Return statement: The function ends with a return statement that outputs the result, and there are several return statements within the verify function as well.

  1. List of Concepts (in order of increasing difficulty)

a) Variable Assignment and Updating: This is a fundamental concept in almost all programming languages. It involves creating variables and updating their values.

b) Function Definition: This involves understanding how to encapsulate a piece of code into a reusable function.

c) String/Data Structure Manipulation: This requires understanding how to interact with and manipulate data in structured formats like strings.

d) Conditional Statements: This involves understanding how to use if and else statements to control the flow of a program based on certain conditions.

e) Loops: This is a slightly more advanced concept and involves understanding how to use loops to repeatedly execute a block of code.

f) Recursion: Understanding recursion can be quite challenging, especially when it involves conditional recursive calls like in this code.

g) Return Statement: Understanding when and how to use the return statement in a function can be considered a more intermediate level concept, as it requires understanding the flow of data in and out of functions.

  1. Problem-solving Approach

The overall approach to solving this problem involves checking if the input string is a palindrome by comparing characters from both ends towards the center. If a pair of characters do not match, the function checks if a character has been deleted before. If not, it tries deleting the character from the left end first, and if that doesn’t result in a palindrome, it tries deleting the character from the right end. If a character has been deleted before and a pair of characters do not match, the function returns False.

The Function Definition concept is used to define the validPalindrome and verify functions.

The String/Data Structure Manipulation concept is utilized to access and compare elements in the string.

Conditional Statements are used to check if characters match, and if a character has been deleted before.

The Loops concept is used to iterate through the string from both ends towards the center.

Recursion is used to try different deletions when characters do not match.

Finally, the Return Statement is used to return the result, which is a boolean indicating whether the string can be made into a palindrome by deleting at most one character.

Targeted Drills in Python

  1. Coding Drills for General Concepts

Here are Python-based drills that demonstrate each of the identified coding concepts:

a) Variable Assignment and Updating

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# Define a variable
variable = 10
print(variable)  # prints: 10

# Update the variable
variable += 5
print(variable)  # prints: 15

b) Function Definition

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# Define a simple function that adds two numbers
def add_numbers(a, b):
    return a + b

print(add_numbers(3, 7))  # prints: 10

c) String/Data Structure Manipulation

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# Define a string
s = "hello"

# Access an element in the string
print(s[1])  # prints: e

# Get the length of the string
print(len(s))  # prints: 5

d) Conditional Statements

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# Simple if-else statement
a = 5
b = 10
if a > b:
    print("a is larger than b")
else:
    print("a is not larger than b")  # prints: a is not larger than b

e) Loops

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# For loop iterating from 1 to 5
for i in range(1, 6):
    print(i)

f) Recursion

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# Function that recursively calculates factorial
def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n - 1)

print(factorial(5))  # prints: 120

g) Return Statement

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# Function that returns the square of a number
def square(n):
    return n ** 2

print(square(4))  # prints: 16
  1. Coding Drills for Problem-Specific Concepts

The problem-specific concept in our case is the palindrome checking and deletion of characters when necessary. Here’s a drill:

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def can_be_palindrome(s):
    left, right = 0, len(s) - 1
    while left < right:
        if s[left] != s[right]:
            s_without_left = s[:left] + s[left + 1:]
            s_without_right = s[:right] + s[right + 1:]
            return s_without_left == s_without_left[::-1] or s_without_right == s_without_right[::-1]
        left += 1
        right -= 1
    return True

print(can_be_palindrome("racecar"))  # prints: True
print(can_be_palindrome("hello"))  # prints: False
  1. Integrating the Drills

To solve the initial problem, these drills can be integrated as follows:

  • Start by defining the main function validPalindrome(s) which is a combination of the Function Definition and Return Statement concepts.

  • Inside the function, define a nested function verify(s, left, right, deleted), which encapsulates the main logic of the solution.

  • In verify, implement a while loop (Loops) to iterate over the string from both ends towards the center (String/Data Structure Manipulation).

  • Inside the loop, use an if condition (Conditional Statements) to check if the characters at left and right indices are not equal.

  • If they’re not equal, use another if condition to check if a character has been deleted before. If it has, return False, otherwise recursively call verify function (Recursion) to try deleting the

character at left first and then at right.

  • If all characters have been processed and the function hasn’t returned yet, return True.

  • Finally, call the verify function from validPalindrome to start the process.

My Notes

title: Valid Palindrome II excerpt: This covers expressing the skipping characters by using logical OR in the return expression. tags: logical-or palindrome

Given a string s, return true if the s can be palindrome after deleting at most one character from it.

Example 1:

Input: s = "aba"
Output: true

Example 2:

Input: s = "abca"
Output: true
Explanation: You could delete the character 'c'.

Example 3:

Input: s = "abc"
Output: false

Constraints

  • 1 <= s.length <= 10^5
  • s consists of lowercase English letters.

Thought Process

Deleting

  • Implement palindrome method.
  • Create a new string by deleting one character
  • Check if the new string is a palindrome.

This will create many string objects.

Skipping

  • Mark which index can be skipped.
  • Let the palindrome method skip this character when comparing characters

Implementation

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def valid_palindrome(s)
  palindrome?(s, 0, s.size - 1)
end

def palindrome?(s, l, r)
  while l < r
    if s[l] == s[r]
      l += 1
      r -= 1
    else
      skip_left = s[l+1..r]
      skip_right = s[l..r-1]

      return skip_left.reverse == skip_left || skip_right.reverse == skip_right
    end
  end

  true
end

The key insight to solve this problem is that we cannot skip more than one character to check if the string is a palindrome or not. This is expressed in the code in the return statement where we have the logical OR condition.

Building Block

  • Logical OR