Palindrome Number

To solve this coding challenge, we need to determine whether a given integer \( x \) is a palindrome. A palindrome is a number that reads the same forward and backward. While it would be simpler to convert the integer to a string and then check if the string is equal to its reverse, the challenge hints at attempting to solve it without string conversion. Hence, we must use pure arithmetic operations to solve this problem. The goal is to devise a method that can yield a boolean result indicating if the number is a palindrome.

Detailed Steps in Pseudocode

# Explanation
  1. Handling Negative Numbers:
    • Any negative number can't be a palindrome due to the negative sign at the beginning.
    • Therefore, if \( x \) is negative, we immediately return false.
  2. Handling Edge Cases:
    • If \( x \) is zero, it is considered a palindrome.
    • If \( x \) ends with zero but is not zero (for example, 10, 100), it can't be a palindrome because it can't start with zero.
  3. Reversing Half of the Number:
    • We can reverse the last half of the number and compare it with the other half.
    • This approach avoids overflow issues that could arise if we tried to reverse the entire number directly.
  4. Reversing Logic:
    • Use a loop to construct the reversed number by extracting the last digit of \( x \) and appending it to the reversed number until the original number \( x \) is less than or equal to the reversed number.
    • In each iteration, update \( x \) by removing its last digit (integer division by 10).
  5. Comparing Halves:
    • Finally, compare the reversed half with the remaining portion of \( x \).
    • Consider cases where reversing might lead to an odd length (ex: \( 12321 \)), where we should also account for removing the middle digit.
    # Pseudocode with Comments
                                                
# Define the main function to check if a number is a palindrome
function isPalindrome(number):
# If the number is negative or ends in zero (but is not zero itself)
if number < 0 or (number % 10 == 0 and number != 0):
# Return false because these cannot be palindromes
return False

# Initialize a variable to hold the reversed number
reversedHalf = 0

# Continue reversing the number until the original number is less than the reversed number
while number > reversedHalf:
# Append the last digit of 'number' to 'reversedHalf'
reversedHalf = reversedHalf * 10 + number % 10
# Remove the last digit from 'number'
number = number // 10

# After the loop, 'number' should be either equal to 'reversedHalf' or equal to 'reversedHalf' // 10
# This handles cases with odd and even lengths of 'number'
return number == reversedHalf or number == reversedHalf // 10
    # Step-by-Step Explanation
  6. Initial Check for Negative and Zero Termination:
    • The first condition checks if the number is negative or ends in zero. Both cases disqualify the number from being a palindrome.
    • For instance, if
      number = -121
      , return
      False
      .
    • If
      number = 10
      , return
      False
      .
  7. Reversing the Number Logic:
    • Initialize
      reversedHalf
      to 0. This will hold the reversed digits from the original number.
    • Use a loop where:
      • Extract the last digit of the number by calculating
        number % 10
        .
      • Update
        reversedHalf
        by shifting its digits left (multiplying by 10) and adding the extracted digit.
      • Remove the last digit from the original number by performing an integer division by 10.
  8. Loop Exit and Check Condition:
    • Continue the loop until the original number
      number
      becomes less than or equal to
      reversedHalf
      .
    • Once the loop exits, compare if the remaining
      number
      equals
      reversedHalf
      or if the integer division of
      reversedHalf
      by 10 equals
      number
      . This accounts for odd-length inputs where one digit may be left in the middle.
  9. Return Result:
    • Based on the comparison, return
      True
      if they match, otherwise return
      False
      .
By following these steps and understanding the logic behind each step, you will be able to implement a solution that identifies if a number is a palindrome without converting the integer to a string, as prescribed by the problem statement.