Palindrome Number

To determine whether an integer is a palindrome, you can compare it to its reversed form. Negative numbers and numbers that end with 0 (except for 0 itself) can be immediately ruled out as palindromes. Here’s the code snippet:

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class Solution:
    def isPalindrome(self, x: int) -> bool:
        # If x is negative or ends with 0 and is not 0, it's not a palindrome
        if x < 0 or (x % 10 == 0 and x != 0):
            return False

        reversed_x = 0
        while x > reversed_x:
            reversed_x = reversed_x * 10 + x % 10
            x //= 10

        # If the length is an odd number, we get rid of the middle digit by dividing by 10
        return x == reversed_x or x == reversed_x // 10

Explanation:

Return False if x is negative or ends with 0 (and is not 0 itself).
Reverse the second half of x and compare it to the first half. This avoids potential overflow issues.
If the length of the number is odd, the middle digit does not affect whether it’s a palindrome, so we divide the reversed second half by 10 and compare.

This approach ensures that the comparison is made without having to reverse the entire integer, thus optimizing for efficiency.

Identifying Problem Isomorphism

“Palindrome Number” can be approximately mapped to the following problems:

“Reverse Integer” (#7) on LeetCode
“Valid Palindrome” (#125) on LeetCode

Reasoning:

These problems revolve around manipulating and checking properties of numerical values or sequences, and particularly they involve reversal of the sequences or numbers.

“Reverse Integer” is a simpler problem, where you are only asked to reverse an integer without worrying about any constraints like palindrome property.
“Valid Palindrome” is a more complex problem. Here, the task is to check whether a given string is palindrome or not, while ignoring case and non-alphanumeric characters. This problem involves checking the palindrome property, similar to “Palindrome Number”, but it has added complexity of handling non-alphanumeric characters and case sensitivity.

These problems are not exact isomorphs of “Palindrome Number”, but they involve similar operations or checks to be performed on the given number or sequence.

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def isPalindrome(self, x: int) -> bool:
	if x < 0:
		return False

	inputNum = x
	newNum = 0
	while x > 0:
		newNum = newNum * 10 + x % 10
		x = x // 10
	return newNum == inputNum

Problem Classification

This problem can be classified under the following categories:

Number Theory: The problem involves working with numbers and properties associated with numbers, such as reversal, negativity, and digits. Hence, it comes under the domain of number theory.
Palindrome: The problem involves identifying whether a given number is a palindrome. A palindrome is a concept from the field of strings or sequences where a sequence remains the same when reversed.
Decision Making: The problem is about making a binary decision - whether the given number is a palindrome or not. Therefore, it involves logical decision making.

Language Agnostic Coding Drills

Understanding Basic Control Flow - Conditional Statements: A key part of this problem is the ability to understand and implement conditional statements (if conditions). The initial part of the problem requires checking if the given number is negative, and if so, return False. This is a basic problem that introduces conditional statements.
Understanding Arithmetic Operations and Variable Updates: The next part of the problem involves extracting digits from the number using modulus and division operations. It also involves updating variables in each iteration of the loop. You should understand how arithmetic operations work, and how variable values are updated in each iteration of a loop.
Understanding Loops: This problem requires the use of a loop to iteratively process each digit of the number until the number becomes 0. The concept of looping is vital here to repeatedly perform an operation (extracting digits and forming the reversed number).
Understanding Problem Solving Approach - Reversing a Number: The specific problem-solving skill that is needed here is the ability to reverse a number. This involves repeatedly taking the remainder when the number is divided by 10 (which gives the last digit), and adding it to the reversed number after shifting its digits one place to the left.
Using Comparison Operators: Finally, the solution requires comparing two numbers for equality. You need to understand how to use comparison operators in your chosen programming language.

When solving the problem, start by handling the edge case of negative numbers. Then, go into a loop where you repeatedly extract the last digit of the number and form the reversed number. Keep updating the original number by removing its last digit. Finally, when the original number becomes 0, the loop ends, and you can compare the reversed number with the original number for equality.

Targeted Drills in Python

Sure, let’s build this step by step using Python:

Understanding Basic Control Flow - Conditional Statements: Create a function that checks if a number is negative, and returns a boolean.

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def is_negative(num):
    if num < 0:
        return True
    return False

print(is_negative(-1))  # Returns: True
print(is_negative(1))   # Returns: False

Understanding Arithmetic Operations and Variable Updates: Write a function that divides a number by 10, updates it, and prints the result. Do the same for modulus operation.

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def divide_by_ten(num):
    num = num // 10
    return num

def modulus_by_ten(num):
    num = num % 10
    return num

print(divide_by_ten(123))  # Returns: 12
print(modulus_by_ten(123)) # Returns: 3

Understanding Loops: Implement a while loop that runs until a given number becomes 0.

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def loop_until_zero(num):
    while num > 0:
        print(num)
        num = num // 10

loop_until_zero(123)  # Prints: 123, 12, 1

Understanding Problem Solving Approach - Reversing a Number: Combine the operations from the above steps to reverse a number.

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def reverse_number(num):
    rev_num = 0
    while num > 0:
        last_digit = num % 10
        rev_num = rev_num * 10 + last_digit
        num = num // 10
    return rev_num

print(reverse_number(123))  # Returns: 321

Using Comparison Operators: Create a function that compares two numbers and returns if they are equal.

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def are_equal(num1, num2):
    if num1 == num2:
        return True
    return False

print(are_equal(123, 123))  # Returns: True
print(are_equal(123, 321))  # Returns: False

Final Drill: Combine everything to check if a number is a palindrome.

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def is_palindrome(num):
    if is_negative(num):
        return False
    input_num = num
    reversed_num = reverse_number(num)
    return are_equal(input_num, reversed_num)

print(is_palindrome(121))  # Returns: True
print(is_palindrome(-121)) # Returns: False
print(is_palindrome(10))   # Returns: False

My Notes

title: Palindrome Number excerpt: This covers the basic building blocks such as modulus, integer division and reversing a number. tags: palindrome modulo-operator integer-division reverse-number

Given an integer x, return true if x is palindrome integer.

An integer is a palindrome when it reads the same backward as forward. For example, 121 is palindrome while 123 is not.

Example 1:

Input: x = 121
Output: true

Example 2:

Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Example 4:

Input: x = -101
Output: false

Constraints

-2^31 <= x <= 2^31 - 1

Follow up: Could you solve it without converting the integer to a string?

Implementation

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def is_palindrome(x)
  if x < 0 
    return false
  end
  
  forward = []
  a = x

  while a != 0
    forward << a % 10
    a = a / 10
  end
  
  forward == forward.reverse
end

Alternative Implementation

The number is reversed and compared with the given number.

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def is_palindrome(x)
  if x < 0 || (x % 10 == 0 && x != 0)
    return false
  end
  
  reversed_number = 0

  while x > reversed_number
    reversed_number = reversed_number * 10 + x % 10
    x = x/10
  end
  
  x == reversed_number || x == reversed_number/10
end

Building Blocks

Modulus
Integer Division
Reverse Number