Palindrome Number in C, Java, and Python

A palindrome number is a number that remains the same when digits are reversed.

For example, the number 12321 is a palindrome number, but 1451 is not a palindrome number.

Problem Statement

Given a positive integer, the task is to check whether the number is palindrome or not.

Example:

• Input: 1221
• Output: True

• Input: 146541
• Output: False

• Input: 5
• Output: True

The idea is to make a number with all the digits of the given number present in reverse order and check if both numbers are equal. Follow the below steps to solve the problem.

• First, Declare a variable reverseNum and initialize it with 0.
• Now, make a while loop till the original number is greater than zero.
• In every loop get the last digit of the number and add that digit at the end of the reverseNum and then, divide the original number by 10.

reverseNum = reverseNum * 10 + (number % 10)

• Lastly, check if the original number and reverseNum number are equal or not.

Dry run of the above approach

• Orginal = 1221
• reverseNum = 0
• tempOrginal = 1221

reverseNum = original

Implementation

C/C++ Implementation

bool checkPalindrome(int original) {

  int reverseNum = 0;
  int tempOriginal = original;

  while (tempOriginal > 0) {

    int lastDigit = tempOriginal % 10;
    reverseNum = reverseNum * 10 + lastDigit;
    tempOriginal = tempOriginal / 10;
  }

  if (original == reverseNum) {
    return true;
  } else {
    return false;
  }
}

Java Implementation

public int checkPalindrome(int original) {
  int reverseNum = 0;
  int tempOriginal = original;

  while (tempOriginal > 0) {

    int lastDigit = tempOriginal % 10;
    reverseNum = reverseNum * 10 + lastDigit;
    tempOriginal = tempOriginal / 10;
  }

  if (original == reverseNum) {
    return 1;
  } else {
    return 0;
  }

}

Python Implementation

def checkPalindrome(original):
    reverseNum = 0
    tempOriginal = original

    while (tempOriginal > 0):
        lastDigit = tempOriginal % 10
        reverseNum = reverseNum * 10 + lastDigit
        tempOriginal = tempOriginal / 10

    if (original == reverseNum):
        return 1
    else:
        return 0

• Time complexity: O(log10(N)), Where N is the given number.
• Space complexity: O(1)

Special Case

The above solution will only work if the number is less than 10¹⁸, but what would happen if the number is greater than 10¹⁸?

Implementation

C Implementation of Special Case

bool checkPalindrome(char original[]) {

  int n = sizeof(original) / sizeof(original[0]);

  for (int i = 0; i < n / 2; i++) {

    if (original[i] != original[n - 1 - i]) {
      return false;
    }
  }

  return true;
}

C++ Implementation of Special Case

bool checkPalindrome(string original) {

  int n = original.size();

  for (int i = 0; i < n / 2; i++) {

    if (original[i] != original[n - 1 - i]) {
      return false;
    }
  }

  return true;
}

Java Implementation of Special Case

public boolean checkPalindrome(String original) {

  int n = original.length();

  for (int i = 0; i < n / 2; i++) {

    if (original.charAt(i) != original.charAt(n - 1 - i)) {
      return false;
    }
  }

  return true;
}

Python Implementation of Special Case

def checkPalindrome(original):
    
    n = len(original)
    for i in range(0, n//2):
        if original[i] != original[n - i - 1]:
            return False
           
    return True

• Time complexity: O(L), Where L is the length of the given string.
• Space complexity: O(1)

Practice Question

• Palindrome Integer
• Commonly Asked Coding Interview Questions

Frequently Asked Questions

Q.1: Can a negative number be palindrome?

Ans: No, negative numbers can not be palindromes.

Q.2: Can we solve this problem without converting an integer to a string?

Ans: Yes but only for integers less than 10¹⁸.

Q.3: Are all single-digit numbers palindrome?

Ans: Yes

Q.4: How is the time complexity of the above algorithm O(log₁₀(N))?

Ans: Because we are dividing the input number N by 10 in every iteration. So, there is going to be a total of log₁₀(N) iterations.