C Program to Check Palindrome Number.

In this C programming tutorial, we will learn how to check if the given number is a palindrome number or not. Before moving to the code section let us understand what is Palindrome number?

Palindrome Number.

A palindrome number is a number that reads the same backward as forward. In other words, the digits of a palindrome number remain unchanged when read from left to right and from right to left.

Example: 121, 454, and 12321 are palindrome numbers.

Algorithm to Check Palindrome Number.

Step 1: Input the number from the user.

Step 2: Create a copy of the original number to compare later.

Step 3: Initialize variables to store the reverse of the number and the remainder.

Step 4: Use a loop to reverse the digits of the number:

a) Extract the last digit of the number using the modulus (%) operator.
b) Add the last digit to the reversed number after moving the previous digits one place left using multiplication (*= 10).
c) Remove the last digit from the original number using integer division ( /= 10).
d) Repeat the above steps until the original number becomes 0.

Step 5: Compare the reversed number with the copy of the original number.

Step 6: If the reversed number is equal to the original number, it is a palindrome. Otherwise, it is not.

Program to Check Palindrome Number.

// c code implementation to check palindrome number
#include <stdio.h>

int main() {
    int num, originalNum, reversedNum = 0, remainder;

    printf("Enter an integer: ");
    scanf("%d", &num);

    originalNum = num;

    // Reverse the number
    while (num != 0) {
        remainder = num % 10;         
        reversedNum = reversedNum * 10 + remainder;   
        num /= 10;                    
    }

    // Compare with the original number
    if (reversedNum == originalNum) {
        printf("%d is a palindrome number.\n", originalNum);
    } else {
        printf("%d is not a palindrome number.\n", originalNum);
    }

    return 0;
}

Output:

Enter an integer: 12321
12321 is a palindrome number.

Time Complexity: O(n) where n is the number of digits present in the given number.

Space Complexity: O(1) no extra space is required to solve this problem.

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