Write a C Program to check whether a given number is Palindrome or not.

A number is said to be a palindrome number if it reads the same forward and backward i.e., on reversing the digits of the number we get the same number.

**Example:** 121, 343.

1. Enter the number.

2. Reverse the number.

3. Check if the reversed number is equal to the original number or not.

4. If they are equal, print the number is a Palindrome number else print, the number is not a palindrome number.

There are several ways to check whether a number is Palindrome or not in C. Let’s look at all different approaches to program this function:

**Palindrome Number Program in C**

- Palindrome Program in C using While Loop
- Palindrome Program in C using Iterative Function
- Palindrome Program in C using Recursion
- Palindrome Program in C using Bitwise Operator

**String Palindrome Program in C**

- String Palindrome Program in C using For Loop
- String Palindrome Program in C using Recursion
- String Palindrome Program in C using Naive Approach
- String Palindrome Program in C using Optimized Approach
- String Palindrome Program in C using Stack

In this approach, we check for Palindrome number using while loop.

Here is the source code of the C program to check whether a number is palindrome or not using while loop. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/* * C Program to Check Number is Palindrome or Not using While Loop. */ #include <stdio.h> int main() { int n, rev = 0; printf("Enter the number: "); scanf("%d", &n); int num= n; //To store the original number in the variable num //Reverse the number and store it in variable rev while (n > 0) { rev = rev * 10 + n % 10; n = n / 10; } // check if original number is same as reversed number or not if (num == rev) printf("%d is a palindrome number.", num); else printf("%d is not a palindrome number.", num); return 0; }

1. Take the number, **n** as input from the user.

2. Initialize the variable num and store the value of **n** in it and also initialize **rev=0**.

3. Run a while loop till **n** is greater than 0.

4. Inside the while loop reverse the number **n** by using the formula **rev=rev*10 + n%10** and in each iteration recalculate the value of n as **n/10**.

5. After reversing the number and storing it in the variable rev check if **num** is equal to **rev** or not.

6. If **num** is equal to rev print the number as palindrome number.

7. Else, print the number as not a palindrome number.

**Example:**

- Consider the number, n entered by the user is 121.
- Assign the number 121 in another variable num. In the while loop check if n>0 or not.
- Since n=121 condition becomes true and enter the loop. Inside the loop calculate rev as rev=rev*10 + n%10 i.e., rev= 0*10 + 121%10 = 0 + 1 = 1.
- Now, calculate the value of n as n=n/10 = 121/10 = 12. Repeat this process till n is greater than 0.
- In the next iteration rev = 1*10+12%10 = 10+2 = 12 and n = 12/10 = 1. Again, n is greater than 0 so enter the while loop and calculate rev. rev= 12*10 + 1%10 = 120+1 = 121 and n=1/10 = 0.
- Now the value of n is 0 so condition in the while loop becomes false hence the loop gets terminated.
- Now check if num is equal to rev or not. The value of num = 121 and the value of rev=121.
- Since both are equal print 121 is a palindrome number.

**Time Complexity: O(log n)**

The above program for checking palindrome number has a time complexity of O(n) as inside the while loop the value of n reduces to n/10 in each iteration.

**Space Complexity: O(1)**

In the above program, space complexity is O(1) as no extra variable has been taken to store the values in the memory. All the variables initialized takes a constant O(1) space.

**Testcase 1:** In this case, we enter the number “121” as input to check whether it is a palindrome number or not.

Enter the number: 121 121 is a palindrome number.

**Testcase 2:** In this case, we enter the number “342” as input to check whether the entered number is palindrome number or not.

Enter the number: 342 342 is not a palindrome number.

In this approach, we check for Palindrome number using while loop.

Here is the source code of the C program to check whether a number is palindrome or not using iterative function. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/* * C Program to Check Number is Palindrome or Not using Iterative. */ #include <stdio.h> // function to reverse digits of num int reverse(int num) { int rev = 0; while (num > 0) { rev = rev * 10 + num % 10; num = num / 10; } return rev; } //Check if the number is palindrome or not int isPalindrome(int n) { //call the function reverse(n) int rev = reverse(n); // Check if rev and n are equal or not. if (rev == n) return 1; else return 0; } //main function int main() { int n,res; printf("Enter the number: "); scanf("%d",&n); res = isPalindrome(n); //Call the function isPalindrome if(res==1) printf("%d is a palindrome number.",n); else printf("%d is not a palindrome number.",n); return 0; }

1. Take the number, **n** as input from the user.

2. Call the function **isPalindrome()** by passing the value of **n** and store the result obtained by the function, in the variable **res**.

3. Inside the **isPalindrome()** function call another function **reverse()** by passing the value n and store the result obtained by the function, in the variable **rev**.

4. Inside the **reverse()** function reverse the number using while loop by using the formula **rev=rev*10 + num%10** and in each iteration recalculate the value of num as num/10 in each iteration.

5. After calculating rev return the value of **rev**. The returned value from the function **reverse()** gets stored in the variable rev in the function **isPalindrome()**.

6. Now, in the function **isPalindrome()** check if **rev** is equal to **n** or not. If they are equal return the value as 1 else return the value as 0.

7. The value returned from **isPalindrome()** function gets stored in the variable res in the main function.

8. In the main function check if res is equal to 1 or not. If it is equal to 1 print the number as palindrome number.

9. Else, print the number as not a palindrome number.

**Time Complexity: O(log n)**

The above program for checking palindrome number has a time complexity of O(n) as inside the while loop the value of n reduces to n/10 in each iteration.

**Space Complexity: O(1)**

In the above program, space complexity is O(1) as no extra variable has been taken to store the values in the memory. All the variables initialized takes a constant O(1) space.

**Testcase 1:** In this case, we enter the number “1001” as input to check whether it is a palindrome number or not.

Enter the number: 1001 1001 is a palindrome number.

**Testcase 2:** In this case, we enter the number “938” as input to check whether the entered number is palindrome number or not.

Enter the number: 938 938 is not a palindrome number.

In this approach, we check for Palindrome number using recursion.

Here is the source code of the C program to check whether a number is palindrome or not using recursion. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/* C Program to check number is Palindrome or Not using recursion*/ #include <stdio.h> //Check if the number is palindrome or not int isPalindrome(int n,int rev) { if (n==0) return rev; //calculate reverse of the number rev = rev * 10 + n % 10; isPalindrome(n/10,rev); } //main function int main() { int n,res,rev=0; printf("Enter the number: "); scanf("%d",&n); res = isPalindrome(n,rev); //Call the function isPalindrome if(res==n) printf("%d is a palindrome number.",n); else printf("%d is not a palindrome number.",n); return 0; }

1. Take the number, **n** as input from the user.

2. Initialize variable **res** to store the result returned from **isPalindrome()** function and variable **rev=0**.

3. Call the function **isPalindrome()** by passing the value of **n** and **rev** and store the result obtained by the function, in the variable **res**.

4. Inside the **isPalindrome()** function check if n = 0 or not, if n=0 return the value of **rev**.

5. Else compute the value of **rev** by using the formula **rev=rev*10 + n%10** and in each recursive call recompute the value of **n** as n=n/10.

6. Recursively call the function **isPalindrome()** by passing the value of **rev** and n/10 until n is not equal to 0.

7. When **n=0** return the value of **rev** i.e., the reversed number and store it in the variable **res** in the main function.

8. Inside the main function then check if **n** is equal to **res** or not.

9. If **n=res**, print It is a palindrome number.

10. Else print, it is not a palindrome number.

**Time Complexity: O(log n)**

The above program for checking palindrome number has a time complexity of O(n) as the recursive function is called the number of times n is divided by 10.

**Space Complexity: – O(1)**

In the above program, space complexity is O(1) as no extra variable has been taken to store the values in the memory. All the variables initialized takes a constant O(1) space.

**Testcase 1:** In this case, we enter the number “121” as input to check whether it is a palindrome number or not.

Enter the number: 121 121 is a palindrome number.

**Testcase 2:** In this case, we enter the number “537” as input to check whether the entered number is palindrome number or not.

Enter the number: 537 537 is not a palindrome number.

In this approach, we checks whether the given number is palindrome or not using bitwise operator.

Here is source code of the C Program to Check whether the given Number is Palindrome or not using Bitwise Operator. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/* * C Program to Check whether the given Number is Palindrome * or not using Bitwise Operator */ #include <stdio.h> #include <string.h> #define SIZE 8 /* Function Prototype */ int is_palindrome(unsigned char[]); void main() { int num, num1 = 0, i = 0, j = SIZE - 1, res; unsigned char c[SIZE]; printf("Enter a number(max 255)"); scanf("%d", &num); num1 = num; while (num != 0) { c[j] = num&1; j--; num = num>>1; /* Shifting right the given number by 1 bit */ } printf("The number %d in binary is:", num1); for (i = 0;i < SIZE;i++) { printf("%d", c[i]); } res = is_palindrome(c); /* Calling Function */ if (res == 0) { printf("\nNUMBER IS PALINDROME\n"); } else { printf("\nNUMBER IS NOT PALINDROME\n"); } } /* Code to check if the number is palindrome or not */ int is_palindrome(unsigned char c[]) { char temp[SIZE]; int i, j, flag = 0; for (i = 0, j = SIZE - 1;i < SIZE, j >= 0;i++, j--) { temp[j] = c[i]; } for (i = 0;i < SIZE;i++) { if (temp[i] != c[i]) { flag = 1; } } return flag; }

1. In this C program, we are reading the value of string using word character array variable. A palindrome is a word, phrase or sentence that reads the same backward or forward.

2. Check the values of a given string and reverse string are equal.

3. If else condition statement is used inside for loop if the condition is true then assign the value of ‘flag’ variable as 1.

4. Otherwise, if the condition is false then execute the else statement. Assign the value of ‘flag’ variable as 0 and exit the loop.

5. If else condition statement is used to check the value of ‘flag’ variable is equal to1.

6. If the condition is true then execute the statement and print the string is palindrome.

7. Otherwise, if the condition is false then execute the else statement and print as not a palindrome.

**Testcase 1:** In this case, we enter the number “153” as input to find binary number and check whether it is a palindrome number or not.

Enter a number(max 255): 153 The number 153 in binary is:10011001 NUMBER IS PALINDROME

**Testcase 2:** In this case, we enter the number “24” as input to find binary number and check whether the entered number is palindrome number or not.

Enter a number(max 255): 24 The number 24 in binary is: 00011000 NUMBER IS PALINDROME

To practice programs on every topic in C, please visit “Programming Examples in C”, “Data Structures in C” and “Algorithms in C”.

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