C Program to Check for Palindrome String

Last Updated : 18 Apr, 2024

A string is said to be palindrome if the reverse of the string is the same as the string. For example, “abba” is a palindrome because the reverse of “abba” will be equal to “abba” so both of these strings are equal and are said to be a palindrome, but “abbc” is not a palindrome.

In this article, we will see a simple program to check whether the string is palindrome or not.

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Algorithm 1 :

1. Initialize 2 variables, l from the start and h from the end of the given string.
2. Now, we will compare the characters at index l and h with each other
3. If the characters are not equal, the string is not palindrome.
4. If the characters are equal, we will increment l and decrement h.
5. Steps 2,3 and 4 will be repeated till ( l < h ) or we find unequal characters.
6. If we reach the condition ( l < h ), it means all the corresponding characters are equal and the string is palindrome.

Basically, we traverse the string in forward and backward directions comparing characters at the same distance from start and end. If we find a pair of distinct characters, the string is not palindrome. If we reach the midpoint, the string is palindrome.

Palindrome String Program in C

C++ ```#include <iostream> #include <string> using namespace std; bool isPalindrome(string str) { int low = 0; int high = str.size() - 1; // Keep comparing characters while they are same while (low < high) { if (str[low] != str[high]) { return false; // not a palindrome. } low++; // move the low index forward high--; // move the high index backwards } return true; // is a palindrome } int main() { string str= "abbba"; string str1 = "abcded"; cout << str << " is palindrome " << isPalindrome(str) << endl; cout << str1 << " is palindrome " << isPalindrome(str1) <<endl; return 0; } ``` C ```#include <stdio.h> #include <string.h> int isPalindrome(char str[]) { int low = 0; int high = strlen(str) - 1; // Keep comparing characters while they are same while (low < high) { if (str[low] != str[high]) { return 0; // not a palindrome. } low++; // move the low index forward high--; // move the high index backwards } return 1; // is a palindrome } int main() { char str[] = "abbba"; char str1[] = "abcded"; printf("%s is palindrome %d\n", str, isPalindrome(str)); printf("%s is palindrome %d\n", str1, isPalindrome(str1)); return 0; } ``` Java ```public class Main { // Function to check if a string is a palindrome static boolean isPalindrome(String str) { int low = 0; int high = str.length() - 1; // Keep comparing characters while they are same while (low < high) { if (str.charAt(low) != str.charAt(high)) { return false; // not a palindrome. } low++; // move the low index forward high--; // move the high index backwards } return true; // is a palindrome } public static void main(String[] args) { String str = "abbba"; String str1 = "abcded"; System.out.println(str + " is palindrome " + isPalindrome(str)); System.out.println(str1 + " is palindrome " + isPalindrome(str1)); } } ```

Output
```abbba is palindrome 1
abcded is palindrome 0
```

Time complexity: O(n), where n is the number of characters in the string.
Auxiliary Space: O(1)

Algorithm 2 : (Recursive Approach)

The approach remains the same. Recursively check if it’s a palindrome or not.
Below is the code for better understanding.

C ```// C implementation to check if a given // string is palindrome or not #include <stdio.h> #include <string.h> #include <stdbool.h> bool isPalindrome(char str[], int l, int h) { // if the string size is 0 or 1 // it is a palindrome if (l >= h) return true; // if both the terminal characters // not matching, not palindrome if (str[l] != str[h]) return false; // call the smaller sub-problem return isPalindrome(str, l + 1, h - 1); } int main() { char str[] = { "abbba" }; // Start from first and // last character of str int l = 0; int h = strlen(str) - 1; bool ans = isPalindrome(str, l, h); if (ans) { printf("%s is a palindrome\n", str); } else { printf("%s is not a palindrome\n", str); } return 0; } ```

Output
```abbba is a palindrome
```

Time complexity: O(n), where n is the number of characters in the string.
Auxiliary Space: O(1)

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