# Check if the given String is Palindrome or not

Problem Statement: “Given a string, check if the string is palindrome or not.”  A string is said to be palindrome if the reverse of the string is the same as the string.

Examples:

```Example 1:
Input: Str =  “ABCDCBA”
Output: Palindrome
Explanation: String when reversed is the same as string.

Example 2:
Input: Str = “TAKE U FORWARD”
Output: Not Palindrome
Explanation: String when reversed is not the same as string.```

### Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Approach:

Run a for loop till half the length of the string in order to check the first and last character of the string.

And check if the first and last elements of the string are equal. And then move both pointers first pointer forward and last pointer backward.

Here we are achieving this with the help of ‘i’ which moves with them for a loop. First element can be get by str[i] and last element by str[str.length() – i – 1]). If this condition gets executed then it is not palindrome and returns false.

If the loop ends after traversing elements till half of the length then, it is Palindrome and returns false.  Code:

## C++ Code

``````#include<bits/stdc++.h>

using namespace std;

bool isPalindrome(string s) {

int left = 0, right = s.length()-1;
while(left<right)
{
if(!isalnum(s[left]))
left++;
else if(!isalnum(s[right]))
right--;
else if(tolower(s[left])!=tolower(s[right]))
return false;
else {
left++;
right--;
}
}
return true;

}
int main() {

string str = "ABCDCBA";
bool ans = isPalindrome(str);

if (ans == true) {
cout << "Palindrome";
} else {
cout << "Not Palindrome";
}
return 0;
}
``````

Output: Palindrome

Time Complexity:  O(N)

Space Complexity: O(1)

## Java Code

``````import java.io.*;
import java.util.Arrays;
class Test {
static private boolean isPalindrome(String s) {
int left = 0, right = s.length()-1;
while(left<right)
{
char l = s.charAt(left), r = s.charAt(right);
if(!Character.isLetterOrDigit(l))
left++;
else if(!Character.isLetterOrDigit(r))
right--;
else if(Character.toLowerCase(l)!=Character.toLowerCase(r))
return false;
else {
left++;
right--;
}
}
return true;
}
public static void main(String[] args) {
String str = "ABCDCBA";
boolean ans = isPalindrome(str);

if (ans == true) {
System.out.println("Palindrome");
} else {
System.out.println("Not Palindrome");
}
}
}
``````

Output: Palindrome

Time Complexity:  O(N)

Space Complexity: O(1)

Special thanks to Rushikesh Adhav for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article