Given a string S consisting of N characters of the English alphabet, the task is to check if the given string is a palindrome. If the given string is a palindrome, then print “Yes“. Otherwise, print “No“.

Note: A string is said to be palindrome if the reverse of the string is the same as the string. 

Examples:

Input: S = “ABCDCBA”
Output: Yes
Explanation:
The reverse of the given string is equal to the (ABCDCBA) which is equal to the given string. Therefore, the given string is palindrome.

Input: S = “GeeksforGeeks”
Output: No
Explanation: 
The reverse of the given string is equal to the (skeeGrofskeeG) which is not equal to the given string. Therefore, the given string is not a palindrome.

Naive Approach: The simplest approach to use the inbuilt reverse function in the STL. Follow the steps below to solve the problem:

• Copy the string S to another string, say P, and then reverse the string S.
• Now check if the string S is equal to the string P and then print “Yes“. Otherwise, print “No“.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check whether
// the string is palindrome
string isPalindrome(string S)
{
	// Stores the reverse of the
	// string S
	string P = S;

	// Reverse the string P
	reverse(P.begin(), P.end());

	// If S is equal to P
	if (S == P) {
		// Return "Yes"
		return "Yes";
	}
	// Otherwise
	else {
		// return "No"
		return "No";
	}
}

// Driver Code
int main()
{
	string S = "ABCDCBA";
	cout << isPalindrome(S);

	return 0;
}

Output

Yes

Efficient Approach: The above approach can be optimized in space complexity by traversing the string and checking whether the character at i^th index is equal to the character at the (N-i-1)^th index for every index in the range [0, N/2]. Follow the steps below to solve the problem:

• Iterate over the range [0, N/2], using the variable i and in each iteration check if the character at index i and N-i-1 are not equal, then print “No” and break.
• If none of the above cases satisfy, then print “Yes“.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check whether string
// is palindrome
string isPalindrome(string S)
{
	// Iterate over the range [0, N/2]
	for (int i = 0; i < S.length() / 2; i++) {

		// If S[i] is not equal to
		// the S[N-i-1]
		if (S[i] != S[S.length() - i - 1]) {
			// Return No
			return "No";
		}
	}
	// Return "Yes"
	return "Yes";
}

// Driver Code
int main()
{
	string S = "ABCDCBA";
	cout << isPalindrome(S);

	return 0;
}

Output: 

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)