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# C++ Program to check if a given String is Palindrome or not

• Difficulty Level : Easy
• Last Updated : 21 Jul, 2021

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:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `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`

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

Efficient Approach: The above approach can be optimized in space complexity by traversing the string and checking whether the character at ith 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++

 `// C++ program for the above approach` `#include ` `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)

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