Mirror String
You are given a string S containing only uppercase English characters. Find whether S is the same as its reflection in the mirror.

For Example, S = “AMAMA” is the same as its reflection in the mirror.

Input Format:

The first line of input contains an integer T, denoting the number of test cases. The first line of each test case contains a string 'S'.

Output Format:

For each test case, print a single line containing “YES” or “NO” depending on whether the string 'S' is the same as its reflection in the mirror or not. The output of each test case is printed in a separate line.

Note:

You do not need to print anything, it has already been taken care of. Just implement the function. The return type of the function is boolean, which returns true if the string is the same as its reflection in the mirror, otherwise returns false.

Constraints

1

Question

Mirror String

You are given a string S containing only uppercase English characters. Find whether S is the same as its reflection in the mirror.

For Example, S = “AMAMA” is the same as its reflection in the mirror.

Input Format:

The first line of input contains an integer T, denoting the number of test cases.

The first line of each test case contains a string 'S'.

Output Format:

For each test case, print a single line containing “YES” or “NO” depending on whether the string 'S' is the same as its reflection in the mirror or not.

The output of each test case is printed in a separate line.

Note:

You do not need to print anything, it has already been taken care of. Just implement the function.

The return type of the function is boolean, which returns true if the string is the same as its reflection in the mirror, otherwise returns false.

Constraints

1 <= T <=10
1 <= Length(S) <= 10 ^ 5

Where ‘T’ is the number of test cases, ‘S’ is the string given in input.

Time limit: 1sec.

Accepted Answer

Check Palindrome and Symmetricity

For a string to be same as its reflection in the mirror, it should satisfy the following conditions:

The given string “S” needs to be a palindrome.
All the characters of the string must be symmetric so that the reflection of the characters remains the same.

The symmetric characters are AHIMOTUVWXY.
Store the symmetric characters in an unordered set, traverse the string, and if the string contains any non-symmetric character, then return false.
If all the characters present in the string are symmetric, then check if the string is a palindrome or not. If the string is a palindrome, then return true, otherwise, return false.

Space Complexity: O(1)Explanation:

O(1).

Only constant additional space is required to store the symmetric characters.

Time Complexity: O(n)Explanation:

O(N), where N is the length of the string.

Finding an element in an unordered set takes constant time, O(1) on average. Thus, traversing the string to check for symmetric characters takes O(N * 1) = O(N) time.
Checking whether the string is a palindrome or not also takes linear time, O(N)

So, the final time complexity is O(N + N) = O(N).

C++ (g++ 5.4)

/*

    Time complexity: O(N)

    Space complexity: O(1)



    Where N is the length of the string.

*/



#include 

#include 



// Function to check if the string is a mirror string.

bool isReflectionEqual(string &s)

{

    // Symmetric characters.

    unordered_set symmetric = {'A', 'H', 'I', 'M', 'O', 'T', 'U', 'V', 'W', 'X', 'Y'};

    int len = s.length();



    // Checking if all the characters in the string are symmetric.

    for (int i = 0; i < len; i++)

    {

        // If the string contains a non-symmetric character return false.

        if (symmetric.find(s[i]) == symmetric.end())

            return false;

    }



    string rev = s;

    reverse(rev.begin(), rev.end());



    // Checking if the string is a palindrome.

    if (s == rev)

        return true;

    else

        return false;

}

Java (SE 1.8)

/*

    Time complexity: O(N)

    Space complexity: O(1)



    Where N is the length of the string.

*/



import java.util.ArrayList;

import java.util.HashSet;



public class Solution {

    public static String Reverse(String s){

        char[] charArray = s.toCharArray();

        reverse(charArray, 0, charArray.length - 1);

        return new String(charArray);

    }



    // Function to check if the string is a mirror string.

    public static boolean isReflectionEqual(String s){

        HashSet symmetric = new HashSet<>();



        // Symmetric characters

        symmetric.add('A');

        symmetric.add('H');

        symmetric.add('I');

        symmetric.add('M');

        symmetric.add('O');

        symmetric.add('T');

        symmetric.add('U');

        symmetric.add('V');

        symmetric.add('W');

        symmetric.add('X');

        symmetric.add('Y');



        int n = s.length();



        // Checking if all the characters in the string are symmetric.

        for (int i = 0; i < n; i++){

            

            // If the string contains a non-symmetric character return false.

            if (symmetric.contains(s.charAt(i)) == false){

                return false;

            }

        }



        String rev = s;

        s = Reverse(s);



        // Check if the String is a palindrome.

        if (rev.equals(s)){

            return true;

        } else {

            return false;

        }

    }



    // Reverse the letters of the word.

    public static void reverse(char str[], int start, int end){



        // Temporary variable to store character.

        char temp;

        while (start <= end){



            // Swapping the first and last character.

            temp = str[start];

            str[start] = str[end];

            str[end] = temp;

            start++;

            end--;

        }

    }

}

Python (3.5)

'''

    Time complexity: O(N)

    Space complexity: O(1)



    Where N is the length of the string.

'''



# Function to check if the string is a mirror string.

def isReflectionEqual(s):

    # Symmetric characters.

    symmetric = {'A', 'H', 'I', 'M', 'O', 'T', 'U', 'V', 'W', 'X', 'Y'}

    n = len(s)



    # Checking if all the characters in the string are symmetric.

    for i in range(n):

        # If the string contains a non-symmetric character return false

        if (s[i] not in symmetric):

            return False



    rev = s[::-1]

    # Checking if the string is a palindrome.

    if (s == rev):

        return True

    else:

        return False

Accepted Answer

def solve(s):

ref="AHIMOTUVWXY"

for i in s:

if i not in ref:

return 'NO'

if s==s[::-1]:

return "YES"

return "NO"

You are given a string S containing only uppercase English characters. Find whether S is the same as its reflection in the mirror.

For Example, S = “AMAMA” is the same as its reflection in the mirror.

Input Format:

Output Format:

Note:

Constraints

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