Check if the given string is the same as its reflection in a mirror

Given a string S containing only uppercase English characters. The task is to find whether S is the same as its reflection in a mirror.

Examples:

Input: str = "AMA"
Output: YES
AMA is same as its reflection in the mirror.

Input: str = "ZXZ"
Output: NO

Approach: The string obviously has to be a palindrome, but that alone is not enough. All characters in the string should be symmetric so that their reflection is also the same. The symmetric characters are AHIMOTUVWXY.



  • Store the symmetric characters in an unordered_set.
  • Traverse the string and check if there is any non-symmetric character present in the string. If yes then return false.
  • Else check if the string is palindrome or not. If the string is palindrome also then return true else return false.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the 
// above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to check reflection
bool isReflectionEqual(string s)
{
    // Symmetric characters
    unordered_set<char> symmetric = { 'A', 'H', 'I', 'M',
                        'O', 'T', 'U', 'V', 'W', 'X', 'Y' };
  
    int n = s.length();
  
    for (int i = 0; i < n; i++)
        // If any non-symmetric character is
        // present, the answer is NO
        if (symmetric.find(s[i]) == symmetric.end())
            return false;
  
    string rev = s;
    reverse(rev.begin(), rev.end());
  
    // Check if the string is a palindrome
    if (rev == s)
        return true;
    else
        return false;
}
  
// Driver code
int main()
{
    string s = "MYTYM";
    if (isReflectionEqual(s))
        cout << "YES";
    else
        cout << "NO";
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of above approach
import java.util.*;
  
class GFG 
{
  
    static String Reverse(String s)
    {
        char[] charArray = s.toCharArray();
        reverse(charArray, 0, charArray.length - 1);
        return new String(charArray);
    }
  
    // Function to check reflection 
    static boolean isReflectionEqual(String s) 
    {
        HashSet<Character> 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();
  
        // If any non-symmetric character is
        for (int i = 0; i < n; i++) 
          
        // present, the answer is NO 
        {
            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 
    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--;
        }
    }
      
    // Driver code 
    public static void main(String[] args) 
    {
        String s = "MYTYM";
        if (isReflectionEqual(s))
        {
            System.out.println("YES");
        
        else 
        {
            System.out.println("NO");
        }
    }
}
  
// This code contributed by Rajput-Ji

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the 
# above approach
  
# Function to check reflection
def isReflectionEqual(s):
  
    # Symmetric characters
    symmetric = dict()
  
    str1 = "AHIMOTUVWXY"
  
    for i in str1:
        symmetric[i] = 1
  
    n = len(s)
  
    for i in range(n):
          
        # If any non-symmetric character 
        # is present, the answer is NO
        if (symmetric[s[i]] == 0):
            return False
  
    rev = s[::-1]
  
    # Check if the is a palindrome
    if (rev == s):
        return True
    else:
        return False
  
# Driver Code
s = "MYTYM"
if (isReflectionEqual(s)):
    print("YES")
else:
    print("NO")
  
# This code is contributed by Mohit Kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach 
using System;
using System.Collections.Generic ;
  
class GFG
{
  
    static string Reverse( string s )
    {
        char[] charArray = s.ToCharArray();
        Array.Reverse( charArray );
        return new string( charArray );
    }
      
    // Function to check reflection 
    static bool isReflectionEqual(string s) 
    
        HashSet<char> symmetric = new HashSet<char>();
          
        // 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; 
      
        for (int i = 0; i < n; i++) 
          
            // If any non-symmetric character is 
            // present, the answer is NO 
            if (symmetric.Contains(s[i]) == false
                return false
      
        string rev = s; 
        s = Reverse(s); 
      
        // Check if the string is a palindrome 
        if (rev == s) 
            return true
        else
            return false
    
  
    // Driver code 
    static public void Main() 
    
        string s = "MYTYM"
        if (isReflectionEqual(s)) 
            Console.WriteLine("YES"); 
        else
            Console.WriteLine("NO"); 
    
}
  
// This code is contributed by Ryuga

chevron_right


Output:

YES

Time Complexity: O(N)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.