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++
// C++ implementation of the // above approach#include <bits/stdc++.h>using namespace std;// Function to check reflectionbool 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 codeint main(){ string s = "MYTYM"; if (isReflectionEqual(s)) cout << "YES"; else cout << "NO";} |
Java
// Java implementation of above approachimport 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 |
Python3
# Python3 implementation of the # above approach# Function to check reflectiondef 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 Codes = "MYTYM"if (isReflectionEqual(s)): print("YES")else: print("NO")# This code is contributed by Mohit Kumar |
C#
// 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 |
Javascript
<script>// JavaScript implementation of above approach function Reverse(s) { let charArray = s.split(""); reverse(charArray, 0, charArray.length - 1); return charArray.join(""); } // Function to check reflection function isReflectionEqual(s) { let symmetric = new Set(); // 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'); let n = s.length; // If any non-symmetric character is for (let i = 0; i < n; i++) // present, the answer is NO { if (symmetric.has(s[i]) == false) { return false; } } let rev = s; s = Reverse(s); // Check if the String is a palindrome if (rev==(s)) { return true; } else { return false; } } // Reverse the letters of the word function reverse(str,start,end) { // Temporary variable to store character let temp; while (start <= end) { // Swapping the first and last character temp = str[start]; str[start] = str[end]; str[end] = temp; start++; end--; } } // Driver code let s = "MYTYM"; if (isReflectionEqual(s)) { document.write("YES"); } else { document.write("NO"); } // This code is contributed by unknown2108</script> |
Output:
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
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