Dada una string S que contiene solo caracteres ingleses en mayúsculas. La tarea es encontrar si S es igual a su reflejo en un espejo.
Ejemplos:
Input: str = "AMA" Output: YES AMA is same as its reflection in the mirror. Input: str = "ZXZ" Output: NO
Enfoque: La cuerda obviamente tiene que ser un palíndromo, pero eso solo no es suficiente. Todos los caracteres de la string deben ser simétricos para que su reflejo también sea el mismo. Los caracteres simétricos son AHIMOTUVWXY .
- Almacene los caracteres simétricos en un conjunto desordenado.
- Recorra la string y verifique si hay algún carácter no simétrico presente en la string. Si es así, devuelve falso.
- De lo contrario, compruebe si la string es palíndromo o no. Si la string también es palíndromo, devuelve verdadero; de lo contrario, devuelve falso.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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";
}
Java
// 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
Python3
# 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
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>
Producción:
YES
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por Abdullah Aslam y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA