Compruebe si la concatenación de substrings divididas de dos strings dadas forma un palíndromo o no

Dadas dos strings a y b de la misma longitud, la tarea es verificar si dividir ambas strings y concatenar sus substrings opuestas, es decir, concatenar la substring izquierda de a con la substring derecha de b o concatenar la substring izquierda de b con la substring derecha de a , forma un palíndromo o no . Si se encuentra que es cierto, escriba «Sí» . De lo contrario, escriba “No” .

Nota: una de las substrings divididas puede estar vacía.

Ejemplos:

Entrada: a = “x”, b = “y”
Salida: Sí
Explicación:
Divida ambas strings en el índice 0. 
Substring izquierda de a (aLeft) = ” “, Substring derecha de a (aRight) = “x” Substring
izquierda de b (bIzquierda) = ” “, Substring derecha de b (bDerecha) = “y”
Dado que aIzquierda + bDerecha = ” ” + “y” = “y”, que es un palíndromo al igual que bIzquierda + aDerecha= ” ” + “x” = “x” también es un palíndromo, escribe Sí.

Entrada: a = “ulacfd”, b = “jizalu”
Salida: Verdadero
Explicación: 
Dividir ambas strings en el índice 3:
Substring izquierda de a (aLeft) = “ula”, Substring derecha de a (aRight) = “cfd”
, Substring izquierda de b (bLeft) = “jiz”, Substring derecha de b (bRight) = “alu”
Dado que aleft + bright = “ula” + “alu” = “ulaalu”, que es un palíndromo, imprime Sí.

Enfoque: La idea es utilizar la técnica Two Pointer para resolver este problema. Siga los pasos a continuación para resolver el problema:

1. Coloque un puntero i en el ^índice 0 de a y «j» en el último índice de b .
2. Itere sobre los caracteres de la string y verifique si a[i] == b[j] , luego incremente i y disminuya j .
3. De lo contrario, simplemente rompa el ciclo ya que no es una secuencia de tipo palíndromo.
4. Concatene aLeft y bRight en una variable de string xa y aRight y bLeft en otra variable de string xb .
5. Compruebe si alguna de las dos strings es un palíndromo o no. Si es cierto, escriba “Sí” . De lo contrario, escriba “No” .

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if concatenating
// opposite substrings after splitting
// two given strings forms a palindrome
// or not
bool checkSplitting(string a, string b)
{
     
    // Length of the string
    int len = a.length();
    int i = 0, j = len - 1;
     
    // Iterate through the strings
    while (i < len)
    {
         
        // If not a palindrome sequence
        if (a[i] != b[j])
        {
            break;
        }
        i += 1;
        j -= 1;
 
        // Concatenate left substring of a
        // and right substring of b in xa
        // Concatenate right substring of a
        // and left substring of b in xb
        string xa = a.substr(i, j + 1);
        string xb = b.substr(i, j + 1);
 
        // Check if either of the two concatenated
        // strings is a palindrome or not
        if (xa == string(xa.rbegin(), xa.rend()) ||
            xb == string(xb.rbegin(), xb.rend()))
            return true;
    }
}
 
// Function to check if concatenation of splitted
// substrings of two given strings forms a palindrome
void isSplitPossible(string a, string b)
{
    if (checkSplitting(a, b) == true)
    {
        cout << "Yes";
    }
    else if (checkSplitting(b, a) == true)
    {
        cout << "Yes";
    }
    else
    {
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    string a = "ulacfd", b = "jizalu";
 
    // Function Call
    isSplitPossible(a, b);
     
    return 0;
}
 
// This code is contributed by pushpendrayadav1057

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if concatenating
// opposite subStrings after splitting
// two given Strings forms a palindrome
// or not
static boolean checkSplitting(String a, String b)
{
     
    // Length of the String
    int len = a.length();
    int i = 0, j = len - 1;
     
    // Iterate through the Strings
    while (i < len)
    {
         
        // If not a palindrome sequence
        if (a.charAt(i) != b.charAt(j))
        {
            break;
        }
        i += 1;
        j -= 1;
 
        // Concatenate left subString of a
        // and right subString of b in xa
        // Concatenate right subString of a
        // and left subString of b in xb
        String xa = a.substring(i, j + 1);
        String xb = b.substring(i, j + 1);
 
        // Check if either of the two concatenated
        // Strings is a palindrome or not
        if (xa.equals(reverse(xa))||xb.equals(reverse(xb)))
            return true;
    }
    return false;
}
 
// Function to check if concatenation of splitted
// subStrings of two given Strings forms a palindrome
static void isSplitPossible(String a, String b)
{
    if (checkSplitting(a, b) == true)
    {
        System.out.print("Yes");
    }
    else if (checkSplitting(b, a) == true)
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
static String reverse(String input) {
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
}
   
// Driver Code
public static void main(String[] args)
{
    String a = "ulacfd", b = "jizalu";
 
    // Function Call
    isSplitPossible(a, b);   
}
}
 
// This code is contributed by 29AjayKumar

# Python3 program for the above approach
 
# Function to check if concatenating
# opposite substrings after splitting
# two given strings forms a palindrome or not
def checkSplitting(a, b, n):
    i, j = 0, n - 1
 
    # Iterate through the strings
    while(i < n):
 
         
        # If not a palindrome sequence
        if(a[i] != b[j]):
            break
        i += 1
        j -= 1
 
        # Concatenate left substring of a
        # and right substring of b in xa
        # Concatenate right substring of a
        # and left substring of b in xb
        xa = a[i:j + 1]
        xb = b[i:j + 1]
 
        # Check if either of the two concatenated
        # strings is a palindrome or not
        if(xa == xa[::-1] or xb == xb[::-1]):
            return True
 
# Function to check if concatenation of splitted
# substrings of two given strings forms a palindrome
def isSplitPossible(a, b):
    if checkSplitting(a, b, len(a)) == True:
        print("Yes")
         
    elif checkSplitting(b, a, len(a)) == True:
        print("Yes")
         
    else:
        print("No")
 
 
# Given string a and b
a = "ulacfd"
b = "jizalu"
 
# Function Call
isSplitPossible(a, b)

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if concatenating
// opposite subStrings after splitting
// two given Strings forms a palindrome
// or not
static bool checkSplitting(String a, String b)
{
     
    // Length of the String
    int len = a.Length;
    int i = 0, j = len - 1;
     
    // Iterate through the Strings
    while (i < len)
    {
         
        // If not a palindrome sequence
        if (a[i] != b[j])
        {
            break;
        }
        i += 1;
        j -= 1;
 
        // Concatenate left subString of a
        // and right subString of b in xa
        // Concatenate right subString of a
        // and left subString of b in xb
        String xa = a.Substring(i, j + 1 - i);
        String xb = b.Substring(i, j + 1 - i);
 
        // Check if either of the two concatenated
        // Strings is a palindrome or not
        if (xa.Equals(reverse(xa)) ||
            xb.Equals(reverse(xb)))
            return true;
    }
    return false;
}
 
// Function to check if concatenation of splitted
// subStrings of two given Strings forms a palindrome
static void isSplitPossible(String a, String b)
{
    if (checkSplitting(a, b) == true)
    {
        Console.Write("Yes");
    }
    else if (checkSplitting(b, a) == true)
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
static String reverse(String input)
{
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--)
    {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
}
   
// Driver Code
public static void Main(String[] args)
{
    String a = "ulacfd", b = "jizalu";
 
    // Function Call
    isSplitPossible(a, b);   
}
}
 
// This code is contributed by 29AjayKumar

<script>
 
// JavaScript program for the above approach
 
// Function to check if the string is palindrome or not
 
function checkPalindrome(str) {
 
    // find the length of a string
    var len = str.length;
 
    // loop through half of the string
    for (var i = 0; i < parseInt(len / 2); i++) {
 
        // check if first and last string are same
        if (str[i] !== str[len - 1 - i]) {
            return false;
        }
    }
    return true;
}
 
// Function to check if concatenating
// opposite substrings after splitting
// two given strings forms a palindrome
// or not
function checkSplitting(a, b)
{
     
    // Length of the string
    var len = a.length;
    var i = 0, j = len - 1;
     
    // Iterate through the strings
    while (i < len)
    {
         
        // If not a palindrome sequence
        if (a[i] != b[j])
        {
            break;
        }
        i += 1;
        j -= 1;
 
        // Concatenate left substring of a
        // and right substring of b in xa
        // Concatenate right substring of a
        // and left substring of b in xb
        var xa = a.substring(i, j + 1);
        var xb = b.substring(i, j + 1);
 
        // Check if either of the two concatenated
        // strings is a palindrome or not
        if (checkPalindrome(xa)==true || checkPalindrome(xb)==true)
            return true;
    }
}
 
// Function to check if concatenation of splitted
// substrings of two given strings forms a palindrome
function isSplitPossible(a, b)
{
    if (checkSplitting(a, b) == true)
    {
       document.write( "Yes");
    }
    else if (checkSplitting(b, a) == true)
    {
        document.write("Yes");
    }
    else
    {
        document.write( "No");
    }
}
 
var a = "ulacfd", b = "jizalu";
 
    // Function Call
    isSplitPossible(a, b);
 
// This code is contributed by SoumikMondal
 
</script>

Yes

Complejidad temporal: O(N)
Espacio auxiliar: O(1)