Dadas dos strings str y str1 , la tarea es verificar si una string se puede convertir en otra usando la siguiente operación:
- Convertir toda la presencia de un personaje por un personaje diferente.
Por ejemplo, si str = “abacd” y la operación es cambiar el carácter ‘a’ a ‘k’, entonces el resultado str = “kbkcd”
Ejemplos:
Entrada: str = “abbcaa”; str1 = “bccdbb”
Salida: Sí
Explicación: Las asignaciones de los caracteres son:
c –> d
b –> c
a –> b
Entrada: str = “abbc”; str1 = “bcca”
Salida: No
Explicación: El mapeo de caracteres es:
a –> b
b –> c
c –> a
Aquí, debido a la presencia de un ciclo, no se puede encontrar un orden específico.
Acercarse:
- De acuerdo con la operación dada, cada carácter único debe asignarse a un carácter único que puede ser igual o diferente.
- Esto se puede verificar fácilmente con un Hashmap .
- Sin embargo, esto falla en los casos en que hay un ciclo en el mapeo y no se puede determinar un orden específico.
- Un ejemplo de tal caso es el Ejemplo 2 anterior.
- Por lo tanto, para el mapeo, los caracteres primero y final se almacenan como bordes en un mapa hash.
- Para encontrar el ciclo con los bordes, estos bordes se asignan uno por uno a un padre y se verifica el ciclo utilizando Union and Find Algorithm .
A continuación se muestra la implementación del enfoque anterior.
CPP
// C++ implementation of the above approach.
#include <bits/stdc++.h>
using namespace std;
int parent[26];
// Function for find
// from Disjoint set algorithm
int find(int x)
{
if (x != parent[x])
return parent[x] = find(parent[x]);
return x;
}
// Function for the union
// from Disjoint set algorithm
void join(int x, int y)
{
int px = find(x);
int pz = find(y);
if (px != pz) {
parent[pz] = px;
}
}
// Function to check if one string
// can be converted to another.
bool convertible(string s1, string s2)
{
// All the characters are checked whether
// it's either not replaced or replaced
// by a similar character using a map.
map<int, int> mp;
for (int i = 0; i < s1.size(); i++) {
if (mp.find(s1[i] - 'a') == mp.end()) {
mp[s1[i] - 'a'] = s2[i] - 'a';
}
else {
if (mp[s1[i] - 'a'] != s2[i] - 'a')
return false;
}
}
// To check if there are cycles.
// If yes, then they are not convertible.
// Else, they are convertible.
for (auto it : mp) {
if (it.first == it.second)
continue;
else {
if (find(it.first) == find(it.second))
return false;
else
join(it.first, it.second);
}
}
return true;
}
// Function to initialize parent array
// for union and find algorithm.
void initialize()
{
for (int i = 0; i < 26; i++) {
parent[i] = i;
}
}
// Driver code
int main()
{
// Your C++ Code
string s1, s2;
s1 = "abbcaa";
s2 = "bccdbb";
initialize();
if (convertible(s1, s2))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
Java
// Java implementation of the above approach.
import java.util.*;
class GFG
{
static int []parent = new int[26];
// Function for find
// from Disjoint set algorithm
static int find(int x)
{
if (x != parent[x])
return parent[x] = find(parent[x]);
return x;
}
// Function for the union
// from Disjoint set algorithm
static void join(int x, int y)
{
int px = find(x);
int pz = find(y);
if (px != pz)
{
parent[pz] = px;
}
}
// Function to check if one String
// can be converted to another.
static boolean convertible(String s1, String s2)
{
// All the characters are checked whether
// it's either not replaced or replaced
// by a similar character using a map.
HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
for (int i = 0; i < s1.length(); i++)
{
if (!mp.containsKey(s1.charAt(i) - 'a'))
{
mp.put(s1.charAt(i) - 'a', s2.charAt(i) - 'a');
}
else
{
if (mp.get(s1.charAt(i) - 'a') != s2.charAt(i) - 'a')
return false;
}
}
// To check if there are cycles.
// If yes, then they are not convertible.
// Else, they are convertible.
for (Map.Entry<Integer, Integer> it : mp.entrySet())
{
if (it.getKey() == it.getValue())
continue;
else
{
if (find(it.getKey()) == find(it.getValue()))
return false;
else
join(it.getKey(), it.getValue());
}
}
return true;
}
// Function to initialize parent array
// for union and find algorithm.
static void initialize()
{
for (int i = 0; i < 26; i++)
{
parent[i] = i;
}
}
// Driver code
public static void main(String[] args)
{
String s1, s2;
s1 = "abbcaa";
s2 = "bccdbb";
initialize();
if (convertible(s1, s2))
System.out.print("Yes" + "\n");
else
System.out.print("No" + "\n");
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the above approach.
parent = [0] * 256
# Function for find
# from Disjoset algorithm
def find(x):
if (x != parent[x]):
parent[x] = find(parent[x])
return parent[x]
return x
# Function for the union
# from Disjoset algorithm
def join(x, y):
px = find(x)
pz = find(y)
if (px != pz):
parent[pz] = px
# Function to check if one string
# can be converted to another.
def convertible(s1, s2):
# All the characters are checked whether
# it's either not replaced or replaced
# by a similar character using a map.
mp = dict()
for i in range(len(s1)):
if (s1[i] in mp):
mp[s1[i]] = s2[i]
else:
if s1[i] in mp and mp[s1[i]] != s2[i]:
return False
# To check if there are cycles.
# If yes, then they are not convertible.
# Else, they are convertible.
for it in mp:
if (it == mp[it]):
continue
else :
if (find(ord(it)) == find(ord(it))):
return False
else:
join(ord(it), ord(it))
return True
# Function to initialize parent array
# for union and find algorithm.
def initialize():
for i in range(256):
parent[i] = i
# Driver code
s1 = "abbcaa"
s2 = "bccdbb"
initialize()
if (convertible(s1, s2)):
print("Yes")
else:
print("No")
# This code is contributed by mohit kumar 29
C#
// C# implementation of the above approach.
using System;
using System.Collections.Generic;
class GFG
{
static int []parent = new int[26];
// Function for find
// from Disjoint set algorithm
static int find(int x)
{
if (x != parent[x])
return parent[x] = find(parent[x]);
return x;
}
// Function for the union
// from Disjoint set algorithm
static void join(int x, int y)
{
int px = find(x);
int pz = find(y);
if (px != pz)
{
parent[pz] = px;
}
}
// Function to check if one String
// can be converted to another.
static bool convertible(String s1, String s2)
{
// All the characters are checked whether
// it's either not replaced or replaced
// by a similar character using a map.
Dictionary<int,int> mp = new Dictionary<int,int>();
for (int i = 0; i < s1.Length; i++)
{
if (!mp.ContainsKey(s1[i] - 'a'))
{
mp.Add(s1[i] - 'a', s2[i] - 'a');
}
else
{
if (mp[s1[i] - 'a'] != s2[i] - 'a')
return false;
}
}
// To check if there are cycles.
// If yes, then they are not convertible.
// Else, they are convertible.
foreach(KeyValuePair<int, int> it in mp)
{
if (it.Key == it.Value)
continue;
else
{
if (find(it.Key) == find(it.Value))
return false;
else
join(it.Key, it.Value);
}
}
return true;
}
// Function to initialize parent array
// for union and find algorithm.
static void initialize()
{
for (int i = 0; i < 26; i++)
{
parent[i] = i;
}
}
// Driver code
public static void Main(String[] args)
{
String s1, s2;
s1 = "abbcaa";
s2 = "bccdbb";
initialize();
if (convertible(s1, s2))
Console.Write("Yes" + "\n");
else
Console.Write("No" + "\n");
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// JavaScript implementation of the above approach.
var parent = new Array(26).fill(0);
// Function for find
// from Disjoint set algorithm
function find(x) {
if (x !== parent[x]) return (parent[x] = find(parent[x]));
return x;
}
// Function for the union
// from Disjoint set algorithm
function join(x, y) {
var px = find(x);
var pz = find(y);
if (px !== pz) {
parent[pz] = px;
}
}
// Function to check if one String
// can be converted to another.
function convertible(s1, s2) {
// All the characters are checked whether
// it's either not replaced or replaced
// by a similar character using a map.
var mp = {};
for (var i = 0; i < s1.length; i++) {
if (!mp.hasOwnProperty(s1[i].charCodeAt(0) -
"a".charCodeAt(0)))
{
mp[s1[i].charCodeAt(0) - "a".charCodeAt(0)] =
s2[i].charCodeAt(0) - "a".charCodeAt(0);
} else {
if (
mp[s1[i].charCodeAt(0) - "a".charCodeAt(0)] !==
s2[i].charCodeAt(0) - "a".charCodeAt(0)
)
return false;
}
}
// To check if there are cycles.
// If yes, then they are not convertible.
// Else, they are convertible.
for (const [key, value] of Object.entries(mp)) {
if (key === value) continue;
else {
if (find(key) == find(value)) return false;
else join(key, value);
}
}
return true;
}
// Function to initialize parent array
// for union and find algorithm.
function initialize() {
for (var i = 0; i < 26; i++) {
parent[i] = i;
}
}
// Driver code
var s1, s2;
s1 = "abbcaa";
s2 = "bccdbb";
initialize();
if (convertible(s1, s2)) document.write("Yes" + "<br>");
else document.write("No" + "<br>");
</script>
Producción:
Yes
Complejidad de tiempo: O(N * logN), donde N es la longitud de la string s1.
Espacio Auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por internprep y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA