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Check if one string can be converted to another
  • Difficulty Level : Hard
  • Last Updated : 30 Jan, 2020

Given two strings str and str1, the task is to check whether one string can be converted to other by using the following operation:

  • Convert all the presence of a character by a different character.

For example, if str = “abacd” and operation is to change character ‘a’ to ‘k’, then the resultant str = “kbkcd”

Examples:

Input: str = “abbcaa”; str1 = “bccdbb”
Output: Yes
Explanation: The mappings of the characters are:
c –> d
b –> c
a –> b

Input: str = “abbc”; str1 = “bcca”
Output: No
Explanation: The mapping of characters are:
a –> b
b –> c
c –> a
Here, due to the presence of a cycle, a specific order cannot be found.



Approach:

  • According to the given operation, every unique character should map to a unique character may be same or different.
  • This can easily be checked by a Hashmap.
  • However, this fails in cases where there is a cycle in mapping and a specific order cannot be determined.
  • One example of such case is Example 2 above.
  • Therefore, for mapping, the first and final characters are stored as edges in a hashmap.
  • For finding cycle with the edges, these edges are mapped one by one to a parent and are checked for cycle using Union and Find Algorithm.

Below is the implementation of the above approach.

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

Python




# 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
Output:
Yes

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