Open In App

Check if one string can be converted to another

Last Updated : 12 Jul, 2025
Comments
Improve
Suggest changes
Like Article
Like
Report

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
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>

Output: 
Yes

 

Time Complexity: O(N * logN), where N is the length of string s1.
Auxiliary Space: O(N)


Next Article
Analysis of Algorithms

I
internprep
Improve
Article Tags :
Practice Tags :

Similar Reads

We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox