Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Print all possible ways to convert one string into another string | Edit-Distance

  • Difficulty Level : Hard
  • Last Updated : 23 Apr, 2021

Prerequisite: Dynamic Programming | Set 5 (Edit Distance) 
Given two strings str1 and str2, the task is to print the all possible ways to convert ‘str1’ into ‘str2’. 
Below are the operations that can be performed on “str1”: 
 

  1. Insert
  2. Remove
  3. Replace

All of the above operations are of equal cost. The task is to print all the various ways to convert ‘str1’ into ‘str2’ using the minimum number of edits (operations) required, where a “way” comprises of the series of all such operations required. 
Examples: 
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: str1 = “abcdef”, str2 = “axcdfdh” 
Output: 
Method 1: 
Add h 
Change f to d 
Change e to f 
Change b to x
Method 2: 
Change f to h 
Add d 
Change e to f 
Change b to x
Method 3: 
Change f to h 
Change e to d 
Add f 
Change b to x 
 



 

Approach for printing one possible way:

The approach for finding the minimum number of edits has been discussed in this post. To print one possible way, iterate from the bottom right corner of the DP matrix formed using Min-Edit Distance method. Check if the character pertaining to that element in both strings is equal or not. If it is, it means it needs no edit, and DP[i][j] was copied from DP[i-1][j-1]. 
 

If str1[i-1] == str2[j-1], proceed diagonally. 

Note that since the DP matrix contains one extra row and column at 0 indices, String indexes will be decreased by one. i.e. DP[i][j] corresponds to i-1 index of str1 and j-1 index of str2.
Now, if the characters were not equal, that means this matrix element DP[i][j] was obtained from the minimum of DP[i-1][j-1], DP[i][j-1] and DP[i-1][j], plus 1. Hence, check from where this element was from. 
 

1. If DP[i][j] == DP[i-1][j-1] + 1 
        It means the character was replaced from str1[i] to str2[j]. Proceed diagonally.
2. If DP[i][j] == DP[i][j-1] + 1
        It means the character was Added from str2[j]. Proceed left.
3. If DP[i][j] == DP[i-1][j] + 1
        It means the character str1[i] was deleted. Proceed up.

Once the end i.e., (i==0 or j==0 ) of either string is reached, converting of one string to other is done. We will have printed all the set of operations required. 
Below is the implementation of the above approach: 
 

C++




// C++ program to print one possible
// way of converting a string to another
#include <bits/stdc++.h>
using namespace std;
 
int DP[100][100];
 
// Function to print the steps
void printChanges(string s1, string s2,
                         int dp[][100])
{
    int i = s1.length();
    int j = s2.length();
 
    // check till the end
    while (i and j)
    {
        // if characters are same
        if (s1[i - 1] == s2[j - 1])
        {
            i--;
            j--;
        }
 
        // Replace
        else if (dp[i][j] == dp[i - 1][j - 1] + 1)
        {
            cout << "change " << s1[i - 1]
                 << " to " << s2[j - 1] << endl;
            i--;
            j--;
        }
 
        // Delete the character
        else if (dp[i][j] == dp[i - 1][j] + 1)
        {
            cout << "Delete " << s1[i - 1] << endl;
            i--;
        }
 
        // Add the character
        else if (dp[i][j] == dp[i][j - 1] + 1)
        {
            cout << "Add " << s2[j - 1] << endl;
            j--;
        }
    }
}
 
// Function to compute the DP matrix
void editDP(string s1, string s2)
{
    int l1 = s1.length();
    int l2 = s2.length();
    DP[l1 + 1][l2 + 1];
 
    // initialize by the maximum edits possible
    for (int i = 0; i <= l1; i++)
        DP[i][0] = i;
    for (int j = 0; j <= l2; j++)
        DP[0][j] = j;
 
    // Compute the DP matrix
    for (int i = 1; i <= l1; i++)
    {
        for (int j = 1; j <= l2; j++)
        {
            // if the characters are same
            // no changes required
            if (s1[i - 1] == s2[j - 1])
                DP[i][j] = DP[i - 1][j - 1];
            else
                // minimum of three operations possible
                DP[i][j] = min(min(DP[i - 1][j - 1],
                                   DP[i - 1][j]),
                                   DP[i][j - 1]) + 1;
        }
    }
 
    // print the steps
    printChanges(s1, s2, DP);
}
// Driver Code
int main()
{
    string s1 = "abcdef";
    string s2 = "axcdfdh";
 
    // calculate the DP matrix
    editDP(s1, s2);
 
    return 0;
}
 
// This code is contributed by
// sanjeev2552

Java




// Java program to print one possible
// way of converting a string to another
import java.util.*;
 
public class Edit_Distance {
    static int dp[][];
 
    // Function to print the steps
    static void printChanges(String s1, String s2)
    {
        int i = s1.length();
        int j = s2.length();
 
        // check till the end
        while (i != 0 && j != 0) {
 
            // if characters are same
            if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
                i--;
                j--;
            }
 
            // Replace
            else if (dp[i][j] == dp[i - 1][j - 1] + 1) {
                System.out.println("change " + s1.charAt(i - 1) + " to " + s2.charAt(j - 1));
                i--;
                j--;
            }
 
            // Delete the character
            else if (dp[i][j] == dp[i - 1][j] + 1) {
                System.out.println("Delete " + s1.charAt(i - 1));
                i--;
            }
 
            // Add the character
            else if (dp[i][j] == dp[i][j - 1] + 1) {
                System.out.println("Add " + s2.charAt(j - 1));
                j--;
            }
        }
    }
 
    // Function to compute the DP matrix
    static void editDP(String s1, String s2)
    {
        int l1 = s1.length();
        int l2 = s2.length();
        int[][] DP = new int[l1 + 1][l2 + 1];
 
        // initialize by the maximum edits possible
        for (int i = 0; i <= l1; i++)
            DP[i][0] = i;
        for (int j = 0; j <= l2; j++)
            DP[0][j] = j;
 
        // Compute the DP matrix
        for (int i = 1; i <= l1; i++) {
            for (int j = 1; j <= l2; j++) {
 
                // if the characters are same
                // no changes required
                if (s1.charAt(i - 1) == s2.charAt(j - 1))
                    DP[i][j] = DP[i - 1][j - 1];
                else {
 
                    // minimum of three operations possible
                    DP[i][j] = min(DP[i - 1][j - 1],
                                   DP[i - 1][j], DP[i][j - 1])
                               + 1;
                }
            }
        }
 
        // initialize to global array
        dp = DP;
    }
 
    // Function to find the minimum of three
    static int min(int a, int b, int c)
    {
        int z = Math.min(a, b);
        return Math.min(z, c);
    }
 
    // Driver Code
    public static void main(String[] args) throws Exception
    {
        String s1 = "abcdef";
        String s2 = "axcdfdh";
 
        // calculate the DP matrix
        editDP(s1, s2);
 
        // print the steps
        printChanges(s1, s2);
    }
}

Python3




# Python3 program to print one possible
# way of converting a string to another
 
# Function to print the steps
def printChanges(s1, s2, dp):
     
    i = len(s1)
    j = len(s2)
     
   # Check till the end
    while(i > 0 and j > 0):
         
        # If characters are same
        if s1[i - 1] == s2[j - 1]:
            i -= 1
            j -= 1
             
        # Replace
        elif dp[i][j] == dp[i - 1][j - 1] + 1:
            print("change", s1[i - 1],
                      "to", s2[j - 1])
            j -= 1
            i -= 1
             
        # Delete
        elif dp[i][j] == dp[i - 1][j] + 1:
            print("Delete", s1[i - 1])
            i -= 1
             
        # Add
        elif dp[i][j] == dp[i][j - 1] + 1:
            print("Add", s2[j - 1])
            j -= 1
             
# Function to compute the DP matrix
def editDP(s1, s2):
     
    len1 = len(s1)
    len2 = len(s2)
    dp = [[0 for i in range(len2 + 1)]
             for j in range(len1 + 1)]
     
    # Initialize by the maximum edits possible
    for i in range(len1 + 1):
        dp[i][0] = i
    for j in range(len2 + 1):
        dp[0][j] = j
     
    # Compute the DP Matrix
    for i in range(1, len1 + 1):
        for j in range(1, len2 + 1):
             
            # If the characters are same
            # no changes required
            if s2[j - 1] == s1[i - 1]:
                dp[i][j] = dp[i - 1][j - 1]
                 
            # Minimum of three operations possible
            else:
                dp[i][j] = 1 + min(dp[i][j - 1],
                                   dp[i - 1][j - 1],
                                   dp[i - 1][j])
                                    
    # Print the steps
    printChanges(s1, s2, dp)
 
# Driver Code
s1 = "abcdef"
s2 = "axcdfdh"
 
# Compute the DP Matrix
editDP(s1, s2)
 
# This code is contributed by Pranav S

C#




// C# program to print one possible
// way of converting a string to another
using System;
 
public class Edit_Distance
{
    static int [,]dp;
 
    // Function to print the steps
    static void printChanges(String s1, String s2)
    {
        int i = s1.Length;
        int j = s2.Length;
 
        // check till the end
        while (i != 0 && j != 0)
        {
 
            // if characters are same
            if (s1[i - 1] == s2[j - 1])
            {
                i--;
                j--;
            }
 
            // Replace
            else if (dp[i, j] == dp[i - 1, j - 1] + 1)
            {
                Console.WriteLine("change " + s1[i - 1] + " to " + s2[j - 1]);
                i--;
                j--;
            }
 
            // Delete the character
            else if (dp[i, j] == dp[i - 1, j] + 1)
            {
                Console.WriteLine("Delete " + s1[i - 1]);
                i--;
            }
 
            // Add the character
            else if (dp[i, j] == dp[i, j - 1] + 1)
            {
                Console.WriteLine("Add " + s2[j - 1]);
                j--;
            }
        }
    }
 
    // Function to compute the DP matrix
    static void editDP(String s1, String s2)
    {
        int l1 = s1.Length;
        int l2 = s2.Length;
        int[,] DP = new int[l1 + 1, l2 + 1];
 
        // initialize by the maximum edits possible
        for (int i = 0; i <= l1; i++)
            DP[i, 0] = i;
        for (int j = 0; j <= l2; j++)
            DP[0, j] = j;
 
        // Compute the DP matrix
        for (int i = 1; i <= l1; i++)
        {
            for (int j = 1; j <= l2; j++)
            {
 
                // if the characters are same
                // no changes required
                if (s1[i - 1] == s2[j - 1])
                    DP[i, j] = DP[i - 1, j - 1];
                else
                {
 
                    // minimu of three operations possible
                    DP[i, j] = min(DP[i - 1, j - 1],
                                DP[i - 1, j], DP[i, j - 1])
                            + 1;
                }
            }
        }
 
        // initialize to global array
        dp = DP;
    }
 
    // Function to find the minimum of three
    static int min(int a, int b, int c)
    {
        int z = Math.Min(a, b);
        return Math.Min(z, c);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        String s1 = "abcdef";
        String s2 = "axcdfdh";
 
        // calculate the DP matrix
        editDP(s1, s2);
 
        // print the steps
        printChanges(s1, s2);
    }
}
 
// This code is contributed by PrinciRaj1992
Output: 
change f to h
change e to d
Add f
change b to x

 

Approach to print all possible ways:

Create a collection of strings that will store the operations required. This collection can be a vector of strings in C++ or a List of strings in Java. Add operations just like printing them before to this collection. Then create a collection of these collections which will store the multiple methods (sets of string operations). 
Else-if was used earlier to check from where we derived the DP[i][j] from. Now, check all If’s to see if there were more than 1 ways you could obtain the element. If there was, we create a new collection from before, remove the last operation, add this new operation and initiate another instance of this function with this new list. In this manner, add new lists whenever there was a new method to change str1 to str2, getting a new method every time.
On reaching the end of either string, add this list to the collection of lists, thus completing the set of all possible operations, and add them. 
Below is the implementation of the above approach: 
 

Java




// Java program to print all the possible
// steps to change a string to another
import java.util.ArrayList;
 
public class Edit_Distance {
    static int dp[][];
 
    // create List of lists that will store all sets of operations
    static ArrayList<ArrayList<String> > arrs =
                              new ArrayList<ArrayList<String> >();
 
    // Function to print all ways
    static void printAllChanges(String s1,
                                String s2, ArrayList<String> changes)
    {
 
        int i = s1.length();
        int j = s2.length();
 
        // Iterate till end
        while (true) {
 
            if (i == 0 || j == 0) {
 
                // Add this list to our List of lists.
                arrs.add(changes);
                break;
            }
 
            // If same
            if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
                i--;
                j--;
            }
 
            else {
                boolean if1 = false, if2 = false;
 
                // Replace
                if (dp[i][j] == dp[i - 1][j - 1] + 1) {
 
                    // Add this step
                    changes.add("Change " + s1.charAt(i - 1)
                                + " to " + s2.charAt(j - 1));
                    i--;
                    j--;
 
                    // note whether this 'if' was true.
                    if1 = true;
                }
 
                // Delete
                if (dp[i][j] == dp[i - 1][j] + 1) {
                    if (if1 == false) {
                        changes.add("Delete " + s1.charAt(i - 1));
                        i--;
                    }
                    else {
                        // If the previous method was true,
                        // create a new list as a copy of previous.
                        ArrayList<String> changes2 = new ArrayList<String>();
                        changes2.addAll(changes);
 
                        // Remove last operation
                        changes2.remove(changes.size() - 1);
 
                        // Add this new operation
                        changes2.add("Delete " + s1.charAt(i));
 
                        // initiate new new instance of this
                        // function with remaining substrings
                        printAllChanges(s1.substring(0, i),
                                        s2.substring(0, j + 1), changes2);
                    }
 
                    if2 = true;
                }
 
                // Add character step
                if (dp[i][j] == dp[i][j - 1] + 1) {
                    if (if1 == false && if2 == false) {
                        changes.add("Add " + s2.charAt(j - 1));
                        j--;
                    }
                    else {
 
                        // Add steps
                        ArrayList<String> changes2 = new ArrayList<String>();
                        changes2.addAll(changes);
                        changes2.remove(changes.size() - 1);
                        changes2.add("Add " + s2.charAt(j));
 
                        // Recursively call for the next steps
                        printAllChanges(s1.substring(0, i + 1),
                                        s2.substring(0, j), changes2);
                    }
                }
            }
        }
    }
 
    // Function to compute the DP matrix
    static void editDP(String s1, String s2)
    {
        int l1 = s1.length();
        int l2 = s2.length();
        int[][] DP = new int[l1 + 1][l2 + 1];
 
        // initialize by the maximum edits possible
        for (int i = 0; i <= l1; i++)
            DP[i][0] = i;
        for (int j = 0; j <= l2; j++)
            DP[0][j] = j;
 
        // Compute the DP matrix
        for (int i = 1; i <= l1; i++) {
            for (int j = 1; j <= l2; j++) {
 
                // if the characters are same
                // no changes required
                if (s1.charAt(i - 1) == s2.charAt(j - 1))
                    DP[i][j] = DP[i - 1][j - 1];
                else {
 
                    // minimum of three operations possible
                    DP[i][j] = min(DP[i - 1][j - 1],
                                   DP[i - 1][j], DP[i][j - 1])
                               + 1;
                }
            }
        }
 
        // initialize to global array
        dp = DP;
    }
 
    // Function to find the minimum of three
    static int min(int a, int b, int c)
    {
        int z = Math.min(a, b);
        return Math.min(z, c);
    }
    static void printWays(String s1, String s2,
                          ArrayList<String> changes)
    {
 
        // Function to print all the ways
        printAllChanges(s1, s2, new ArrayList<String>());
 
        int i = 1;
 
        // print all the possible ways
        for (ArrayList<String> ar : arrs) {
            System.out.println("\nMethod " + i++ + " : \n");
            for (String s : ar) {
                System.out.println(s);
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args) throws Exception
    {
        String s1 = "abcdef";
        String s2 = "axcdfdh";
 
        // calculate the DP matrix
        editDP(s1, s2);
 
        // Function to print all ways
        printWays(s1, s2, new ArrayList<String>());
    }
}
Output: 
Method 1 : 

Add h
Change f to d
Change e to f
Change b to x

Method 2 : 

Change f to h
Add d
Change e to f
Change b to x

Method 3 : 

Change f to h
Change e to d
Add f
Change b to x

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :