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

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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”:

Insert
Remove
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:

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 and j==0 ) of both strings 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

import java.util.*;
 
class FollowUp_PrintOneWayToConvert {
 
    public static void main(String[] args) {
        var o = new FollowUp_PrintOneWayToConvert();
        o.minDistance("abcdef", "axcdfdh");
        for(var r : o.dp) System.out.println(Arrays.toString(r));
        o.printChanges("abcdef", "axcdfdh");
 
        // Only add: GeeksForGeeks impl won't print added chars
        o = new FollowUp_PrintOneWayToConvert();
        o.minDistance("", "abc");
        for(var r : o.dp) System.out.println(Arrays.toString(r));
        o.printChanges("", "abc");
 
        // Only remove: GeeksForGeeks impl won't print removed chars
        o = new FollowUp_PrintOneWayToConvert();
        o.minDistance("abc", "");
        for(var r : o.dp) System.out.println(Arrays.toString(r));
        o.printChanges("abc", "");
    }
 
    int[][] dp;
 
    int minDistance(String s, String targ) {
        int m = s.length(), n = targ.length();
        dp = new int[m + 1][n + 1];
 
        // initialize by the maximum edits possible
        for (int i = 0; i <= m; i++) dp[i][0] = i;
        for (int j = 0; j <= n; j++) dp[0][j] = j;
 
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                if (s.charAt(i - 1) == targ.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1];
                else dp[i][j] = 1 + Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1]));
 
        return dp[m][n];
    }
 
    void printChanges(String s, String t) {
        int i = s.length(), j = t.length();
        while (i > 0 && j > 0) {
            // Match(if characters same):
            if (s.charAt(i - 1) == t.charAt(j - 1)) { i--; j--; }
            // No match:
            // Replace
            else if (dp[i][j] == dp[i - 1][j - 1] + 1) { System.out.println("Change " + s.charAt(i - 1) + " to " + t.charAt(j - 1)); i--; j--; }
            // Delete the character
            else if (dp[i][j] == dp[i - 1][j] + 1) { System.out.println("Delete " + s.charAt(i - 1)); i--; }
            // Add character
            else if (dp[i][j] == dp[i][j - 1] + 1) { System.out.println("Add " + t.charAt(j - 1)); j--; }
        } // i or j == 0, so we need to only remove or add chars
        while(i > 0) System.out.println("Delete " + s.charAt(--i)); // 't' matched but some 's' chars left not removed - rm all 's' chars
        while(j > 0) System.out.println("Add " + t.charAt(--j)); // 's' chars finished but not all 't' chars matched - add missing
    }
}

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 GFG
{
    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
                {
 
                    // 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)
    {
        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

Javascript

// Javascript code for the above approach
let DP = Array(100).fill().map(() => Array(100).fill(0));
 
function printChanges(s1, s2, dp) {
    let i = s1.length;
    let j = s2.length;
 
    // check till the end
    while (i && 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) {
            console.log(`change ${s1[i - 1]} to ${s2[j - 1]<br>}`);
            i--;
            j--;
        }
        // Delete the character
        else if (dp[i][j] === dp[i - 1][j] + 1) {
            console.log(`Delete ${s1[i - 1]}<br>`);
            i--;
        }
        // Add the character
        else if (dp[i][j] === dp[i][j - 1] + 1) {
            console.log(`Add ${s2[j - 1]}<br>`);
            j--;
        }
    }
}
 
function editDP(s1, s2) {
    let l1 = s1.length;
    let l2 = s2.length;
 
    // initialize by the maximum edits possible
    for (let i = 0; i <= l1; i++) {
        DP[i][0] = i;
    }
    for (let j = 0; j <= l2; j++) {
        DP[0][j] = j;
    }
 
    // Compute the DP matrix
    for (let i = 1; i <= l1; i++) {
        for (let 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] = Math.min(Math.min(DP[i - 1][j - 1], DP[i - 1][j]), DP[i][j - 1]) + 1;
            }
        }
    }
 
    // print the steps
    printChanges(s1, s2, DP);
}
 
// Driver Code
let s1 = "abcdef";
let s2 = "axcdfdh";
 
// calculate the DP matrix
editDP(s1, s2);
 
// This code is contributed by lokeshpotta20.

Output

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

The time complexity of the given implementation of the edit distance algorithm is O(m*n), where m and n are the lengths of the input strings s1 and s2, respectively. This is because the DP matrix is computed in a nested loop that iterates over each position in the matrix exactly once.

The space complexity of the implementation is O(m*n) as well, since the DP matrix is stored in a two-dimensional array of size (m+1) x (n+1). This is because the algorithm needs to store the minimum edit distance values for each prefix of s1 and s2, including the empty string prefixes.

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:

C++

using System;
using System.Collections.Generic;
 
class Program
{
    // Create List of lists that will store all sets of operations
    static List<List<string>> arrs = new List<List<string>>();
    static int[,] dp = new int[1001, 1001];
 
    // Function to print all ways
    static void PrintAllChanges(string s1, string s2, List<string> changes)
    {
        int i = s1.Length;
        int j = s2.Length;
 
        // Iterate till the end
        while (true)
        {
            if (i == 0 || j == 0)
            {
                // Add this list to our List of lists.
                arrs.Add(new List<string>(changes));
                break;
            }
 
            // If same
            if (s1[i - 1] == s2[j - 1])
            {
                i--;
                j--;
            }
            else
            {
                bool if1 = false, if2 = false;
 
                // Replace
                if (dp[i, j] == dp[i - 1, j - 1] + 1)
                {
                    // Add this step
                    changes.Add("Change " + s1[i - 1] + " to " + s2[j - 1]);
                    i--;
                    j--;
                    if1 = true;
                }
 
                // Delete
                if (dp[i, j] == dp[i - 1, j] + 1)
                {
                    if (!if1)
                    {
                        changes.Add("Delete " + s1[i - 1]);
                        i--;
                    }
                    else
                    {
                        // If the previous method was true,
                        // create a new list as a copy of the previous.
                        List<string> changes2 = new List<string>(changes);
                        changes2.RemoveAt(changes2.Count - 1);
 
                        // Add this new operation
                        changes2.Add("Delete " + s1[i]);
                        // Initiate a 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 && !if2)
                    {
                        changes.Add("Add " + s2[j - 1]);
                        j--;
                    }
                    else
                    {
                        // Add steps
                        List<string> changes2 = new List<string>(changes);
                        changes2.RemoveAt(changes2.Count - 1);
                        changes2.Add("Add " + s2[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;
 
        // 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 the 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] = Math.Min(dp[i - 1, j - 1], Math.Min(dp[i - 1, j], dp[i, j - 1])) + 1;
            }
        }
    }
 
    static void PrintWays(string s1, string s2, List<string> changes)
    {
        // Function to print all the ways
        PrintAllChanges(s1, s2, new List<string>());
        int i = 1;
        // Print all the possible ways
        foreach (var ar in arrs)
        {
            Console.WriteLine("\nMethod " + i + " : ");
            foreach (var s in ar)
            {
                Console.WriteLine(s);
            }
            i++;
        }
    }
 
    // Driver Code
    static void Main()
    {
        string s1 = "abcdef";
        string s2 = "axcdfdh";
 
        // Calculate the DP matrix
        EditDP(s1, s2);
 
        // Function to print all ways
        PrintWays(s1, s2, new List<string>());
    }
}

Java

import java.util.*;
 
class FollowUp_PrintAllPossibleChanges {
    int[][] dp;
    // List of changes stores all possible ways, each way - list of operations
    List<List<String> > allWays = new ArrayList<>();
 
    int minDistance(String s, String targ) {
        int m = s.length(), n = targ.length();
        dp = new int[m + 1][n + 1];
 
        // initialize by the maximum edits possible
        for (int i = 0; i <= m; i++) dp[i][0] = i;
        for (int j = 0; j <= n; j++) dp[0][j] = j;
 
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                if (s.charAt(i - 1) == targ.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1];
                else dp[i][j] = 1 + Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1]));
 
        return dp[m][n];
    }
 
    void printAllChanges(String s, String t, List<String> changes) {
        int i = s.length(), j = t.length();
 
        while (i > 0 && j > 0) {
 
            if (s.charAt(i - 1) == t.charAt(j - 1)) { i--; j--; } // Match
            else { // Not match:
                boolean replaced = false, deleted = false;
                int nextI = i, nextJ = j; // We dont want 1st if influence other ifs
                // Don't just pick single way, try all 3 ways if they all minimal
                if (dp[i][j] == dp[i - 1][j - 1] + 1) { // Replace
                    changes.add(s.charAt(i - 1) + "->" + t.charAt(j - 1));
                    nextI = i - 1; nextJ = j - 1; replaced = true;
                }
 
                if (dp[i][j] == dp[i - 1][j] + 1) { // Delete
                    if (!replaced) { changes.add("Del " + s.charAt(i - 1)); nextI = i - 1; } // Reuse 'changes' if only single operation possible
                    else {
                        // If also was replaced -> we have at least 2 ways for edit,
                        // so create new changes(like new path) & remove 'Replace' operation coz here we chosen to 'Delete'
                        // (Changes should have one of 3 operations, can't replace and remove same char, that doesnt make sense)
                        var changes2 = new ArrayList<>(changes);
                        changes2.remove(changes.size() - 1); // Remove last operation
                        changes2.add("Del " + s.charAt(i - 1)); // Add this new operation
                        // initiate new instance of this fun with remaining substrings
                        printAllChanges(s.substring(0, i - 1), t.substring(0, j), changes2);
                    }
                    deleted = true;
                }
 
                if (dp[i][j] == dp[i][j - 1] + 1) { // Add ch
                    if (!replaced && !deleted) { changes.add("Add " + t.charAt(j - 1)); nextJ = j - 1; } // One op only if not Replaced & Deleted
                    else {
                        var changes2 = new ArrayList<>(changes);
                        changes2.remove(changes.size() - 1);
                        changes2.add("Add " + t.charAt(j - 1));
                        printAllChanges(s.substring(0, i), t.substring(0, j - 1), changes2);
                    }
                }
                i = nextI; j = nextJ;
            }
        }
        while(i > 0) changes.add("Del " + s.charAt(--i)); // 't' matched but some 's' chars left not removed - rm all 's' chars
        while(j > 0) changes.add("Add " + t.charAt(--j)); // 's' chars finished but not all 't' chars matched - add missing
        allWays.add(changes);
    }
 
    void printWays(String s1, String s2) {
        printAllChanges(s1, s2, new ArrayList<>());
        int i = 1;
        for (var way : allWays) {
            System.out.print("Method #" + i++ + ": ");
            System.out.println(String.join(", ", way));
        }
    }
 
    public static void main(String[] args) throws Exception {
        var o = new FollowUp_PrintAllPossibleChanges();
        o.minDistance("abcdef", "axcdfdh");
        o.printWays("abcdef", "axcdfdh");
 
        // Only add: GeeksForGeeks impl won't print added chars
        o = new FollowUp_PrintAllPossibleChanges();
        o.minDistance("", "abc");
        o.printWays("", "abc");
 
        // Only remove: GeeksForGeeks impl won't print removed chars
        o = new FollowUp_PrintAllPossibleChanges();
        o.minDistance("abc", "");
        o.printWays("abc", "");
    }
}

Python3

import numpy as np
 
# create List of lists that will store all sets of operations
arrs = []
 
# Function to print all ways
def printAllChanges(s1, s2, changes):
 
    i = len(s1)
    j = len(s2)
 
    # Iterate till end
    while True:
        if i == 0 or j == 0:
            # Add this list to our List of lists.
            arrs.append(changes)
            break
 
        # If same
        if s1[i - 1] == s2[j - 1]:
            i -= 1
            j -= 1
        else:
            if1 = False
            if2 = False
 
            # Replace
            if dp[i][j] == dp[i - 1][j - 1] + 1:
 
                # Add this step
                changes.append(f"Change {s1[i-1]} to {s2[j-1]}")
                i -= 1
                j -= 1
                if1 = True
 
            # Delete
            if dp[i][j] == dp[i - 1][j] + 1:
                if not if1:
                    changes.append(f"Delete {s1[i-1]}")
                    i -= 1
                else:
                    # If the previous method was true,
                    # create a new list as a copy of previous.
                    changes2 = changes.copy()
                    changes2.pop()
 
                    # Add this new operation
                    changes2.append(f"Delete {s1[i]}")
 
                    # initiate new new instance of this function with remaining substrings
                    printAllChanges(s1[:i], s2[:j + 1], changes2)
 
                if2 = True
 
            # Add character step
            if dp[i][j] == dp[i][j - 1] + 1:
                if not if1 and not if2:
                    changes.append(f"Add {s2[j-1]}")
                    j -= 1
                else:
                    # Add steps
                    changes2 = changes.copy()
                    changes2.pop()
                    changes2.append(f"Add {s2[j]}")
 
                    # Recursively call for the next steps
                    printAllChanges(s1[:i + 1], s2[:j], changes2)
 
            # Move to previous subproblem
            i -= 1
            j -= 1
 
# Function to compute the DP matrix
def editDP(s1, s2):
 
    l1 = len(s1)
    l2 = len(s2)
 
    # initialize by the maximum edits possible
    dp = np.zeros((l1+1, l2+1))
    for i in range(l1+1):
        dp[i][0] = i
    for j in range(l2+1):
        dp[0][j] = j
 
    # Compute the DP matrix
    for i in range(1, l1+1):
        for j in range(1, l2+1):
            # 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(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
 
    return dp
 
def printWays(s1, s2, changes):
    # Function to print all the ways
    printAllChanges(s1, s2, [])
 
    i = 1
    # print all the possible ways
    for ar in arrs:
        print("\nMethod {}: ".format(i))
        for s in ar:
            print(s)
        i += 1
 
# Driver Code
if __name__ == "__main__":
    s1 = "abcdef"
    s2 = "axcdfdh"
 
    # calculate the DP matrix
    editDP(s1, s2)
 
    # Function to print all ways
    printWays(s1, s2, [])

C#

using System;
using System.Collections.Generic;
 
class Program
{
    // Create List of lists that will store all sets of operations
    static List<List<string>> arrs = new List<List<string>>();
    static int[,] dp = new int[1001, 1001];
 
    // Function to print all ways
    static void PrintAllChanges(string s1, string s2, List<string> changes)
    {
        int i = s1.Length;
        int j = s2.Length;
 
        // Iterate till the end
        while (true)
        {
            if (i == 0 || j == 0)
            {
                // Add this list to our List of lists.
                arrs.Add(new List<string>(changes));
                break;
            }
 
            // If same
            if (s1[i - 1] == s2[j - 1])
            {
                i--;
                j--;
            }
            else
            {
                bool if1 = false, if2 = false;
 
                // Replace
                if (dp[i, j] == dp[i - 1, j - 1] + 1)
                {
                    // Add this step
                    changes.Add("Change " + s1[i - 1] + " to " + s2[j - 1]);
                    i--;
                    j--;
                    if1 = true;
                }
 
                // Delete
                if (dp[i, j] == dp[i - 1, j] + 1)
                {
                    if (!if1)
                    {
                        changes.Add("Delete " + s1[i - 1]);
                        i--;
                    }
                    else
                    {
                        // If the previous method was true,
                        // create a new list as a copy of the previous.
                        List<string> changes2 = new List<string>(changes);
                        changes2.RemoveAt(changes2.Count - 1);
 
                        // Add this new operation
                        changes2.Add("Delete " + s1[i]);
                        // Initiate a 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 && !if2)
                    {
                        changes.Add("Add " + s2[j - 1]);
                        j--;
                    }
                    else
                    {
                        // Add steps
                        List<string> changes2 = new List<string>(changes);
                        changes2.RemoveAt(changes2.Count - 1);
                        changes2.Add("Add " + s2[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;
 
        // 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 the 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] = Math.Min(dp[i - 1, j - 1], Math.Min(dp[i - 1, j], dp[i, j - 1])) + 1;
            }
        }
    }
 
    static void PrintWays(string s1, string s2, List<string> changes)
    {
        // Function to print all the ways
        PrintAllChanges(s1, s2, new List<string>());
        int i = 1;
        // Print all the possible ways
        foreach (var ar in arrs)
        {
            Console.WriteLine("\nMethod " + i + " : ");
            foreach (var s in ar)
            {
                Console.WriteLine(s);
            }
            i++;
        }
    }
 
    // Driver Code
    static void Main()
    {
        string s1 = "abcdef";
        string s2 = "axcdfdh";
 
        // Calculate the DP matrix
        EditDP(s1, s2);
 
        // Function to print all ways
        PrintWays(s1, s2, new List<string>());
    }
}

Javascript

// JavaScript program to print all the possible
// steps to change a string to another
 
// create List of lists that will store all sets of
// operations
let arrs = [];
let dp = new Array(1001).fill(null).map(() => new Array(1001).fill(0));
 
// Function to print all ways
function printAllChanges(s1, s2, changes) {
    let i = s1.length;
    let j = s2.length;
 
    // Iterate till end
    while (true) {
        if (i === 0 || j === 0) {
            // Add this list to our List of lists.
            arrs.push(changes.slice());
            break;
        }
 
        // If same
        if (s1[i - 1] === s2[j - 1]) {
            i--;
            j--;
        } else {
            let if1 = false,
                if2 = false;
 
            // Replace
            if (dp[i][j] === dp[i - 1][j - 1] + 1) {
                // Add this step
                changes.push(`Change ${s1[i - 1]} to ${s2[j - 1]}`);
                i--;
                j--;
                if1 = true;
            }
            // Delete
            if (dp[i][j] === dp[i - 1][j] + 1) {
                if (!if1) {
                    changes.push(`Delete ${s1[i - 1]}`);
                    i--;
                } else {
                    // If the previous method was true,
                    // create a new list as a copy of
                    // previous.
                    let changes2 = changes.slice(0, -1);
 
                    // Add this new operation
                    changes2.push(`Delete ${s1[i]}`);
 
                    // initiate new new instance of this
                    // function with remaining substrings
                    printAllChanges(s1.slice(0, i), s2.slice(0, j + 1), changes2);
                }
                if2 = true;
            }
 
            // Add character step
            if (dp[i][j] === dp[i][j - 1] + 1) {
                if (!if1 && !if2) {
                    changes.push(`Add ${s2[j - 1]}`);
                    j--;
                } else {
                    // Add steps
                    let changes2 = changes.slice(0, -1);
                    changes2.push(`Add ${s2[j]}`);
                    // Recursively call for the next steps
                    printAllChanges(s1.slice(0, i + 1), s2.slice(0, j), changes2);
                }
            }
        }
    }
}
 
// Function to compute the DP matrix
function editDP(s1, s2) {
    let l1 = s1.length;
    let l2 = s2.length;
 
    // initialize by the maximum edits possible
    for (let i = 0; i <= l1; i++) {
        dp[i][0] = i;
    }
 
 
    for (let j = 0; j <= l2; j++) {
        dp[0][j] = j;
    }
 
    // Compute the DP matrix
    for (let i = 1; i <= l1; i++) {
        for (let 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] = Math.min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1;
            }
        }
    }
}
 
function printWays(s1, s2, changes) {
    // Function to print all the ways
    printAllChanges(s1, s2, []);
    let i = 1;
 
    // print all the possible ways
    for (let ar of arrs) {
        console.log(`\nMethod ${i++} : `);
 
        for (let s of ar) {
            console.log(s);
        }
    }
}
 
// Driver Code
let s1 = "abcdef";
let s2 = "axcdfdh";
 
// calculate the DP matrix
editDP(s1, s2);
 
// Function to print all ways
printWays(s1, s2, []);
 
// Contributed by adityasharmadev01

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

Time Complexity: O(m*n*n*k) m – length of s1, n – length of s2, second n caused by fact that we copy ‘changes’ every time there is more than one way found, k – num of all possible ways, upper-bound for k= 3^n

Space Complexity: O(n^2) due to the use of the dp array

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