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 stepsvoid 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 matrixvoid 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 Codeint 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 stepsdef 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 MatrixeditDP(s1, s2)# This code is contributed by Pranav S |
C#
// C# program to print one possible// way of converting a string to anotherusing 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 approachlet 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 Codelet s1 = "abcdef";let s2 = "axcdfdh";// calculate the DP matrixeditDP(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 operationsarrs = []# Function to print all waysdef 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 matrixdef 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 dpdef 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 Codeif __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// operationslet arrs = [];let dp = new Array(1001).fill(null).map(() => new Array(1001).fill(0));// Function to print all waysfunction 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 matrixfunction 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 Codelet s1 = "abcdef";let s2 = "axcdfdh";// calculate the DP matrixeditDP(s1, s2);// Function to print all waysprintWays(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
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
