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 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]; // initilize 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(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]; // initilize 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 { // 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) 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)] # Initilize 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]; // initilize 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 |
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 charater 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]; // initilize 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 { // 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); } 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>()); } } |
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
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