Edit Distance | DP-5

Given two strings str1 and str2 and below operations that can performed on str1. Find minimum number of edits (operations) required to convert ‘str1’ into ‘str2’.

  1. Insert
  2. Remove
  3. Replace

All of the above operations are of equal cost.

Examples:

Input:   str1 = "geek", str2 = "gesek"
Output:  1
We can convert str1 into str2 by inserting a 's'.

Input:   str1 = "cat", str2 = "cut"
Output:  1
We can convert str1 into str2 by replacing 'a' with 'u'.

Input:   str1 = "sunday", str2 = "saturday"
Output:  3
Last three and first characters are same.  We basically
need to convert "un" to "atur".  This can be done using
below three operations. 
Replace 'n' with 'r', insert t, insert a


What are the subproblems in this case?
The idea is process all characters one by one staring from either from left or right sides of both strings.
Let us traverse from right corner, there are two possibilities for every pair of character being traversed.

m: Length of str1 (first string)
n: Length of str2 (second string)
  1. If last characters of two strings are same, nothing much to do. Ignore last characters and get count for remaining strings. So we recur for lengths m-1 and n-1.
  2. Else (If last characters are not same), we consider all operations on ‘str1’, consider all three operations on last character of first string, recursively compute minimum cost for all three operations and take minimum of three values.
    1. Insert: Recur for m and n-1
    2. Remove: Recur for m-1 and n
    3. Replace: Recur for m-1 and n-1

Below is C++ implementation of above Naive recursive solution.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Naive recursive C++ program to find minimum number
// operations to convert str1 to str2
#include <bits/stdc++.h>
using namespace std;
  
// Utility function to find minimum of three numbers
int min(int x, int y, int z)
{
    return min(min(x, y), z);
}
  
int editDist(string str1, string str2, int m, int n)
{
    // If first string is empty, the only option is to
    // insert all characters of second string into first
    if (m == 0)
        return n;
  
    // If second string is empty, the only option is to
    // remove all characters of first string
    if (n == 0)
        return m;
  
    // If last characters of two strings are same, nothing
    // much to do. Ignore last characters and get count for
    // remaining strings.
    if (str1[m - 1] == str2[n - 1])
        return editDist(str1, str2, m - 1, n - 1);
  
    // If last characters are not same, consider all three
    // operations on last character of first string, recursively
    // compute minimum cost for all three operations and take
    // minimum of three values.
    return 1 + min(editDist(str1, str2, m, n - 1), // Insert
                   editDist(str1, str2, m - 1, n), // Remove
                   editDist(str1, str2, m - 1, n - 1) // Replace
                   );
}
  
// Driver program
int main()
{
    // your code goes here
    string str1 = "sunday";
    string str2 = "saturday";
  
    cout << editDist(str1, str2, str1.length(), str2.length());
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Naive recursive Java program to find minimum number
// operations to convert str1 to str2
class EDIST {
    static int min(int x, int y, int z)
    {
        if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
    }
  
    static int editDist(String str1, String str2, int m, int n)
    {
        // If first string is empty, the only option is to
        // insert all characters of second string into first
        if (m == 0)
            return n;
  
        // If second string is empty, the only option is to
        // remove all characters of first string
        if (n == 0)
            return m;
  
        // If last characters of two strings are same, nothing
        // much to do. Ignore last characters and get count for
        // remaining strings.
        if (str1.charAt(m - 1) == str2.charAt(n - 1))
            return editDist(str1, str2, m - 1, n - 1);
  
        // If last characters are not same, consider all three
        // operations on last character of first string, recursively
        // compute minimum cost for all three operations and take
        // minimum of three values.
        return 1 + min(editDist(str1, str2, m, n - 1), // Insert
                       editDist(str1, str2, m - 1, n), // Remove
                       editDist(str1, str2, m - 1, n - 1) // Replace
                       );
    }
  
    public static void main(String args[])
    {
        String str1 = "sunday";
        String str2 = "saturday";
  
        System.out.println(editDist(str1, str2, str1.length(), str2.length()));
    }
}
/*This code is contributed by Rajat Mishra*/

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# A Naive recursive Python program to fin minimum number
# operations to convert str1 to str2
def editDistance(str1, str2, m, n):
  
    # If first string is empty, the only option is to
    # insert all characters of second string into first
    if m == 0:
         return n
  
    # If second string is empty, the only option is to
    # remove all characters of first string
    if n == 0:
        return m
  
    # If last characters of two strings are same, nothing
    # much to do. Ignore last characters and get count for
    # remaining strings.
    if str1[m-1]== str2[n-1]:
        return editDistance(str1, str2, m-1, n-1)
  
    # If last characters are not same, consider all three
    # operations on last character of first string, recursively
    # compute minimum cost for all three operations and take
    # minimum of three values.
    return 1 + min(editDistance(str1, str2, m, n-1),    # Insert
                   editDistance(str1, str2, m-1, n),    # Remove
                   editDistance(str1, str2, m-1, n-1)    # Replace
                   )
  
# Driver program to test the above function
str1 = "sunday"
str2 = "saturday"
print editDistance(str1, str2, len(str1), len(str2))
  
# This code is contributed by Bhavya Jain

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Naive recursive C# program to
// find minimum numberoperations
// to convert str1 to str2
using System;
  
class GFG {
    static int min(int x, int y, int z)
    {
        if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
    }
  
    static int editDist(String str1, String str2, int m, int n)
    {
        // If first string is empty, the only option is to
        // insert all characters of second string into first
        if (m == 0)
            return n;
  
        // If second string is empty, the only option is to
        // remove all characters of first string
        if (n == 0)
            return m;
  
        // If last characters of two strings are same, nothing
        // much to do. Ignore last characters and get count for
        // remaining strings.
        if (str1[m - 1] == str2[n - 1])
            return editDist(str1, str2, m - 1, n - 1);
  
        // If last characters are not same, consider all three
        // operations on last character of first string, recursively
        // compute minimum cost for all three operations and take
        // minimum of three values.
        return 1 + min(editDist(str1, str2, m, n - 1), // Insert
                       editDist(str1, str2, m - 1, n), // Remove
                       editDist(str1, str2, m - 1, n - 1) // Replace
                       );
    }
  
    // Driver code
    public static void Main()
    {
        String str1 = "sunday";
        String str2 = "saturday";
        Console.WriteLine(editDist(str1, str2, str1.Length,
                                   str2.Length));
    }
}
  
// This Code is Contributed by Sam007

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php 
// A Naive recursive Python program  
// to find minimum number operations 
// to convert str1 to str2 
function editDistance($str1, $str2
                      $m, $n)
    // If first string is empty, 
    // the only option is to insert.
    // all characters of second
    // string into first 
    if ($m == 0)
        return $n
  
    // If second string is empty,
    // the only option is to 
    // remove all characters of 
    // first string 
    if ($n == 0) 
        return $m
  
    // If last characters of two 
    // strings are same, nothing 
    // much to do. Ignore last 
    // characters and get count 
    // for remaining strings. 
    if ($str1[$m - 1] == $str2[$n - 1])
    
        return editDistance($str1, $str2
                            $m - 1, $n - 1); 
    }
      
    // If last characters are not same, 
    // consider all three operations on 
    // last character of first string, 
    // recursively compute minimum cost 
    // for all three operations and take 
    // minimum of three values. 
  
    return 1 + min(editDistance($str1, $str2
                                $m, $n - 1), // Insert 
                   editDistance($str1, $str2
                                $m - 1, $n), // Remove 
                   editDistance($str1, $str2
                                $m - 1, $n - 1)); // Replace 
  
// Driver Code
$str1 = "sunday";
$str2 = "saturday";
echo editDistance($str1, $str2, strlen($str1), 
                                strlen($str2)); 
  
// This code is contributed 
// by Shivi_Aggarwal
?>

chevron_right



Output:



3

The time complexity of above solution is exponential. In worst case, we may end up doing O(3m) operations. The worst case happens when none of characters of two strings match. Below is a recursive call diagram for worst case.
EditDistance

We can see that many subproblems are solved, again and again, for example, eD(2, 2) is called three times. Since same suproblems are called again, this problem has Overlapping Subprolems property. So Edit Distance problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array that stores results of subproblems.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Dynamic Programming based C++ program to find minimum
// number operations to convert str1 to str2
#include <bits/stdc++.h>
using namespace std;
  
// Utility function to find the minimum of three numbers
int min(int x, int y, int z)
{
    return min(min(x, y), z);
}
  
int editDistDP(string str1, string str2, int m, int n)
{
    // Create a table to store results of subproblems
    int dp[m + 1][n + 1];
  
    // Fill d[][] in bottom up manner
    for (int i = 0; i <= m; i++) {
        for (int j = 0; j <= n; j++) {
            // If first string is empty, only option is to
            // insert all characters of second string
            if (i == 0)
                dp[i][j] = j; // Min. operations = j
  
            // If second string is empty, only option is to
            // remove all characters of second string
            else if (j == 0)
                dp[i][j] = i; // Min. operations = i
  
            // If last characters are same, ignore last char
            // and recur for remaining string
            else if (str1[i - 1] == str2[j - 1])
                dp[i][j] = dp[i - 1][j - 1];
  
            // If the last character is different, consider all
            // possibilities and find the minimum
            else
                dp[i][j] = 1 + min(dp[i][j - 1], // Insert
                                   dp[i - 1][j], // Remove
                                   dp[i - 1][j - 1]); // Replace
        }
    }
  
    return dp[m][n];
}
  
// Driver program
int main()
{
    // your code goes here
    string str1 = "sunday";
    string str2 = "saturday";
  
    cout << editDistDP(str1, str2, str1.length(), str2.length());
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Dynamic Programming based Java program to find minimum
// number operations to convert str1 to str2
class EDIST {
    static int min(int x, int y, int z)
    {
        if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
    }
  
    static int editDistDP(String str1, String str2, int m, int n)
    {
        // Create a table to store results of subproblems
        int dp[][] = new int[m + 1][n + 1];
  
        // Fill d[][] in bottom up manner
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                // If first string is empty, only option is to
                // insert all characters of second string
                if (i == 0)
                    dp[i][j] = j; // Min. operations = j
  
                // If second string is empty, only option is to
                // remove all characters of second string
                else if (j == 0)
                    dp[i][j] = i; // Min. operations = i
  
                // If last characters are same, ignore last char
                // and recur for remaining string
                else if (str1.charAt(i - 1) == str2.charAt(j - 1))
                    dp[i][j] = dp[i - 1][j - 1];
  
                // If the last character is different, consider all
                // possibilities and find the minimum
                else
                    dp[i][j] = 1 + min(dp[i][j - 1], // Insert
                                       dp[i - 1][j], // Remove
                                       dp[i - 1][j - 1]); // Replace
            }
        }
  
        return dp[m][n];
    }
  
    public static void main(String args[])
    {
        String str1 = "sunday";
        String str2 = "saturday";
        System.out.println(editDistDP(str1, str2, str1.length(), str2.length()));
    }
} /*This code is contributed by Rajat Mishra*/

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# A Dynamic Programming based Python program for edit
# distance problem
def editDistDP(str1, str2, m, n):
    # Create a table to store results of subproblems
    dp = [[0 for x in range(n + 1)] for x in range(m + 1)]
  
    # Fill d[][] in bottom up manner
    for i in range(m + 1):
        for j in range(n + 1):
  
            # If first string is empty, only option is to
            # insert all characters of second string
            if i == 0:
                dp[i][j] = j    # Min. operations = j
  
            # If second string is empty, only option is to
            # remove all characters of second string
            elif j == 0:
                dp[i][j] = i    # Min. operations = i
  
            # If last characters are same, ignore last char
            # and recur for remaining string
            elif str1[i-1] == str2[j-1]:
                dp[i][j] = dp[i-1][j-1]
  
            # If last character are different, consider all
            # possibilities and find minimum
            else:
                dp[i][j] = 1 + min(dp[i][j-1],        # Insert
                                   dp[i-1][j],        # Remove
                                   dp[i-1][j-1])    # Replace
  
    return dp[m][n]
  
# Driver program
str1 = "sunday"
str2 = "saturday"
  
print(editDistDP(str1, str2, len(str1), len(str2)))
# This code is contributed by Bhavya Jain

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Dynamic Programming based
// C# program to find minimum
// number operations to
// convert str1 to str2
using System;
  
class GFG {
    static int min(int x, int y, int z)
    {
        if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
    }
  
    static int editDistDP(String str1, String str2, int m, int n)
    {
        // Create a table to store
        // results of subproblems
        int[, ] dp = new int[m + 1, n + 1];
  
        // Fill d[][] in bottom up manner
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                // If first string is empty, only option is to
                // insert all characters of second string
                if (i == 0)
  
                    // Min. operations = j
                    dp[i, j] = j;
  
                // If second string is empty, only option is to
                // remove all characters of second string
                else if (j == 0)
  
                    // Min. operations = i
                    dp[i, j] = i;
  
                // If last characters are same, ignore last char
                // and recur for remaining string
                else if (str1[i - 1] == str2[j - 1])
                    dp[i, j] = dp[i - 1, j - 1];
  
                // If the last character is different, consider all
                // possibilities and find the minimum
                else
                    dp[i, j] = 1 + min(dp[i, j - 1], // Insert
                                       dp[i - 1, j], // Remove
                                       dp[i - 1, j - 1]); // Replace
            }
        }
  
        return dp[m, n];
    }
  
    // Driver code
    public static void Main()
    {
        String str1 = "sunday";
        String str2 = "saturday";
        Console.Write(editDistDP(str1, str2, str1.Length,
                                 str2.Length));
    }
}
// This Code is Contributed by Sam007

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// A Dynamic Programming based 
// Python program for edit 
// distance problem 
function editDistDP($str1, $str2
                    $m, $n)
// Fill d[][] in bottom up manner 
for ($i = 0; $i <= $m; $i++) 
{  
    for ($j = 0; $j <= $n; $j++) 
    
  
        // If first string is empty, 
        // only option is to insert 
        // all characters of second string 
        if ($i == 0) 
            $dp[$i][$j] = $j ; // Min. operations = j 
  
        // If second string is empty, 
        // only option is to remove 
        // all characters of second string 
        else if($j == 0) 
            $dp[$i][$j] = $i; // Min. operations = i 
  
        // If last characters are same, 
        // ignore last char and recur 
        // for remaining string 
        else if($str1[$i - 1] == $str2[$j - 1]) 
            $dp[$i][$j] = $dp[$i - 1][$j - 1];
  
        // If last character are different, 
        // consider all possibilities and 
        // find minimum 
        else
        
            $dp[$i][$j] = 1 + min($dp[$i][$j - 1],     // Insert 
                                  $dp[$i - 1][$j],     // Remove 
                                  $dp[$i - 1][$j - 1]); // Replace 
        }
    }
}
return $dp[$m][$n] ;
}
  
// Driver Code 
$str1 = "sunday";
$str2 = "saturday";
  
echo editDistDP($str1, $str2, strlen($str1), 
                              strlen($str2));
  
// This code is contributed 
// by Shivi_Aggarwal
?>

chevron_right


Output:

3

Time Complexity: O(m x n)
Auxiliary Space: O(m x n)

Space Complex Solution: In the above-given method we require O(m x n) space. This will not be suitable if the length of strings is greater than 2000 as it can only create 2D array of 2000 x 2000. To fill a row in DP array we require only one row the upper row. For example, if we are filling the i = 10 rows in DP array we require only values of 9th row. So we simply create a DP array of 2 x str1 length. This approach reduces the space complexity. Here is the C++ implementation of the above-mentioned problem.

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Space efficient Dynamic Programming
// based C++ program to find minimum
// number operations to convert str1 to str2
#include <bits/stdc++.h>
using namespace std;
  
void EditDistDP(string str1, string str2)
{
    int len1 = str1.length();
    int len2 = str2.length();
  
    // Create a DP array to memoize result
    // of previous computations
    int DP[2][len1 + 1];
  
    // To fill the DP array with 0
    memset(DP, 0, sizeof DP);
  
    // Base condition when second string
    // is empty then we remove all characters
    for (int i = 0; i <= len1; i++)
        DP[0][i] = i;
  
    // Start filling the DP
    // This loop run for every
    // character in second string
    for (int i = 1; i <= len2; i++) {
        // This loop compares the char from
        // second string with first string
        // characters
        for (int j = 0; j <= len1; j++) {
            // if first string is empty then
            // we have to perform add character
            // operation to get second string
            if (j == 0)
                DP[i % 2][j] = i;
  
            // if character from both string
            // is same then we do not perform any
            // operation . here i % 2 is for bound
            // the row number.
            else if (str1[j - 1] == str2[i - 1]) {
                DP[i % 2][j] = DP[(i - 1) % 2][j - 1];
            }
  
            // if character from both string is
            // not same then we take the minimum
            // from three specified operation
            else {
                DP[i % 2][j] = 1 + min(DP[(i - 1) % 2][j],
                                       min(DP[i % 2][j - 1],
                                           DP[(i - 1) % 2][j - 1]));
            }
        }
    }
  
    // after complete fill the DP array
    // if the len2 is even then we end
    // up in the 0th row else we end up
    // in the 1th row so we take len2 % 2
    // to get row
    cout << DP[len2 % 2][len1] << endl;
}
  
// Driver program
int main()
{
    string str1 = "food";
    string str2 = "money";
    EditDistDP(str1, str2);
    return 0;
}

chevron_right


Output:

4

Time Complexity: O(m x n)
Auxiliary Space: O( m )

Applications: There are many practical applications of edit distance algorithm, refer Lucene API for sample. Another example, display all the words in a dictionary that are near proximity to a given wordincorrectly spelled word.

Thanks to Vivek Kumar for suggesting updates.

Thanks to Venki for providing initial post. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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.




My Personal Notes arrow_drop_up