Home Data Structures and Algorithms Edit Distance Using Dynamic Programming (Memoization)

Edit Distance Using Dynamic Programming (Memoization)

Edit Distance using Recursion:

Edit Distance Using Dynamic Programming (Bottom-Up Approach):

In the above recursive approach, there are several overlapping subproblems:
Edit_Distance(M-1, N-1) is called Three times
Edit_Distance(M-1, N-2) is called Two times
Edit_Distance(M-2, N-1) is called Two times. And so on…

So, we can use Memoization technique to store the result of each subproblems to avoid recalculating the result again and again.

Below are the illustration of overlapping subproblems during the recursion.

Below is the implementation of Edit Distance Using Dynamic Programming (Memoization):

C++ 
#include <bits/stdc++.h>
using namespace std;
int minDis(string s1, string s2, int n, int m,
           vector<vector<int> >& dp)
{

    // If any string is empty,
    // return the remaining characters of other string

    if (n == 0)
        return m;

    if (m == 0)
        return n;

    // To check if the recursive tree
    // for given n & m has already been executed

    if (dp[n][m] != -1)
        return dp[n][m];

    // If characters are equal, execute
    // recursive function for n-1, m-1

    if (s1[n - 1] == s2[m - 1]) {
        return dp[n][m] = minDis(s1, s2, n - 1, m - 1, dp);
    }
    // If characters are nt equal, we need to
    // find the minimum cost out of all 3 operations.
    // 1. insert 2.delete 3.replace
    else {
        int insert, del, replace; // temp variables

        insert = minDis(s1, s2, n, m - 1, dp);
        del = minDis(s1, s2, n - 1, m, dp);
        replace = minDis(s1, s2, n - 1, m - 1, dp);
        return dp[n][m]
               = 1 + min(insert, min(del, replace));
    }
}

// Driver program
int main()
{

    string str1 = "GEEXSFRGEEKKS";
    string str2 = "w3wiki";

    int n = str1.length(), m = str2.length();
    vector<vector<int> > dp(n + 1, vector<int>(m + 1, -1));

    cout << minDis(str1, str2, n, m, dp);
    return 0;
}
C 
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int min(int x, int y, int z)
{
    if (x < y && x < z)
        return x;
    else if (y < x && y < z)
        return y;
    else
        return z;
}

int minDis(char* s1, char* s2, int n, int m)
{
    int dp[n + 1][m + 1];

    // Fill dp[][] table in bottom up manner
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            // If first string is empty, only option is to
            // insert all characters of second string
            if (i == 0)
                dp[i][j] = j;
            // If second string is empty, only option is to
            // remove all characters of second string
            else if (j == 0)
                dp[i][j] = i;
            // If last characters are same, ignore last char
            // and recur for remaining string
            else if (s1[i - 1] == s2[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[n][m];
}

int main()
{
    char* str1 = "GEEXSFRGEEKKS";
    char* str2 = "w3wiki";

    int n = strlen(str1), m = strlen(str2);

    printf("%d", minDis(str1, str2, n, m));

    return 0;
}
Java 
import java.util.*;
class GFG {

    static int minDis(String s1, String s2, int n, int m,
                      int[][] dp)
    {

        // If any String is empty,
        // return the remaining characters of other String
        if (n == 0)
            return m;
        if (m == 0)
            return n;

        // To check if the recursive tree
        // for given n & m has already been executed
        if (dp[n][m] != -1)
            return dp[n][m];

        // If characters are equal, execute
        // recursive function for n-1, m-1
        if (s1.charAt(n - 1) == s2.charAt(m - 1)) {
            return dp[n][m]
                = minDis(s1, s2, n - 1, m - 1, dp);
        }
        // If characters are nt equal, we need to
        // find the minimum cost out of all 3 operations.
        else {

            int insert, del, replace; // temp variables

            insert = minDis(s1, s2, n, m - 1, dp);
            del = minDis(s1, s2, n - 1, m, dp);
            replace = minDis(s1, s2, n - 1, m - 1, dp);
            return dp[n][m]
                = 1
                  + Math.min(insert,
                             Math.min(del, replace));
        }
    }
    // Driver program
    public static void main(String[] args)
    {

        String str1 = "GEEXSFRGEEKKS";
        String str2 = "w3wiki";

        int n = str1.length(), m = str2.length();
        int[][] dp = new int[n + 1][m + 1];
        for (int i = 0; i < n + 1; i++)
            Arrays.fill(dp[i], -1);
        System.out.print(minDis(str1, str2, n, m, dp));
    }
}
Python3 
def minDis(s1, s2, n, m, dp):

    # If any string is empty,
    # return the remaining characters of other string
    if(n == 0):
        return m
    if(m == 0):
        return n

    # To check if the recursive tree
    # for given n & m has already been executed
    if(dp[n][m] != -1):
        return dp[n][m]

    # If characters are equal, execute
    # recursive function for n-1, m-1
    if(s1[n - 1] == s2[m - 1]):
        if(dp[n - 1][m - 1] == -1):
            dp[n][m] = minDis(s1, s2, n - 1, m - 1, dp)
            return dp[n][m]
        else:
            dp[n][m] = dp[n - 1][m - 1]
            return dp[n][m]

    # If characters are nt equal, we need to
    # find the minimum cost out of all 3 operations.
    else:
        if(dp[n - 1][m] != -1):
            m1 = dp[n - 1][m]
        else:
            m1 = minDis(s1, s2, n - 1, m, dp)

        if(dp[n][m - 1] != -1):
            m2 = dp[n][m - 1]
        else:
            m2 = minDis(s1, s2, n, m - 1, dp)
        if(dp[n - 1][m - 1] != -1):
            m3 = dp[n - 1][m - 1]
        else:
            m3 = minDis(s1, s2, n - 1, m - 1, dp)

        dp[n][m] = 1 + min(m1, min(m2, m3))
        return dp[n][m]


# Driver code
str1 = "GEEXSFRGEEKKS"
str2 = "w3wiki"

n = len(str1)
m = len(str2)
dp = [[-1 for i in range(m + 1)] for j in range(n + 1)]

print(minDis(str1, str2, n, m, dp))
C# 
using System;
using System.Collections.Generic;
class GFG {

    static int minDis(string s1, string s2, int n, int m,
                      List<List<int> > dp)
    {

        // If any string is empty,
        // return the remaining characters of other string
        if (n == 0)
            return m;

        if (m == 0)
            return n;

        // To check if the recursive tree
        // for given n & m has already been executed
        if (dp[n][m] != -1)
            return dp[n][m];

        // If characters are equal, execute
        // recursive function for n-1, m-1
        if (s1[n - 1] == s2[m - 1]) {
            if (dp[n - 1][m - 1] == -1) {
                return dp[n][m]
                    = minDis(s1, s2, n - 1, m - 1, dp);
            }
            else
                return dp[n][m] = dp[n - 1][m - 1];
        }

        // If characters are nt equal, we need to
        // find the minimum cost out of all 3 operations.
        else {
            int m1, m2, m3; // temp variables

            if (dp[n - 1][m] != -1) {
                m1 = dp[n - 1][m];
            }
            else {
                m1 = minDis(s1, s2, n - 1, m, dp);
            }

            if (dp[n][m - 1] != -1) {
                m2 = dp[n][m - 1];
            }
            else {
                m2 = minDis(s1, s2, n, m - 1, dp);
            }

            if (dp[n - 1][m - 1] != -1) {
                m3 = dp[n - 1][m - 1];
            }
            else {
                m3 = minDis(s1, s2, n - 1, m - 1, dp);
            }
            return dp[n][m]
                = 1 + Math.Min(m1, Math.Min(m2, m3));
        }
    }

    // Driver code
    static void Main()
    {
        string str1 = "GEEXSFRGEEKKS";
        string str2 = "w3wiki";

        int n = str1.Length, m = str2.Length;
        List<List<int> > dp = new List<List<int> >();
        for (int i = 0; i < n + 1; i++) {
            dp.Add(new List<int>());
            for (int j = 0; j < m + 1; j++) {
                dp[i].Add(-1);
            }
        }
        Console.WriteLine(minDis(str1, str2, n, m, dp));
    }
}
JavaScript 
function minDis(s1,s2,n,m,dp)
{
    // If any String is empty,
  // return the remaining characters of other String
  if(n == 0)   
    return m; 
  if(m == 0)   
    return n;
              
  // To check if the recursive tree
  // for given n & m has already been executed
  if(dp[n][m] != -1)   
    return dp[n][m];
                 
  // If characters are equal, execute
  // recursive function for n-1, m-1
  if(s1[n - 1] == s2[m - 1])
  {          
    if(dp[n - 1][m - 1] == -1)
    {              
      return dp[n][m] = minDis(s1, s2, n - 1, m - 1, dp);          
    }       
    else
      return dp[n][m] = dp[n - 1][m - 1];  
  }
   
  // If characters are nt equal, we need to
           
  // find the minimum cost out of all 3 operations.     
  else
  {          
    let m1, m2, m3;        // temp variables  
    if(dp[n-1][m] != -1)
    {   
      m1 = dp[n - 1][m];     
    }          
    else
    {  
      m1 = minDis(s1, s2, n - 1, m, dp);     
    }           
             
    if(dp[n][m - 1] != -1)
    {               
      m2 = dp[n][m - 1];           
    }           
    else
    {   
      m2 = minDis(s1, s2, n, m - 1, dp);     
    }                                  
    
    if(dp[n - 1][m - 1] != -1)
    {   
      m3 = dp[n - 1][m - 1];     
    }  
    else
    {  
      m3 = minDis(s1, s2, n - 1, m - 1, dp);      
    }    
    return dp[n][m] = 1 + Math.min(m1, Math.min(m2, m3));       
  }
}

// Driver program

let str1 = "GEEXSFRGEEKKS";
let str2 = "w3wiki";

let n= str1.length, m = str2.length;   
let dp = new Array(n + 1);
for(let i = 0; i < n + 1; i++)
{
    dp[i]=new Array(m+1);
    for(let j=0;j<m+1;j++)
        dp[i][j]=-1;
}

console.log(minDis(str1, str2, n, m, dp));

Output
3

Time Complexity: O(m x n) 
Auxiliary Space: O( m *n)+O(m+n) , (m*n) extra array space and (m+n) recursive stack space.

Edit Distance

Given two strings str1 and str2 of length M and N respectively and below operations that can be performed on str1. Find the minimum number of edits (operations) to convert ‘str1‘ into ‘str2‘.  

  • Operation 1 (INSERT): Insert any character before or after any index of str1
  • Operation 2 (REMOVE): Remove a character of str1
  • Operation 3 (Replace): Replace a character at any index of str1 with some other character.

Note: All of the above operations are of equal cost. 

Examples: 

Input:   str1 = “geek”, str2 = “gesek”
Output:  1
Explanation: We can convert str1 into str2 by inserting a ‘s’ between two consecutive ‘e’ in str2.

Input:   str1 = “cat”, str2 = “cut”
Output:  1
Explanation: We can convert str1 into str2 by replacing ‘a’ with ‘u’.

Input:   str1 = “sunday”, str2 = “saturday”
Output:  3
Explanation: 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

Illustration of Edit Distance:

Let’s suppose we have str1=”GEEXSFRGEEKKS” and str2=”w3wiki”
Now to convert str1 into str2 we would require 3 minimum operations:
Operation 1: Replace ‘X‘ to ‘K‘
Operation 2: Insert ‘O‘ between ‘F‘ and ‘R‘
Operation 3: Remove second last character i.e. ‘K‘

Refer the below image for better understanding.

Recommended Practice
Edit Distance
Try It!

Similar Reads

Edit Distance using Recursion:

Subproblems in Edit Distance:...

Edit Distance Using Dynamic Programming (Memoization):

In the above recursive approach, there are several overlapping subproblems:Edit_Distance(M-1, N-1) is called Three times Edit_Distance(M-1, N-2) is called Two times Edit_Distance(M-2, N-1) is called Two times. And so on… So, we can use Memoization technique to store the result of each subproblems to avoid recalculating the result again and again. Below are the illustration of overlapping subproblems during the recursion....

Edit Distance Using Dynamic Programming (Bottom-Up Approach):

Use a table to store solutions of subproblems to avoiding recalculate the same subproblems multiple times. By doing this, if same subproblems repeated during, we retrieve the solutions from the table itself....

Edit Distance Using Dynamic Programming (Optimization in Space Complexity):

Optimized Space Complexity Solution: In the above bottom up approach we require O(m x n) space. Let’s take an observation and try to optimize our space complexity: To fill a row in DP array we require only one row i.e. the upper row. For example, if we are filling the row where i=10 in DP array then 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 from O(N*M) to O(2*N)....

Edit Distance Using Dynamic Programming (Further Optimization in Space Complexity):

As discussed the above approach uses two 1-D arrays, now the question is can we achieve our task by using only a single 1-D array? The answer is Yes and it requires a simple observation as mentioned below:In the previous approach The curr[] array is updated using 3 values only : Value 1: curr[j] = prev[j-1] when str1[i-1] is equal to str2[j-1]Value 2: curr[j] = prev[j] when str1[i-1] is not equal to str2[j-1] Value 3: curr[j] = curr[j-1] when str1[i-1] is not equal to str2[j-1] By keeping the track of these three values we can achiever our task using only a single 1-D array...

Real-World Applications of Edit Distance:

Spell Checking and Auto-CorrectionDNA Sequence AlignmentPlagiarism DetectionNatural Language ProcessingVersion Control SystemsString Matching...

Tags:
#Amazon #edit-distance #DSA #Dynamic Programming #Amazon #Dynamic Programming

Edit Distance using Recursion:

Edit Distance Using Dynamic Programming (Bottom-Up Approach):

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