Minimum number of deletions and insertions to transform one string into another

Given two strings ‘str1’ and ‘str2’ of size m and n respectively. The task is to remove/delete and insert minimum number of characters from/in str1 so as to transform it into str2. It could be possible that the same character needs to be removed/deleted from one point of str1 and inserted to some another point.

Examples:

Input : str1 = "heap", str2 = "pea" 
Output : Minimum Deletion = 2 and
         Minimum Insertion = 1
p and h deleted from heap
Then, p is inserted at the beginning
One thing to note, though p was required yet
it was removed/deleted first from its position and
then it is inserted to some other position.
Thus, p contributes one to the deletion_count
and one to the insertion_count.

Input : str1 = "geeksforgeeks", str2 = "geeks"
Output : Minimum Deletion = 8
         Minimum Insertion = 0       

A simple solution is to consider all subsequences of str1 and for each subsequence calculate minimum deletions and insertions so as to transform it into str2. A very complex method and the time complexity of this solution is exponential.

An efficient approach uses the concept of finding the length of the longest common subsequence of the given two sequences.



Algorithm:

-->str1 and str2 be the given strings.
-->m and n be their lengths respectively.
-->len be the length of the longest 
   common subsequence of str1 and str2
-->// minimum number of deletions 
   minDel = m - len
-->// minimum number of Insertions 
   minInsert = n - len

C++

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// Dynamic Programming C++ implementation to find 
// minimum number of deletions and insertions
#include <bits/stdc++.h>
    
using namespace std;  
    
// Returns length of length common subsequence
// for str1[0..m-1], str2[0..n-1]
int lcs(string str1, string str2, int m, int n )
{
   int L[m+1][n+1];
   int i, j;
    
   // Following steps build L[m+1][n+1] in bottom
   // up fashion. Note that L[i][j] contains 
   // length of LCS of str1[0..i-1] and str2[0..j-1] 
   for (i=0; i<=m; i++)
   {
     for (j=0; j<=n; j++)
     {
       if (i == 0 || j == 0)
         L[i][j] = 0;
    
       else if (str1.at(i-1) == str2.at(j-1))
         L[i][j] = L[i-1][j-1] + 1;
    
       else
         L[i][j] = max(L[i-1][j], L[i][j-1]);
     }
   }
      
   // L[m][n] contains length of LCS 
   // for X[0..n-1] and Y[0..m-1] 
   return L[m][n];
}
    
// function to find minimum number 
// of deletions and insertions 
void printMinDelAndInsert(string str1, string str2)  
{
    int m = str1.size();
    int n = str2.size();
      
    int len = lcs(str1, str2, m, n);
      
    cout << "Minimum number of deletions = "
         << (m - len) << endl;
        
    cout << "Minimum number of insertions = "   
         << (n - len) << endl;  
}
    
// Driver program to test above
int main()
{
  string str1 = "heap";
  string str2 = "pea";  
  printMinDelAndInsert(str1, str2); 
  return 0;

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Java

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// Dynamic Programming Java implementation 
// to find minimum number of deletions and 
// insertions
import java.io.*;
  
class GFG {
      
    // Returns length of length common 
    // subsequence for str1[0..m-1], 
    // str2[0..n-1]
    static int lcs(String str1, String str2,
                           int m, int n )
    {
        int L[][] = new int[m+1][n+1];
        int i, j;
  
    // Following steps build L[m+1][n+1] in
    // bottom up fashion. Note that L[i][j]
    // contains length of LCS of str1[0..i-1]
    // and str2[0..j-1] 
    for (i = 0; i <= m; i++)
    {
        for (j = 0; j <= n; j++)
        {
            if (i == 0 || j == 0)
            L[i][j] = 0;
  
        else if (str1.charAt(i-1) == str2.charAt(
                                       j-1))
            L[i][j] = L[i-1][j-1] + 1;
  
        else
            L[i][j] = Math.max(L[i-1][j], 
                               L[i][j-1]);
        }
    }
      
    // L[m][n] contains length of LCS 
    // for X[0..n-1] and Y[0..m-1] 
    return L[m][n];
    }
  
    // function to find minimum number 
    // of deletions and insertions 
    static void printMinDelAndInsert(String str1, 
                                    String str2) 
    {
        int m = str1.length();
        int n = str2.length();
      
        int len = lcs(str1, str2, m, n);
      
        System.out.println("Minimum number of "+
                           "deletions = ");
        System.out.println(m - len);
      
        System.out.println("Minimum number of "+
                           "insertions = "); 
        System.out.println(n - len); 
    }
  
    // Driver program to test above
    public static void main(String[] args)
    {
    String str1 = new String("heap");
    String str2 = new String("pea"); 
    printMinDelAndInsert(str1, str2); 
    
}
// This code is contributed by Prerna Saini

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Python3

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# Dynamic Programming Python3 
# implementation to find minimum 
# number of deletions and insertions
  
# Returns length of length 
# common subsequence for
# str1[0..m-1], str2[0..n-1]
def lcs(str1, str2, m, n):
      
    L = [[0 for i in range(n + 1)]
            for i in range(m + 1)]
              
    # Following steps build L[m+1][n+1]
    # in bottom up fashion. Note that 
    # L[i][j] contains length of LCS 
    # of str1[0..i-1] and str2[0..j-1] 
    for i in range(m + 1):
        for j in range(n + 1):
            if (i == 0 or j == 0):
                L[i][j] = 0
            elif(str1[i - 1] == str2[j - 1]):
                L[i][j] = L[i - 1][j - 1] + 1
            else:
                L[i][j] = max(L[i - 1][j], 
                              L[i][j - 1])
                                
    # L[m][n] contains length of LCS
    # for X[0..n-1] and Y[0..m-1] 
    return L[m][n]
  
# function to find minimum number 
# of deletions and insertions
def printMinDelAndInsert(str1, str2):
    m = len(str1)
    n = len(str2)
    leng = lcs(str1, str2, m, n)
    print("Minimum number of deletions = "
                       m - leng, sep = ' ')
    print("Minimum number of insertions = "
                        n - leng, sep = ' ')
                          
# Driver Code
str1 = "heap"
str2 = "pea"
printMinDelAndInsert(str1, str2)
  
# This code is contributed 
# by sahilshelangia

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C#

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// Dynamic Programming C# implementation 
// to find minimum number of deletions and 
// insertions
using System;
  
class GFG {
      
    // Returns length of length common 
    // subsequence for str1[0..m-1], 
    // str2[0..n-1]
    static int lcs(string str1, string str2,
                              int m, int n )
    {
        int [,]L = new int[m+1,n+1];
        int i, j;
  
        // Following steps build L[m+1][n+1] in
        // bottom up fashion. Note that L[i][j]
        // contains length of LCS of str1[0..i-1]
        // and str2[0..j-1] 
        for (i = 0; i <= m; i++)
        {
            for (j = 0; j <= n; j++)
            {
                if (i == 0 || j == 0)
                    L[i,j] = 0;
      
                else if (str1[i-1] == str2[j-1])
                    L[i,j] = L[i-1,j-1] + 1;
      
                else
                    L[i,j] = Math.Max(L[i-1,j], 
                                       L[i,j-1]);
            }
        }
          
        // L[m][n] contains length of LCS 
        // for X[0..n-1] and Y[0..m-1] 
        return L[m,n];
    }
  
    // function to find minimum number 
    // of deletions and insertions 
    static void printMinDelAndInsert(string str1, 
                                     string str2) 
    {
        int m = str1.Length;
        int n = str2.Length;
      
        int len = lcs(str1, str2, m, n);
      
        Console.Write("Minimum number of "+
                                "deletions = ");
        Console.WriteLine(m - len);
      
        Console.Write("Minimum number of "+
                               "insertions = "); 
        Console.Write(n - len); 
    }
  
    // Driver program to test above
    public static void Main()
    {
        string str1 = new string("heap");
        string str2 = new string("pea"); 
        printMinDelAndInsert(str1, str2); 
    
}
  
// This code is contributed by nitin mittal.

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Output:

Minimum number of deletions = 2
Minimum number of insertions = 1

Time Complexity: O(m * n)

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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