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

Difficulty Level : Medium
Last Updated : 27 May, 2021

Given two strings ‘str1’ and ‘str2’ of size m and n respectively. The task is to remove/delete and insert the minimum number of characters from/in str1 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.

Example 1:

Input : 
str1 = "heap", str2 = "pea" 
Output : 
Minimum Deletion = 2 and
Minimum Insertion = 1
Explanation:
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.

Pictorial Representation of pea

Example 2:

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

Pictorial Representation of geeksforgeeks

Simple Approach:

A simple approach 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.

Efficient Approach:

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

Below is the implementation of the above code:

C++

// 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 Code
int main()
{
    string str1 = "heap";
    string str2 = "pea";
   
      // Function Call
    printMinDelAndInsert(str1, str2);
    return 0;
}

Java

// 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 code
    public static void main(String[] args)
    {
        String str1 = new String("heap");
        String str2 = new String("pea");
       
          // Function Call
        printMinDelAndInsert(str1, str2);
    }
}
// This code is contributed by Prerna Saini

Python3

# 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"
 
# Function Call
printMinDelAndInsert(str1, str2)
 
# This code is contributed
# by sahilshelangia

C#

// 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 code
    public static void Main()
    {
        string str1 = new string("heap");
        string str2 = new string("pea");
       
          // Function Call
        printMinDelAndInsert(str1, str2);
    }
}
 
// This code is contributed by nitin mittal.

Javascript

<script>
    // Dynamic Programming Javascript implementation
    // to find minimum number of deletions and
    // insertions
 
    // Returns length of length common
    // subsequence for str1[0..m-1],
    // str2[0..n-1]
    function lcs(str1, str2, m, n)
    {
        let L = new Array(m + 1);
             
        let i, j;
         
        for (i = 0; i <= m; i++)
        {
            L[i] = new Array(n + 1);
            for (j = 0; j <= n; j++)
            {
                L[i][j] = 0;
            }
        }
  
        // 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
    function printMinDelAndInsert(str1, str2)
    {
        let m = str1.length;
        let n = str2.length;
  
        let len = lcs(str1, str2, m, n);
  
        document.write("Minimum number of "
                           + "deletions = ");
        document.write((m - len) + "</br>");
  
        document.write("Minimum number of "
                           + "insertions = ");
        document.write((n - len) + "</br>");
    }
     
    let str1 = "heap";
    let str2 = "pea";
 
    // Function Call
    printMinDelAndInsert(str1, str2);
     
    // This code is contributed by rameshtravel07.
</script>

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 write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes