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

Difficulty Level : Medium
Last Updated : 02 Oct, 2020

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.

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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