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# Edit distance and LCS (Longest Common Subsequence)

In standard Edit Distance where we are allowed 3 operations, insert, delete, and replace. Consider a variation of edit distance where we are allowed only two operations insert and delete, find edit distance in this variation.

Examples:

Input : str1 = "cat", st2 = "cut"
Output : 2
We are allowed to insert and delete. We delete 'a'
from "cat" and insert "u" to make it "cut".

Input : str1 = "acb", st2 = "ab"
Output : 1
We can convert "acb" to "ab" by removing 'c'.

One solution is to simply modify the Edit Distance Solution by making two recursive calls instead of three. An interesting solution is based on LCS.

1. Find LCS of two strings. Let the length of LCS be x
2. Let the length of the first string be m and the length of the second string be n. Our result is (m – x) + (n – x). We basically need to do (m – x) delete operations and (n – x) insert operations.

Implementation:

## C++

 // CPP program to find Edit Distance (when only two// operations are allowed, insert and delete) using LCS.#includeusing namespace std; int editDistanceWith2Ops(string &X, string &Y){    // Find LCS    int m = X.length(), n = Y.length();    int L[m+1][n+1];    for (int i=0; i<=m; i++)    {        for (int j=0; j<=n; j++)        {            if (i == 0 || j == 0)                L[i][j] = 0;            else if (X[i-1] == Y[j-1])                L[i][j] = L[i-1][j-1] + 1;            else                L[i][j] = max(L[i-1][j], L[i][j-1]);        }    }       int lcs  = L[m][n];     // Edit distance is delete operations +    // insert operations.    return (m - lcs) + (n - lcs);} /* Driver program to test above function */int main(){    string X = "abc", Y = "acd";    cout << editDistanceWith2Ops(X, Y);    return 0;}

## Java

 //Java program to find Edit Distance (when only two// operations are allowed, insert and delete) using LCS. class GFG {     static int editDistanceWith2Ops(String X, String Y) {        // Find LCS        int m = X.length(), n = Y.length();        int L[][] = new int[m + 1][n + 1];        for (int i = 0; i <= m; i++) {            for (int j = 0; j <= n; j++) {                if (i == 0 || j == 0) {                    L[i][j] = 0;                } else if (X.charAt(i - 1) == Y.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]);                }            }        }        int lcs = L[m][n];         // Edit distance is delete operations +        // insert operations.        return (m - lcs) + (n - lcs);    }     /* Driver program to test above function */    public static void main(String[] args) {        String X = "abc", Y = "acd";        System.out.println(editDistanceWith2Ops(X, Y));     }}/* This Java code is contributed by 29AjayKumar*/

## Python 3

 # Python 3 program to find Edit Distance# (when only two operations are allowed,# insert and delete) using LCS. def editDistanceWith2Ops(X, Y):     # Find LCS    m = len(X)    n = len(Y)    L = [[0 for x in range(n + 1)]            for y in range(m + 1)]    for i in range(m + 1):        for j in range(n + 1):            if (i == 0 or j == 0):                L[i][j] = 0            elif (X[i - 1] == Y[j - 1]):                L[i][j] = L[i - 1][j - 1] + 1            else:                L[i][j] = max(L[i - 1][j],                              L[i][j - 1])     lcs = L[m][n]     # Edit distance is delete operations +    # insert operations.    return (m - lcs) + (n - lcs) # Driver Codeif __name__ == "__main__":         X = "abc"    Y = "acd"    print(editDistanceWith2Ops(X, Y)) # This code is contributed by ita_c

## C#

 // C# program to find Edit Distance// (when only two operations are// allowed, insert and delete) using LCS.using System; class GFG{ static int editDistanceWith2Ops(String X,                                String Y){    // Find LCS    int m = X.Length, n = Y.Length;    int [ , ]L = new int[m + 1 , n + 1];    for (int i = 0; i <= m; i++)    {        for (int j = 0; j <= n; j++)        {            if (i == 0 || j == 0)            {                L[i , j] = 0;            }            else if (X[i - 1] == Y[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]);            }        }    }    int lcs = L[m , n];     // Edit distance is delete operations +    // insert operations.    return (m - lcs) + (n - lcs);} // Driver Codepublic static void Main(){    String X = "abc", Y = "acd";    Console.Write(editDistanceWith2Ops(X, Y));}} // This code is contributed// by 29AjayKumar



## Javascript



Output

2

Complexity Analysis:

• Time Complexity: O(m * n)
• Auxiliary Space: O(m * n)