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 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 at 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.

Example 2: 

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

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 (as we only need to delete from str1 because we are reducing it to str2)
• minimum number of Insertions minInsert = n – len (as we are inserting x in str1 , x is a number of characters in str2 which are not taking part in the string which is longest common subsequence )

Below is the implementation of the above code:

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

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

Method 2: Top-down approach using memoization

This method also uses the idea of finding the length of the longest common subsequence of the given two sequences. But this approach is using a top-down approach using memoization.

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

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

Efficient approach: Space optimization

In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

• Create a 1D vector dp of size n+1.
• Set a base case by initializing the values of DP .
• Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
• Now initialize a variable prev used to store the previous computations and a variable temp to store the current computation.
• After every iteration assign the value of prev to temp for further iteration.
• At last return and print the final answer stored in dp[0].

Implementation: 

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

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

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