Edit Distance

To solve this coding challenge of finding the minimum number of operations required to convert one string into another, we can make use of dynamic programming (DP). The operations allowed are insertion, deletion, and replacement of characters. The goal is to build a 2D DP table where the cell \( dp[i][j] \) will store the minimum number of operations required to transform the first \( i \) characters of

word1

into the first \( j \) characters of

word2

. Let's break the problem down step-by-step.

Explanation

1. Initialization :
• Construct a 2D array
  dp
  where
  dp[i][j]
  will represent the minimum number of operations needed to convert the first
  i
  characters of
  word1
  to the first
  j
  characters of
  word2
  .
• The size of this array will be
  (len(word1)+1) x (len(word2)+1)
  since we need to account for the transforming process from zero-length substrings to the full length of both words.
2. Base Cases :
• dp[i][0] = i
  : If
  word2
  is empty, the only option is to delete all characters of
  word1
  . Therefore,
  dp[i][0]
  equals the number of characters in
  word1
  up to
  i
  .
• dp[0][j] = j
  : Similarly, if
  word1
  is empty, we need
  j
  insertions to transform it into the first
  j
  characters of
  word2
  .
3. Filling the DP Table :
• For each character pair
  i
  and
  j
  (where
  i
  ranges from 1 to
  len(word1)
  and
  j
  ranges from 1 to
  len(word2)
  ):
• If the characters are the same (
  word1[i-1] == word2[j-1]
  ), no new operation is needed, so the value at
  dp[i][j]
  will just carry over from
  dp[i-1][j-1]
  .
• If the characters are different, we consider the minimum value from three possible operations:
• Replacement: 1 +
  dp[i-1][j-1]
  (replace
  word1[i-1]
  with
  word2[j-1]
  )
• Insertion: 1 +
  dp[i][j-1]
  (insert
  word2[j-1]
  into
  word1
  )
• Deletion: 1 +
  dp[i-1][j]
  (delete
  word1[i-1]
  )
4. Return the Result :
• The value at
  dp[len(word1)][len(word2)]
  will hold the minimum number of operations required to transform
  word1
  into
  word2
  .

Detailed Steps in Pseudocode

5. Initialize the DP table :

                                            
Create a 2D array dp of size (len(word1) + 1) x (len(word2) + 1)

                                        

6. Base Cases :

                                            
for i from 0 to len(word1):
dp[i][0] = i  # Minimum operations to convert word1[0:i] to an empty string

for j from 0 to len(word2):
dp[0][j] = j  # Minimum operations to convert an empty string to word2[0:j]

                                        

7. DP Table population :

                                            
for i from 1 to len(word1):
for j from 1 to len(word2):
if word1[i-1] == word2[j-1]:
dp[i][j] = dp[i-1][j-1]  # Characters match, carry forward the value
else:
dp[i][j] = 1 + minimum(dp[i-1][j-1], dp[i][j-1], dp[i-1][j])  # Calculate minimum operations

                                        

8. Result :

                                            
return dp[len(word1)][len(word2)]

                                        

Pseudocode with detailed comments

                                            
# Create a 2D array to keep track of the minimum operations
dp = Create a 2D array of size (len(word1)+1) x (len(word2)+1)

# Initialize base cases for transforming `word1` to and from an empty string
for i from 0 to len(word1):
    dp[i][0] = i  # All deletions to transform `word1[0:i]` to an empty string

for j from 0 to len(word2):
    dp[0][j] = j  # All insertions to transform an empty string to `word2[0:j]`

# Iterate through each character pair in `word1` and `word2`
for i from 1 to len(word1):
    for j from 1 to len(word2):
        if word1[i-1] == word2[j-1]:  # Characters match
            dp[i][j] = dp[i-1][j-1]  # No new operation
        else:  # Characters do not match
            dp[i][j] = 1 + minimum(dp[i-1][j-1], dp[i][j-1], dp[i-1][j])  # Choose the minimum operation cost

# The result is the value at dp[len(word1)][len(word2)]
return dp[len(word1)][len(word2)]

                                        

This pseudocode provides a structured and detailed explanation of how to tackle the Edit Distance problem using dynamic programming. The comments clarify the rationale behind each step and help in understanding the flow of the solution.