Edit Distance

Problem Statement

• Example 1

Input: s1 = "bat"
       s2 = "but"
Output: 1
Explanation: We just need to replace 'a' with 'u' to transform s1 to s2.

• Example 2

Input: s1 = "abdca"
       s2 = "cbda"
Output: 2
Explanation: We can replace first 'a' with 'c' and delete second 'c'.

• Example 3

Input: s1 = "passpot"
       s2 = "ppsspqrt"
Output: 3 
Explanation: Replace 'a' with 'p', 'o' with 'q', and insert 'r'.

Basic Solution

A basic brute-force solution could be to try all operations (one by one) on each character of s1. We can iterate through s1 and s2 together. Let’s assume index1 and index2 point to the current indexes of s1 and s2 respectively, so we have two options at every step:

• If the strings have a matching character, we can recursively match for the remaining lengths.

• If the strings don’t match, we start three new recursive calls representing the three edit operations. Whichever recursive call returns the minimum count of operations will be our answer.

Code

Here is the recursive implementation:

def find_min_operations(s1, s2):
    return find_min_operations_recursive(s1, s2, 0, 0)


def find_min_operations_recursive(s1, s2, i1, i2):

    n1, n2 = len(s1), len(s2)
    # if we have reached the end of s1, then we have to insert all the remaining characters of s2
    if i1 == n1:
        return n2 - i2

    # if we have reached the end of s2, then we have to delete all the remaining characters of s1
    if i2 == n2:
        return n1 - i1

    # If the strings have a matching character, we can recursively match for the remaining lengths
    if s1[i1] == s2[i2]:
        return find_min_operations_recursive(s1, s2, i1 + 1, i2 + 1)

    # perform deletion
    c1 = 1 + find_min_operations_recursive(s1, s2, i1 + 1, i2)
    # perform insertion
    c2 = 1 + find_min_operations_recursive(s1, s2, i1, i2 + 1)
    # perform replacement
    c3 = 1 + find_min_operations_recursive(s1, s2, i1 + 1, i2 + 1)

    return min(c1, min(c2, c3))


def main():
    print(find_min_operations("bat", "but"))
    print(find_min_operations("abdca", "cbda"))
    print(find_min_operations("passpot", "ppsspqrt"))


main()

Top-down Dynamic Programming with Memoization

We can use an array to store the already solved subproblems.

The two changing values in our recursive function are the two indexes, i1 and i2. Therefore, we can store the results of all the subproblems in a two-dimensional array. (Another alternative could be to use a hash-table whose key would be a string (i1 + “|” + i2)).

Code

Here is the code for Top-down DP approach:

def find_min_operations(s1, s2):
    dp = [[-1 for _ in range(len(s2)+1)] for _ in range(len(s1)+1)]
    return find_min_operations_recursive(dp, s1, s2, 0, 0)


def find_min_operations_recursive(dp, s1,  s2,  i1,  i2):
    n1, n2 = len(s1), len(s2)
    if dp[i1][i2] == -1:
        # if we have reached the end of s1, then we have to insert all the remaining  characters of s2
        if i1 == n1:
            dp[i1][i2] = n2 - i2

        # if we have reached the end of s2, then we have to delete all the remaining characters of s1
        elif i2 == n2:
            dp[i1][i2] = n1 - i1

        # If the strings have a matching character, we can recursively match for the remaining lengths
        elif s1[i1] == s2[i2]:
            dp[i1][i2] = find_min_operations_recursive(
                dp, s1, s2, i1 + 1, i2 + 1)
        else:
            c1 = find_min_operations_recursive(
                dp, s1, s2, i1 + 1, i2)  # delete
            c2 = find_min_operations_recursive(
                dp, s1, s2, i1, i2 + 1)  # insert
            c3 = find_min_operations_recursive(
                dp, s1, s2, i1 + 1, i2 + 1)  # replace
            dp[i1][i2] = 1 + min(c1, min(c2, c3))

    return dp[i1][i2]


def main():
    print(find_min_operations("bat", "but"))
    print(find_min_operations("abdca", "cbda"))
    print(find_min_operations("passpot", "ppsspqrt"))


main()

What is the time and space complexity of the above solution?

Bottom-up Dynamic Programming

Since we want to match all the characters of the given two strings, we can use a two-dimensional array to store our results. The lengths of the two strings will define the size of the two dimensions of the array. So for every index ‘i1’ in string ‘s1’ and ‘i2’ in string ‘s2’, we will choose one of the following options:

• If the character s1[i1] matches s2[i2], the count of the edit operations will be equal to the count of the edit operations for the remaining strings.

• If the character s1[i1] does not match s2[i2], we will take the minimum count from the remaining strings after performing any of the three edit operations.

So our recursive formula would be:

if s1[i1] == s2[i2] 
    dp[i1][i2] = dp[i1-1][i2-1]
else 
    dp[i1][i2] = 1 + min(dp[i1-1][i2],  // delete
                        dp[i1][i2-1],   // insert 
                        dp[i1-1][i2-1]) // replace

Code

Here is the code for our bottom-up dynamic programming approach:

def find_min_operations(s1, s2):
  n1, n2 = len(s1), len(s2)
  dp = [[-1 for _ in range(n2+1)] for _ in range(n1+1)]

  # if s2 is empty, we can remove all the characters of s1 to make it empty too
  for i1 in range(n1+1):
    dp[i1][0] = i1

  # if s1 is empty, we have to insert all the characters of s2
  for i2 in range(n2+1):
    dp[0][i2] = i2

  for i1 in range(1, n1+1):
    for i2 in range(1, n2+1):
      # If the strings have a matching character, we can recursively match for the remaining lengths
      if s1[i1 - 1] == s2[i2 - 1]:
        dp[i1][i2] = dp[i1 - 1][i2 - 1]
      else:
        dp[i1][i2] = 1 + min(dp[i1 - 1][i2],  # delete
                             min(dp[i1][i2 - 1],  # insert
                                 dp[i1 - 1][i2 - 1]))  # replace

  return dp[n1][n2]


def main():
  print(find_min_operations("bat", "but"))
  print(find_min_operations("abdca", "cbda"))
  print(find_min_operations("passpot", "ppsspqrt"))


main()

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