Longest Common Subsequence

Problem Statement

• Example 1

Input: s1 = "abdca"
       s2 = "cbda"
Output: 3
Explanation: The longest common subsequence is "bda".

• Example 2

Input: s1 = "passport"
       s2 = "ppsspt"
Output: 5
Explanation: The longest common subsequence is "psspt".

Basic Solution

A basic brute-force solution could be to try all subsequences of s1 and s2 to find the longest one. We can match both the strings one character at a time. So for every index i in s1 and j in s2 we must choose between:

• If the character s1[i] matches s2[j], we can recursively match for the remaining lengths.

• If the character s1[i] does not match s2[j], we will start two new recursive calls by skipping one character separately from each string.

Code

Here is the code:

def find_LCS_length(s1, s2):
  return find_LCS_length_recursive(s1, s2, 0, 0)


def find_LCS_length_recursive(s1, s2, i1, i2):
  if i1 == len(s1) or i2 == len(s2):
    return 0

  if s1[i1] == s2[i2]:
    return 1 + find_LCS_length_recursive(s1, s2, i1 + 1, i2 + 1)

  c1 = find_LCS_length_recursive(s1, s2, i1, i2 + 1)
  c2 = find_LCS_length_recursive(s1, s2, i1 + 1, i2)

  return max(c1, c2)


def main():
  print(find_LCS_length("abdca", "cbda"))
  print(find_LCS_length("passport", "ppsspt"))


main()

Top-down Dynamic Programming with Memoization

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

The two changing values to 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 this:

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


def find_LCS_length_recursive(dp, s1, s2, i1, i2):
    if i1 == len(s1) or i2 == len(s2):
        return 0

    if dp[i1][i2] == -1:
        if s1[i1] == s2[i2]:
            dp[i1][i2] = 1 + \
                find_LCS_length_recursive(dp, s1, s2, i1 + 1, i2 + 1)
        else:
            c1 = find_LCS_length_recursive(dp, s1, s2, i1, i2 + 1)
            c2 = find_LCS_length_recursive(dp, s1, s2, i1 + 1, i2)
            dp[i1][i2] = max(c1, c2)

    return dp[i1][i2]


def main():
    print(find_LCS_length("abdca", "cbda"))
    print(find_LCS_length("passport", "ppsspt"))


main()

Bottom-up Dynamic Programming

Since we want to match all the subsequences 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 array’s two dimensions. So for every index i in string s1 and j in string s2, we will choose one of the following two options:

• If the character s1[i] matches s2[j], the length of the common subsequence would be one plus the length of the common subsequence till the i-1 and j-1 indexes in the two respective strings.

• If the character s1[i] does not match s2[j], we will take the longest subsequence by either skipping ith or jth character from the respective strings.

So our recursive formula would be:

if s1[i] == s2[j] 
    dp[i][j] = 1 + dp[i-1][j-1]
else 
    dp[i][j] = max(dp[i-1][j], dp[i][j-1])

Let’s draw this visually for “abcda” and “cbda”. Starting with a subsequence of zero lengths, if any string has zero length then the common subsequence will be of zero length:

From the above visualization, we can clearly see that the longest common subsequence is of length 3 – as shown by dp[4][5].

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

def find_LCS_length(s1, s2):
    n1, n2 = len(s1), len(s2)
    dp = [[0 for _ in range(n2+1)] for _ in range(n1+1)]
    maxLength = 0
    for i in range(1, n1+1):
        for j in range(1, n2+1):
            if s1[i - 1] == s2[j - 1]:
                dp[i][j] = 1 + dp[i - 1][j - 1]
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

            maxLength = max(maxLength, dp[i][j])
    return maxLength


def main():
    print(find_LCS_length("abdca", "cbda"))
    print(find_LCS_length("passport", "ppsspt"))


main()

Challenge

def find_LCS_length(s1, s2):
    n1, n2 = len(s1), len(s2)
    dp = [[0 for _ in range(n2+1)] for _ in range(2)]
    maxLength = 0
    for i in range(1, n1+1):
        for j in range(1, n2+1):
            if s1[i - 1] == s2[j - 1]:
                dp[i % 2][j] = 1 + dp[(i - 1) % 2][j - 1]
            else:
                dp[i % 2][j] = max(dp[(i - 1) % 2][j], dp[i % 2][j - 1])

            maxLength = max(maxLength, dp[i % 2][j])

    return maxLength


def main():
    print(find_LCS_length("abdca", "cbda"))
    print(find_LCS_length("passport", "ppsspt"))


main()

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