Longest Common Substring

Problem Statement

• Example 1

Input: s1 = "abdca"
       s2 = "cbda"
Output: 2
Explanation: The longest common substring is "bd".

• Example 2

Input: s1 = "passport"
       s2 = "ppsspt"
Output: 3
Explanation: The longest common substring is "ssp".

Basic Solution

A basic brute-force solution could be to try all substrings of s1 and s2 to find the longest common one. We can start matching both the strings one character at a time, so we have two options at any step:

• If the strings have a matching character, we can recursively match for the remaining lengths and keep a track of the current matching length.

• If the strings don’t match, we start two new recursive calls by skipping one character separately from each string and reset the matching length.

The length of the Longest Common Substring (LCS) will be the maximum number returned by the three recurse calls in the above two options.

Code

Here is the code:

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


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

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

  c1 = find_LCS_length_recursive(s1, s2, i1, i2 + 1, 0)
  c2 = find_LCS_length_recursive(s1, s2, i1 + 1, i2, 0)

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

Code

Here is the code for this:

def find_LCS_length(s1, s2):
  n1, n2 = len(s1), len(s2)
  maxLength = min(n1, n2)
  dp = [[[-1 for _ in range(maxLength)] for _ in range(n2)]
        for _ in range(n1)]
  return find_LCS_length_recursive(dp, s1, s2, 0, 0, 0)


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

  if dp[i1][i2][count] == -1:
    c1 = count
    if s1[i1] == s2[i2]:
      c1 = find_LCS_length_recursive(
        dp, s1, s2, i1 + 1, i2 + 1, count + 1)
    c2 = find_LCS_length_recursive(dp, s1, s2, i1, i2 + 1, 0)
    c3 = find_LCS_length_recursive(dp, s1, s2, i1 + 1, i2, 0)
    dp[i1][i2][count] = max(c1, max(c2, c3))

  return dp[i1][i2][count]


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 substrings 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 i in string s1 and j in string s2, we have two options:

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

• If the character at the s1[i] does not match s2[j], we don’t have any common substring.

So our recursive formula would be:

if s1[i] == s2[j] 
    dp[i][j] = 1 + dp[i-1][j-1]
else 
    dp[i][j] = 0

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

From the above visualization, we can clearly see that the longest common substring is of length 2– as shown by dp[3][3]. 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]
                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):
            dp[i % 2][j] = 0
            if s1[i - 1] == s2[j - 1]:
                dp[i % 2][j] = 1 + dp[(i - 1) % 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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