Longest common subsequence (LCS)

Time: O(MxN); Space: O(min(M,N)); medium

Given two strings text1 and text2, return the length of their longest common subsequence.

A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

Example 1:

Input: text1 = “abcde”, text2 = “ace”

Output: 3

Explanation:

  • The longest common subsequence is “ace” and its length is 3.

Example 2:

Input: text1 = “abc”, text2 = “abc”

Output: 3

Explanation:

  • The longest common subsequence is “abc” and its length is 3.

Example 3:

Input: text1 = “abc”, text2 = “def”

Output: 0

Explanation:

  • There is no such common subsequence, so the result is 0.

Constraints:

  • 1 <= len(text1) <= 1000

  • 1 <= len(text2) <= 1000

  • The input strings consist of lowercase English characters only.

Hints:

  1. Try dynamic programming. DP[i][j] represents the longest common subsequence of text1[0 … i] & text2[0 … j].

  2. DP[i][j] = DP[i - 1][j - 1] + 1 , if text1[i] == text2[j] DP[i][j] = max(DP[i - 1][j], DP[i][j - 1]) , otherwise

1. Dynamic programming [O(MxN), O(min(M,N))]

[1]:

class Solution1(object):
    """
    Time: O(M*N)
    Space: O(MIN(M,N))
    """
    def longestCommonSubsequence(self, text1, text2):
        """
        :type text1: str
        :type text2: str
        :rtype: int
        """
        if len(text1) < len(text2):
            return self.longestCommonSubsequence(text2, text1)

        dp = [[0 for _ in range(len(text2)+1)] for _ in range(2)]

        for i in range(1, len(text1)+1):
            for j in range(1, len(text2)+1):
                dp[i%2][j] = dp[(i-1)%2][j-1]+1 if text1[i-1] == text2[j-1] \
                                              else max(dp[(i-1)%2][j], dp[i%2][j-1])

        return dp[len(text1)%2][len(text2)]

[2]:

s = Solution1()

text1 = "abcde"
text2 = "ace"
assert s.longestCommonSubsequence(text1, text2) == 3

text1 = "abc"
text2 = "abc"
assert s.longestCommonSubsequence(text1, text2) == 3

text1 = "abc"
text2 = "def"
assert s.longestCommonSubsequence(text1, text2) == 0