1143. Longest Common Subsequence

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.

Solution

Approach 1: DP.

Code (Python)

Approach 1:

    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        if not text1 or not text2:
            return 0
        lcs = [[0 for _ in range(len(text2) + 1)] for _ in range(len(text1) + 1)]
        for i in range(1, len(lcs)):
            for j in range(1, len(lcs[0])):
                if text1[i-1] == text2[j-1]:
                    lcs[i][j] = max(lcs[i-1][j-1] + 1, lcs[i-1][j], lcs[i][j-1])
                else:
                    lcs[i][j] = max(lcs[i-1][j-1], lcs[i-1][j], lcs[i][j-1])
        return lcs[-1][-1]

        # same idea, but saves space by keeping only 1 row in the table
        ```
        lcs = [0 for _ in range(len(text2) + 1)]
        for i in range(1, len(text1) + 1):
            diagonal, vertical, horizontal = lcs[0], lcs[1], 0
            for j in range(1, len(lcs)):
                new_diagonal = lcs[j]
                if text1[i-1] == text2[j-1]:
                    lcs[j] = max(diagonal + 1, vertical, horizontal)
                else:
                    lcs[j] = max(diagonal, vertical, horizontal)
                if j < len(lcs) - 1:
                    diagonal, vertical, horizontal = new_diagonal, lcs[j+1], lcs[j]
        return lcs[-1]
        ```

Code (C++)

Approach 1:

Approach 2: