readme.md

1143. Longest Common Subsequence

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

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.

• For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

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 <= text1.length, text2.length <= 1000
• text1 and text2 consist of only lowercase English characters.

Solution

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if text1[i - 1] == text2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

        return dp[m][n]

Thoughts

Time Complexity

O(m * n): Where m is the length of text1 and n is the length of text2. The nested loops over the dimensions of text1 and text2 determine the time complexity.

Space Complexity

O(m * n): We use a 2D array dp of size (m + 1) x (n + 1) to store the lengths of LCS up to each pair of indices. This represents the memory usage for storing the LCS values.

Space optimised solution

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        if len(text1) < len(text2):
            text1, text2 = text2, text1
        m, n = len(text1), len(text2)
        dp = [0] * (n + 1)

        for i in range(1, m + 1):
            prev = 0
            for j in range(1, n + 1):
                temp = dp[j]
                if text1[i - 1] == text2[j - 1]:
                    dp[j] = prev + 1
                else:
                    dp[j] = max(dp[j], dp[j - 1])
                prev = temp

        return dp[n]

Now the space complexity becomes O(min(m, n)) and time complexity remains the same.