1143. Longest Common Subsequence

1. Description

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.

2. Example

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.

3. Constraints

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

4. Solutions

Dynamic Programming

m = text1.size(), n = text2.size()
Time complexity: O(mn)
Space complexity: O(min(m, n))

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class Solution {
public:
    int longestCommonSubsequence(const string &text1, const string &text2) {
        bool text1_shorter = text1.size() < text2.size();
        const string &t1 = text1_shorter ? text2 : text1;
        const string &t2 = text1_shorter ? text1 : text2;
        const int m = t1.size(), n = t2.size();

        vector<int> sequence(n, 0);
        for (int i = 0; i < m; ++i) {
            int prev = sequence[0];
            sequence[0] = max(sequence[0], t1[i] == t2[0] ? 1 : 0);
            for (int j = 1; j < n; ++j) {
                int temp = sequence[j];
                if (t1[i] == t2[j]) {
                    sequence[j] = prev + 1;
                } else {
                    sequence[j] = max(sequence[j], sequence[j - 1]);
                }

                prev = temp;
            }
        }

        return sequence.back();
    }
};