Longest Common Subsequence

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.

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.

Code

Time Complexity: $O(n^2)$, Space Complexity: $O(n^2)$

重點就是把註解的 DP 關係式列出來，填好表格即可（注意 base case）。

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.size(), n = text2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
 
        for(int i = 1; i < m + 1; i++) {
            for(int j = 1; j < n + 1; j++) {
                if(text1[i - 1] == text2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = max(dp[i][j], max(dp[i - 1][j], dp[i][j - 1]));
                }
            }
        }
        return dp[m][n];
    }
};
 
/*
dp[i][j]: lcs from text1[0...i] and text2[0...j]
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]);
 
*/

Source

• Longest Common Subsequence - LeetCode