LeetCode 1143 - Longest Common Subsequence

Difficulty: medium

Similar problem: LeetCode 718 - Maximum Length of Repeated Subarray

Problem Description

English (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.

Chinese (最长公共子序列)

给定两个字符串 text1 和 text2，返回这两个字符串的最长 公共子序列 的长度。如果不存在 公共子序列，返回 0。

一个字符串的 子序列 是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。

例如，"ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。
两个字符串的 公共子序列 是这两个字符串所共同拥有的子序列。

示例 1：

输入：text1 = "abcde", text2 = "ace"
输出：3
解释：最长公共子序列是 "ace" ，它的长度为 3 。

示例 2：

输入：text1 = "abc", text2 = "abc"
输出：3
解释：最长公共子序列是 "abc" ，它的长度为 3 。

示例 3：

输入：text1 = "abc", text2 = "def"
输出：0
解释：两个字符串没有公共子序列，返回 0 。

提示：

• 1 <= text1.length, text2.length <= 1000
• text1 和 text2 仅由小写英文字符组成。

Solution

C++

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int text1_size = text1.size(), text2_size = text2.size();

        vector<vector<int>> dp(text1_size + 1, vector<int>(text2_size + 1));
        for (int i = 0; i < text1_size; ++i) {
            for (int j = 0; j < text2_size; ++j) {
                if (text1[i] == text2[j]) {
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                } else {
                    dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);
                }
            }
        }

        return dp[text1_size][text2_size];
    }
};

Python

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        text1_len, text2_len = len(text1), len(text2)

        dp = [[0 for j in range(text2_len+1)] for i in range(text1_len+1)]
        for i in range(text1_len):
            for j in range(text2_len):
                if text1[i] == text2[j]:
                    dp[i+1][j+1] = dp[i][j] + 1
                else:
                    dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])

        return dp[-1][-1]

Advanced

It’s available to compress the dp array into one dimension.