Leetcode 1143 - Longest Common Subsequence

Link to problem

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.

Solution

If we are finding the longest common subsequence of two strings, "abcde" and "ace" and starting from the first character:

• (a) 'a' == 'a', therefore, our problem becomes a sub problem of finding the longest common subsequence of "bcde" and "ce".
• (b) Then we compare 'b' and 'c', which are not equal. Therefore, the longest common subsequence might be longestCommonSubsequence("bcde", "e") or longestCommonSubsequence("cde", "ce"). We will be using the longer one.

By the above observation we could solve this problem using dynamic programming with a 2D DP table where:

1. x and y-axis are the two string’s characters
2. If str[i] == str[j], then dp[i][j] will be dp[i-1][j-1] + 1. (a)
3. If str[i] != str[j], then dp[i][j] will be the maximum of dp[i][j-1] and dp[i-1][j]. (longest common subsequence of str1[0:i], str2[0:j-1] and str1[0:i-1], str2[0:j]. (b)

Implementation

Did a little trick to remove the boundary checks in this implementation. The DP table’s dimention is one larger than the inputs’ sizes. This way, the i=0 row and j=0 column are the initial state 0. See the table below:

		a	b	c	d	e
	0	0	0	0	0	0
a	0	1	1	1	1	1
c	0	1	1	2	2	2
e	0	1	1	2	2	3

class Solution {
public:
    int longestCommonSubsequence(const string& text1, const string& text2) {
        size_t m = text1.size() + 1;
        size_t n = text2.size() + 1;
        vector<vector<int>> dp(m, vector<int>(n));

        for (size_t i = 1; i < m; i++) {
            for (size_t j = 1; j < n; j++) {
                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.back().back();
    }
};

impl Solution {
    pub fn longest_common_subsequence(text1: String, text2: String) -> i32 {
        let (m, n) = (text1.len()+1, text2.len()+1);
        let mut dp: Vec<Vec<i32>> = vec![vec![0; n]; m];

        for (i, c1) in text1.chars().enumerate() {
            for (j, c2) in text2.chars().enumerate() {
                dp[i+1][j+1] = if c1 == c2 {
                    dp[i][j] + 1
                } else {
                    std::cmp::max(dp[i][j+1], dp[i+1][j])
                }
            }
        }

        dp[m-1][n-1]
    }
}