1143

1143. Longest Common Subsequence (Medium)

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.

Companies:
Amazon, Google, Karat, Indeed, Adobe, VMware

Related Topics:
String, Dynamic Programming

Similar Questions:

• Longest Palindromic Subsequence (Medium)
• Delete Operation for Two Strings (Medium)
• Shortest Common Supersequence (Hard)

Solution 1. DP

Let dp[i+1][j+1] be the length of the longest common subsequence of s[0..i] and t[0..j].

dp[i+1][j+1] = 1 + dp[i][j]                    // If s[i-1] == t[j-1]
             = max(dp[i+1][j], dp[i][j+1])     // If s[i-1] != t[j-1]
dp[i][0] = dp[0][i] = 0

// OJ: https://leetcode.com/problems/longest-common-subsequence/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int longestCommonSubsequence(string s, string t) {
        int M = s.size(), N = t.size();
        vector<vector<int>> dp(M + 1, vector<int>(N + 1));
        for (int i = 1; i <= M; ++i) {
            for (int j = 1; j <= N; ++j) {
                if (s[i - 1] == t[j - 1]) dp[i][j] = 1 + dp[i - 1][j - 1];
                else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[M][N];
    }
};

Solution 2. DP + Space Optimization

Since dp[i+1][j+1] is only dependent on dp[i][j], dp[i+1][j] and dp[i][j+1], we can reduce the space of dp array from N * N to 1 * N with a temporary variable storing dp[i][j].

// OJ: https://leetcode.com/problems/longest-common-subsequence/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(min(M, N))
class Solution {
public:
    int longestCommonSubsequence(string s, string t) {
        int M = s.size(), N = t.size();
        if (M < N) swap(M, N), swap(s, t);
        vector<int> dp(N + 1);
        for (int i = 0; i < M; ++i) {
            int prev = 0;
            for (int j = 0; j < N; ++j) {
                int cur = dp[j + 1];
                if (s[i] == t[j]) dp[j + 1] = prev + 1;
                else dp[j + 1] = max(dp[j], dp[j + 1]);
                prev = cur; 
            }
        }
        return dp[N];
    }
};