Formatted question description: https://leetcode.ca/all/1143.html

1143. Longest Common Subsequence (Medium)

Given two strings text1 and text2, return the length of their longest common subsequence.

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. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.

 

If there is no common subsequence, return 0.

 

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 <= 1000
  • 1 <= text2.length <= 1000
  • The input strings consist of lowercase English characters only.

Related Topics:
Dynamic Programming

Similar Questions:

Solution 1. DP

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

dp[i][j] = 1 + dp[i-1][j-1]                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/

// 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][j] is only dependent on dp[i-1][j-1], 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-1][j-1].

// OJ: https://leetcode.com/problems/longest-common-subsequence/

// 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 = 1; i <= M; ++i) {
            int prev = 0;
            for (int j = 1; j <= N; ++j) {
                int cur = dp[j];
                if (s[i - 1] == t[j - 1]) dp[j] = 1 + prev;
                else dp[j] = max(dp[j], dp[j - 1]);
                prev = cur;
            }
        }
        return dp[N];
    }
};

Java

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int length1 = text1.length(), length2 = text2.length();
        int[][] dp = new int[length1 + 1][length2 + 1];
        for (int i = 1; i <= length1; i++) {
            char c1 = text1.charAt(i - 1);
            for (int j = 1; j <= length2; j++) {
                char c2 = text2.charAt(j - 1);
                if (c1 == c2)
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                else
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        int commonLength = dp[length1][length2];
        return commonLength;
    }
}

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