1143. Longest Common Subsequence.md

1143. Longest Common Subsequence

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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.

Solution

• java
  • mine Runtime: 9 ms, faster than 89.03%, Memory Usage: 42.9 MB, less than 100.00% of Java online submissions

    //O(m*n) time O(m*n)space
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length();
        int n = text2.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
    

  • the most votes Runtime: 9 ms, faster than 89.03%, Memory Usage: 36.9 MB, less than 100.00% of Java online submissions

    //O(m*n) time O(2*min(m,n))space
    public int longestCommonSubsequence(String s1, String s2) {
        int m = s1.length(), n = s2.length();
        if (m < n) return longestCommonSubsequence(s2, s1);
        int[][] dp = new int[2][n + 1];
        for (int i = 0, k = 1; i < m; ++i, k ^= 1)
            for (int j = 0; j < n; ++j)
                if (s1.charAt(i) == s2.charAt(j)) dp[k][j + 1] = 1 + dp[k ^ 1][j];
                else dp[k][j + 1] = Math.max(dp[k ^ 1][j + 1], dp[k][j]);
        return dp[m % 2][n];
    }