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LongestCommonSubsequence.md

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[1143]Longest Common Subsequence

Explanation

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.

Tag

dynamic-programming

Corner Cases

兩字串長度不一 || 子序列相反

Input: text1 = "abcde", text2 = "ace" 

Input: text1 = "adefb", text2 = "bcgan"

Solution

Approach 1 - Dynamic-Programming

Algo Goal

找到兩字串的LCS(Longest Common Subsequence)

dp_array[i][j] = char_array1[i - 1] == char_array2[j - 1] ? 
                    dp_array[i - 1][j - 1] + 1 : Math.max(dp_array[i][j - 1], dp_array[i - 1][j]);

Processing

定義dp[i][j] 為text1前i個字元與text2前j個字元的LCS

當掃描到text1第i個字元與text2第j個字元時:

如兩字元相等 -> 可判斷LCS為text1前(i - 1)個字元與text2前(j - 1)的LCS + 1;

如兩字元不相等 -> 則需退一步取text1前i個字元與text2前(j - 1)個字元的LCS && text1前(i - 1)個字元與text2前j個字元的LCS之最大值

Code

    public int longestCommonSubsequence(String text1, String text2) {
        int[][] dp_array = new int[1000][1000];
        int len1 = text1.length(), len2 = text2.length();
        char[] char_array1 = text1.toCharArray(), char_array2 = text2.toCharArray();

        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++){
                if(char_array1[i - 1] == char_array2[j - 1]){
                    dp_array[i][j] = dp_array[i - 1][j - 1] + 1;
                }else{
                    dp_array[i][j] = Math.max(dp_array[i][j - 1], dp_array[i - 1][j]);
                }
            }
        }
        return dp_array[len1][len2];
    }

Analysis

  • 37/37 cases passed (17 ms)

  • Your runtime beats 45.08 % of java submissions

  • Your memory usage beats 100 % of java submissions (49.8 MB)

    Time Complexity: $O(n^2)$

    Space Complexity: $O(n^2)$