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LeetCode

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Problem

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.

Code

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int len1 = text1.length();
        int len2 = text2.length();
        if(len1 < len2) return longestCommonSubsequence(text2, text1);

        int[][] dp = new int[2][len2 + 1];
        int currRow = 1;

        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                char a = text1.charAt(i - 1);
                char b = text2.charAt(j - 1);
                if (a == b) {
                    dp[currRow][j] = dp[currRow ^ 1][j - 1] + 1;
                } else {
                    dp[currRow][j] = Math.max(dp[currRow ^ 1][j], dp[currRow][j - 1]);
                }
            }
            currRow ^= 1;
        }

        return dp[currRow ^ 1][len2];
    }
}

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int len1 = text1.length();
        int len2 = text2.length();

        int[][] dp = new int[len1 + 1][len2 + 1];

        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                char a = text1.charAt(i - 1);
                char b = text2.charAt(j - 1);
                if (a == b) {
                    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[len1][len2];
    }
}