[LeetCode] 115. Distinct Subsequences

Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.

Thought process:
  1. Sub-problem: count the number of distinct subsequences of a substring of T and a substring of S.
  2. Function: 
    1. If the last character of S and T are not the same, the last character of S will not contribute to the subsequences. f[i][j] = f[i - 1][j].
    2. Otherwise, we add the number of subsequences where the last character of S is in. f[i][j] = f[i - 1][j] + f[i - 1][j - 1].
  3. Initialization: 
    1. If T is empty, there is one way to get an empty string out of S, which is to choose nothing. f[i][0] = 1.
    2. If S is empty and T is not, there is no way to get T out of S. f[0][j] = 0.
  4. Answer: f[s.length][t.length].

Solution:

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public class Solution {
    public int numDistinct(String s, String t) {
        int sLength = s.length();
        int tLength = t.length();
        
        int[][] f = new int[sLength + 1][tLength + 1];
        
        for (int i = 0; i <= sLength; i++) {
            f[i][0] = 1;
        }
        
        for (int i = 1; i <= sLength; i++) {
            for (int j = 1; j <= tLength; j++) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    f[i][j] = f[i - 1][j] + f[i - 1][j - 1];
                } else {
                    f[i][j] = f[i - 1][j];
                }
            }
        }
        
        return f[sLength][tLength];
    }
}
Time complexity: 
Say m = t.length and n = s.length, the overall time complexity is O(mn).
Location: San Jose, CA, USA

Comments

  1. UnknownDecember 23, 2021 at 5:39 AM

    It is a best solution found that very popular and helpful:
    https://www.youtube.com/watch?v=6ngJWBA-nZk&ab_channel=EricProgramming

    ReplyDelete

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