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.

Analysis

  • 注意subsequences可以是不连续的子串
  • 一般有两个字符串的情况要想到用二维数组dp[i][j], 表示string 1的前i的字符和string 2的前j个字符相比较 DP problem
  • State: dp[i][j]: Number of distinct subsequences of t.substr(1....i) in s(1....j)
  • Transfer: dp[i][j] = dp[i][j-1] + (dp[i-1][j-1] or 0)
    • 如果s和t最后一个字母一样，那么出现的次数是s除去最后一个字母(不match最后一个)+同时除去最后一个字母(match最后一个)
    • 如果s和t最后一个字母不一样，那么t在s中出现的次数和t在s除去最后一个字母出现的次数一样多
  • 如何想到这样用DP见yuanbin分析

Note

易犯错误：s和t哪个是横左边，哪个做纵坐标容易犯错。这道题s做j，t做i，分析起来会比较方便。

Code

public class Solution {
    public int numDistinct(String s, String t) {
        int m = t.length();
        int n = s.length();
        if (m > n) return 0;

        int[][] dp = new int[m+1][n+1];

        for (int i=0; i<=n; i++) dp[0][i]=1;

        for (int j=1; j<=n; j++){
            for (int i=1; i<=m; i++){
                dp[i][j] = dp[i][j-1] + (t.charAt(i-1)==s.charAt(j-1) ? dp[i-1][j-1] : 0);
            }
        }

        return dp[m][n];
    }
}

Note

  • DP经常用在这种看似需要用DFS，实际上不需要求具体的变换过程，只需要count number的题目，用来大大减小时间复杂度

Reference

Leetcode Yuanbin：分析清楚到位！