[LintCode/LeetCode] Distinct Subsequences [一维DP]

Problem

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).

Example

Given S = "rabbbit", T = "rabbit", return 3.

Challenge

Do it in O(n2) time and O(n) memory.

O(n2) memory is also acceptable if you do not know how to optimize memory.

Note

用动规方法做：
建立长度为m+1和n+1的二维dp数组，dp[i][j]表示S的第0到i位子串包含不同的T的第0到j位子串的个数。
初始化：当T的子串长度为0时，dp[i][0] = 1；当S的子串长度为0时，dp[0][i] = 0；当S和T子串都为0时，0包含0，故dp[0][0] = 1。
两次循环S和T，若S的最后一位和T的最后一位不同，那么S增加一位不会增加更多的包含关系，即dp[i][j]仍然等于dp[i-1][j]。若S的最后一位和T的最后一位相等，最后的结果一定也包含S和T各减一位的结果，如S="abb"，T="ab"，所以dp[i][j]包含dp[i-1][j-1]，再与与dp[i-1][j]相加。

见下面的例子。

     0    A    B    C

0    1    0    0    0

C    1    0    0    0
    
A    1    1    0    0  

B    1    1    1    0   

B    1    1    2    0
 
B    1    1    3    0

C    1    1    3    3

Solution

public class Solution {
    public int numDistinct(String S, String T) {
        int m = S.length(), n = T.length();
        if (m < n) return 0;
        int[][] dp = new int[m+1][n+1];
        for (int i = 0; i <= m; i++) dp[i][0] = 1;
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = dp[i-1][j];
                if (S.charAt(i-1) == T.charAt(j-1)) dp[i][j] += dp[i-1][j-1];
            }
        }
        return dp[m][n];
    }
}

One-dimension DP, O(n) space

public class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length();
        int n = t.length();
        int[] dp = new int[n+1];
        dp[0] = 1;
        for (int i = 1; i <= m; i++) {
            for (int j = n; j >= 1; j--) {                   
                if (s.charAt(i-1) == t.charAt(j-1)) {
                    dp[j] += dp[j-1];
                }
            }
        }
        return dp[n];
    }
}