[LeetCode] 115. Distinct Subsequences

[grandyang]

 

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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 1:

Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^

Example 2:

Input: S = "babgbag", T = "bag"
Output: 5
Explanation:

As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)

babgbag
^^ ^
babgbag
^^    ^
babgbag
^    ^^
babgbag
  ^  ^^
babgbag
    ^^^

 

看到有关字符串的子序列或者配准类的问题，首先应该考虑的就是用动态规划 Dynamic Programming 来求解，这个应成为条件反射。而所有 DP 问题的核心就是找出状态转移方程，想这道题就是递推一个二维的 dp 数组，其中 dp[i][j] 表示s中范围是 [0, i] 的子串中能组成t中范围是 [0, j] 的子串的子序列的个数。下面我们从题目中给的例子来分析，这个二维 dp 数组应为：

 

  Ø r a b b b i t
Ø 1 1 1 1 1 1 1 1
r 0 1 1 1 1 1 1 1
a 0 0 1 1 1 1 1 1
b 0 0 0 1 2 3 3 3
b 0 0 0 0 1 3 3 3
i 0 0 0 0 0 0 3 3
t 0 0 0 0 0 0 0 3 

 

首先，若原字符串和子序列都为空时，返回1，因为空串也是空串的一个子序列。若原字符串不为空，而子序列为空，也返回1，因为空串也是任意字符串的一个子序列。而当原字符串为空，子序列不为空时，返回0，因为非空字符串不能当空字符串的子序列。理清这些，二维数组 dp 的边缘便可以初始化了，下面只要找出状态转移方程，就可以更新整个 dp 数组了。我们通过观察上面的二维数组可以发现，当更新到 dp[i][j] 时，dp[i][j] >= dp[i][j - 1] 总是成立，再进一步观察发现，当 T[i - 1] == S[j - 1] 时，dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]，若不等， dp[i][j] = dp[i][j - 1]，所以，综合以上，递推式为：

dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)

根据以上分析，可以写出代码如下：

 

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.size(), n = t.size();
        vector<vector<long>> dp(n + 1, vector<long>(m + 1));
        for (int j = 0; j <= m; ++j) dp[0][j] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                dp[i][j] = dp[i][j - 1] + (t[i - 1] == s[j - 1] ? dp[i - 1][j - 1] : 0);
            }
        }
        return dp[n][m];
    }
};

 

Github 同步地址：

#115

 

参考资料：

https://leetcode.com/problems/distinct-subsequences/

https://leetcode.com/problems/distinct-subsequences/discuss/37327/Easy-to-understand-DP-in-Java

https://leetcode.com/problems/distinct-subsequences/discuss/37412/Any-better-solution-that-takes-less-than-O(n2)-space-while-in-O(n2)-time

 

LeetCode All in One 题目讲解汇总(持续更新中...)

[lld2006]

no need to use a two dimensional array. just use two one dimensional vectors of s.size()+1

[lld2006]

可以用一维数组来solve这个题目，写起来也意外的很简单， 另外dp可能会数组越界， 用long比较好。

class Solution {
public:
    int numDistinct(string s, string t) {
      if(t.size() > s.size()) return 0;
      if(t.size()== s.size()) return s==t;
      vector < long> dp(t.size()+1,0);  
      dp[0] = 1;
      for(int i = 0; i < s.size(); ++i){
        for(int j =static_cast(t.size())-1; j>=0; --j){
          //dp[j] means number of subsequence that has matched j char in t, need to compare with t[j]
          if(t[j]==s[i]){
            dp[j+1] += dp[j];//use this character, so dp[j+1] should be modified
            //at the same time keep dp[j] unchanged, which means skip this character
          }
        }  
      }
      return dp.back();
    }
};