[LeetCode] Distinct Subsequences

Well, a dynamic programming problem. Let's first define its state dp[i][j] to be the number of distinct subsequences of t[0..i - 1] in s[0..j - 1]. Then we have the following state equations:

1. General case 1: dp[i][j] = dp[i][j - 1] if t[i - 1] != s[j - 1];
2. General case 2: dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1] if t[i - 1] == s[j - 1];
3. Boundary case 1: dp[0][j] = 1 for all j;
4. Boundary case 2: dp[i][0] = 0 for all positive i.

Now let's give brief explanations to the four equations above.

1. If t[i - 1] != s[j - 1], the distinct subsequences will not include s[j - 1] and thus all the number of distinct subsequences will simply be those in s[0..j - 2], which corresponds to dp[i][j - 1];
2. If t[i - 1] == s[j - 1], the number of distinct subsequences include two parts: those withs[j - 1] and those without;
3. An empty string will have exactly one subsequence in any string :-)
4. Non-empty string will have no subsequences in an empty string.

Putting these together, we will have the following simple codes (just like translation :-)):

 1 class Solution {
 2 public:
 3     int numDistinct(string s, string t) {
 4         int m = t.length(), n = s.length();
 5         vector<vector<int>> dp(m + 1, vector<int> (n + 1, 0));
 6         for (int j = 0; j <= n; j++) dp[0][j] = 1;
 7         for (int j = 1; j <= n; j++)
 8             for (int i = 1; i <= m; i++)
 9                 dp[i][j] = dp[i][j - 1] + (t[i - 1] == s[j - 1] ? dp[i - 1][j - 1] : 0);
10         return dp[m][n];
11     }
12 };

Notice that we keep the whole m*n matrix simply for dp[i - 1][j - 1]. So we can simply store that value in a single variable and further optimize the space complexity. The final code is as follows.

 1 class Solution {
 2 public:
 3     int numDistinct(string s, string t) {
 4         int m = t.length(), n = s.length();
 5         vector<int> cur(m + 1, 0);
 6         cur[0] = 1;
 7         for (int j = 1; j <= n; j++) { 
 8             int pre = 1;
 9             for (int i = 1; i <= m; i++) {
10                 int temp = cur[i];
11                 cur[i] = cur[i] + (t[i - 1] == s[j - 1] ? pre : 0);
12                 pre = temp;
13             }
14         }
15         return cur[m];
16     }
17 };