[LeetCode] 115. different sequences

Topic links: https://leetcode-cn.com/problems/distinct-subsequences/

Subject description:

Given a string and a string S T, calculate the number of the partial sequences S T appears.

A string is a sequence of sub-means, by eliminating some of (and may not be deleted) does not interfere with the new character string consisting of the relative position of the remaining characters. (E.g., "ACE" is "ABCDE" of a sequence, and "AEC" instead)

Example:

Example 1:

输入: S = "rabbbit", T = "rabbit"
输出: 3
解释:

如下图所示, 有 3 种可以从 S 中得到 "rabbit" 的方案。
(上箭头符号 ^ 表示选取的字母)

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

Example 2:

输入: S = "babgbag", T = "bag"
输出: 5
解释:

如下图所示, 有 5 种可以从 S 中得到 "bag" 的方案。 
(上箭头符号 ^ 表示选取的字母)

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

Ideas:

Dynamic Programming

dp[i][j]Representative Tbefore ithe string may be made Sbefore the jstrings the maximum number.

So dynamic equation:

When S[j] == T[i], dp[i][j] = dp[i-1][j-1] + dp[i][j-1];

When S[j] != T[i],dp[i][j] = dp[i][j-1]

For example, as in the example of

For the first line, Tit is empty, because empty set is a subset of all the strings, so we are the first line1

For the first column, Sis empty, so that the composition Tnumber is of course 0the

As for how to below, we can by dynamic equations, simulate it on their own!

Code:

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        n1 = len(s)
        n2 = len(t)
        dp = [[0] * (n1 + 1) for _ in range(n2 + 1)]
        for j in range(n1 + 1):
            dp[0][j] = 1
        for i in range(1, n2 + 1):
            for j in range(1, n1 + 1):
                if t[i - 1] == s[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]  + dp[i][j - 1]
                else:
                    dp[i][j] = dp[i][j - 1]
        #print(dp)
        return dp[-1][-1]

java

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