package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
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 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
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 * - 版权所有，转发请注明出处
 */
public class _115_DistinctSubsequences {
    /**
     * 115. Distinct Subsequences
     * 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).

     Here is an example:
     S = "rabbbit", T = "rabbit"

     Return 3.

     int[][] dp

     1, S.charAt(i) != T.charAt(j)
        dp[i][j] = dp[i-1][j]

     2, S.charAt(i) == T.charAt(j)
        dp[i][j] = dp[i-1][j] + dp[i-1][j-1]

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

     time : O(m * n)
     space : O(m * n)


     * @param S
     * @param T
     * @return
     */
    public static int numDistinct(String S, String T) {
        int[][] dp = new int[S.length() + 1][T.length() + 1];
        for (int i = 0; i < S.length(); i++) {
            dp[i][0] = 1;
        }

        for (int i = 1; i <= S.length(); i++) {
            for (int j = 1; j <= T.length(); j++) {
                if (S.charAt(i - 1) == T.charAt(j - 1)) {
                    dp[i][j] = dp[i-1][j] + dp[i-1][j-1];
                } else {
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        return dp[S.length()][T.length()];
    }
}