Question

Formatted question description: https://leetcode.ca/all/115.html

Given two strings s and t, return the number of distinct subsequences of s which equals t.

The test cases are generated so that the answer fits on a 32-bit signed integer.

Example 1:

Input: s = "rabbbit", t = "rabbit"
Output: 3
Explanation:
As shown below, there are 3 ways you can generate "rabbit" from s.
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.
babgbag
babgbag
babgbag
babgbag
babgbag

Constraints:

1 <= s.length, t.length <= 1000
s and t consist of English letters.

Algorithm

	r a b b b i t
	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

dp[0][0] = 1; // T and S are both empty strings.
dp[0][1 ... S.length()-1] = 1; // T is an empty string, S has only one subsequence match.
dp[1 ... T.length()-1][0] = 0; // S is an empty string, T is not an empty string, and S has no subsequence matching.
dp[i][j] = dp[i][j-1] + (T[i-1] == S[j-1]? dp[i-1][j-1]: 0)

Code

public class Distinct_Subsequences {

	public class Solution_iteration {
	    public int numDistinct(String s, String t) {

	        if (s == null || t == null)     return 0;
	        if (s.length() < t.length())    return 0;

	        int TLength = t.length();
	        int SLength = s.length();

	        int[][] dp = new int[TLength + 1][SLength + 1];
	        dp[0][0] = 1;

	        for (int i = 1; i <= SLength; i++) {
	            dp[0][i] = 1;
	        }
	        for (int i = 1; i <= TLength; i++) {
	            dp[i][0] = 0;
	        }

	        for (int i = 1; i <= TLength; i++) {
	            for (int j = 1; j <= SLength; j++) {
	                if (t.charAt(i - 1) == s.charAt(j - 1)) {
	                    dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1];
	                } else {
	                    dp[i][j] = dp[i][j - 1];
	                }
	            }
	        }

	        return dp[TLength][SLength];
	    }
	}

	/*
	Last executed input:
	"aabdbaabeeadcbbdedacbbeecbabebaeeecaeabaedadcbdbcdaabebdadbbaeabdadeaabbabbecebbebcaddaacccebeaeedababedeacdeaaaeeaecbe"
	"bddabdcae"
	*/
	public class Solution_recursion {
	    public int numDistinct(String s, String t) {

	        if (s.length() == 0) {
	            return t.length() == 0 ? 1 : 0;
	        }

	        if (t.length() == 0) {
	            return s.length() >= 0 ? 1 : 0; // @note: t is empty, so delete all in s is one solution
	        }

	        if (s.length() < t.length()) {
	            return 0;
	        }

	        int sum = 0;

	        if (s.charAt(0) == t.charAt(0)) {
	            sum += numDistinct(s.substring(1), t.substring(1));
	        }

	        sum += numDistinct(s.substring(1), t); // @note: anyway, have to try this one: eg, s='aa' and t='a', result=2

	        return sum;

	    }
	}

}

############

class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length(), n = t.length();
        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];
    }
}

// OJ: https://leetcode.com/problems/distinct-subsequences
// Time: O(MN)
// Space: O(N)
class Solution {
public:
    int numDistinct(string s, string t) {
        int M = s.size(), N = t.size();
        vector<long> dp(N + 1);
        dp[0] = 1;
        for (int i = 0; i < M; ++i) {
            for (int j = N - 1; j >= 0; --j) {
                if (s[i] == t[j]) dp[j + 1] += dp[j];
                if (dp[j + 1] > INT_MAX) dp[j + 1] = 0;
            }
        }
        return dp[N];
    }
};

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

############

class Solution(object):
  # space O(n^2)
  def _numDistinct(self, s, t):
    """
    :type s: str
    :type t: str
    :rtype: int
    """
    dp = [[0] * (len(t) + 1) for _ in range(0, len(s) + 1)]

    for i in range(0, len(s) + 1):
      dp[i][0] = 1

    for i in range(1, len(s) + 1):
      for j in range(1, len(t) + 1):
        dp[i][j] += dp[i - 1][j]
        if t[j - 1] == s[i - 1]:
          dp[i][j] += dp[i - 1][j - 1]

    return dp[-1][-1]

  def numDistinct(self, s, t):
    """
    :type s: str
    :type t: str
    :rtype: int
    """
    dp = [0] * (len(t) + 1)
    for i in range(1, len(s) + 1):
      pre = 1
      for j in range(1, len(t) + 1):
        tmp = dp[j]
        if t[j - 1] == s[i - 1]:
          dp[j] += pre
        pre = tmp
    return dp[-1]

func numDistinct(s string, t string) int {
	m, n := len(s), len(t)
	dp := make([][]int, m+1)
	for i := 0; i <= m; i++ {
		dp[i] = make([]int, n+1)
		dp[i][0] = 1
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			dp[i][j] = dp[i-1][j]
			if s[i-1] == t[j-1] {
				dp[i][j] += dp[i-1][j-1]
			}
		}
	}
	return dp[m][n]
}

function numDistinct(s: string, t: string): number {
    const m = s.length;
    const n = t.length;
    const dp: number[][] = new Array(m + 1)
        .fill(0)
        .map(() => new Array(n + 1).fill(0));
    for (let i = 0; i <= m; ++i) {
        dp[i][0] = 1;
    }
    for (let i = 1; i <= m; ++i) {
        for (let 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];
}

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