README_EN.md

115. Distinct Subsequences

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Description

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

A string's subsequence is a new string formed from the original string by deleting some (can be none) of the characters without disturbing the remaining characters' relative positions. (i.e., "ACE" is a subsequence of "ABCDE" while "AEC" is not).

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.

Solutions

Python3

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]

Java

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

Go

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]
}

C++

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

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