# Distinct Subsequences Problem

## Description

LeetCode Problem 115.

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:

``````1
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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:

``````1
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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.

## Sample C++ Code

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class Solution {
public:
// Define 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:
// General case 1: dp[i][j] = dp[i][j - 1] if t[i - 1] != s[j - 1];
// General case 2: dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1] if t[i - 1] == s[j - 1];
// Boundary case 1: dp[j] = 1 for all j;
// Boundary case 2: dp[i] = 0 for all positive i.
int numDistinct(string s, string t) {
int m = t.length(), n = s.length();
vector<vector<int>> dp(m + 1, vector<int> (n + 1, 0));
for (int j = 0; j <= n; j++) dp[j] = 1;
for (int j = 1; j <= n; j++)
for (int i = 1; i <= m; i++)
dp[i][j] = dp[i][j - 1] + (t[i - 1] == s[j - 1] ? dp[i - 1][j - 1] : 0);
return dp[m][n];
}
};
``````