# Distinct Subsequences LeetCode Solution | Easy Approach Share:

Distinct Subsequences 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.

Time: O(mn)
Space: O(mn)

### C++

`````` class Solution {
public:
int numDistinct(string s, string t) {
const int m = s.length();
const int n = t.length();
vector<vector<long>> dp(m + 1, vector<long>(n + 1));

for (int i = 0; i <= m; ++i)
dp[i] = 1;

for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
if (s[i - 1] == t[j - 1])
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
else
dp[i][j] = dp[i - 1][j];

return dp[m][n];
}
};
``````

### Java

``````class Solution {
public int numDistinct(String s, String t) {
final int m = s.length();
final int n = t.length();
long[][] dp = new long[m + 1][n + 1];

for (int i = 0; i <= m; ++i)
dp[i] = 1;

for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
if (s.charAt(i - 1) == t.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
else
dp[i][j] = dp[i - 1][j];

return (int) dp[m][n];
}
}
``````

### Python

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

for i in range(m + 1):
dp[i] = 1

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

return dp[m][n]
``````

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