# 115. Distinct Subsequences

### Description

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).

It's guaranteed the answer fits on a 32-bit signed integer.

### Constraints

### Approach

### Links

* [GeeksforGeeks](https://www.geeksforgeeks.org/count-distinct-occurrences-as-a-subsequence/)
* [Leetcode](https://leetcode.com/problems/distinct-subsequences/)
* [ProgramCreek](https://www.programcreek.com/2013/01/leetcode-distinct-subsequences-total-java/)
* YouTube

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** S = "rabbbit", T = "rabbit"

**Output:** 3

**Explanation:** As shown below, there are 3 ways you can generate "rabbit" from S.

(The caret symbol ^ means the chosen letters)

<div align="left"><img src="/files/-MG2EyexYt206N-KirXL" alt=""></div>
{% endtab %}

{% tab title="Example 2" %}
**Input:** S = "babgbag", T = "bag"

**Output:** 5

**Explanation:** As shown below, there are 5 ways you can generate "bag" from S.

(The caret symbol ^ means the chosen letters)

<div align="left"><img src="/files/-MG2FJ5tMlPEXhrsU-L1" alt=""></div>
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(M*N) where M is the length of S and N is the length of T.
 * Space complexity : O(M*N) which is occupied by the 2D dp array that we create.
 */

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++) {
                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[m][n];
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : O(M*N) where M is the length of S and N is the length of T.
 * Space complexity : O(N) since we are using a single array which is the 
 *    size of the string T.
 */

class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length(), n = t.length();
        int[] dp = new int[n+1];
        dp[0] = 1;
        
        for(int i = 1; i <= m; i++) {
            for(int j = n; j > 0; j--) {
                if(s.charAt(i-1) == t.charAt(j-1)) {
                    dp[j] += dp[j-1];
                }
            }
        }
        
        return dp[n];
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://code-snippets.hbamithkumara.com/leetcode/problems/101-200/distinct-subsequences.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.