> For the complete documentation index, see [llms.txt](https://docs-57.gitbook.io/data-structure-and-algorithms/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://docs-57.gitbook.io/data-structure-and-algorithms/problems/string/is-subsequence.md).

# Is Subsequence

{% hint style="info" %}
String, Two Pointers
{% endhint %}

Given two strings `s` and `t`, return `true` *if* `s` *is a **subsequence** of* `t`*, or* `false` *otherwise*.

A **subsequence** of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., `"ace"` is a subsequence of `"abcde"` while `"aec"` is not).

 

**Example 1:**

<pre><code><strong>Input: s = "abc", t = "ahbgdc"
</strong><strong>Output: true
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: s = "axc", t = "ahbgdc"
</strong><strong>Output: false
</strong></code></pre>

&#x20;

**Constraints:**

* `0 <= s.length <= 100`
* `0 <= t.length <= 104`
* `s` and `t` consist only of lowercase English letters.<br>

### Solutions

#### Approach - Two Pointers Technique

This algorithm checks if string `s` is a subsequence of string `t`. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

**Steps**

1. Initialize two pointers `i` and `j` to 0. `i` is used to traverse string `s` and `j` is used to traverse string `t`. Also, initialize a counter `found` to 0 which keeps track of the number of characters of `s` found in `t`.
2. Start a `while` loop that continues as long as `i` is less than the length of `s` and `j` is less than the length of `t`.
3. Inside the loop, check if the `j`-th character of `t` is equal to the `i`-th character of `s`. If it is, increment both `i` and `found`.
4. Regardless of whether the characters match, increment `j` to move to the next character in `t`.
5. After the loop ends, check if `found` is equal to the length of `s`. If it is, return `true`, indicating that `s` is a subsequence of `t`. If not, return `false`.

```csharp
public class Solution {
    public bool IsSubsequence(string s, string t) {
        
        int j =0;
        int i=0;
         int found=0;
            while(i<s.Length && j<t.Length){
                if(t[j]==s[i]){
                    i++;
                    found++;
                }
                j++;
            }
        return found==s.Length;
    }
}
```

> Complexity
>
> * **Time complexity**: O(n) where `n` is the length of `t` string.
> * **Space complexity**: O(1)