---
title: '392. Is Subsequence'
description: Given two strings s and t, return true if s is a subsequence of t, or false otherwise
icon: dot
topics:
  - { name: "Two Pointers", slug: "two-pointers" }
  - { name: "String", slug: "string" }
  - { name: "Dynamic Programming", slug: "dynamic-programming" }
video: "99RVfqklbCE"
issue: "https://github.com/prdlk/leetcode/issues/45"
sidebar:
  label: 'Is Subsequence'
  badge: 'Easy'
---

::::warning
Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 10^9, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?
::::

### Example 1:
- Input: `s = "abc", t = "ahbgdc"`
- Output: `true`

### Example 2:
- Input: `s = "axc", t = "ahbgdc"`
- Output: `false`

### Constraints:

- `0 <= s.length <= 100`
- `0 <= t.length <= 10^4`
- `s` and `t` consist only of lowercase English letters.

## Approach

```mermaid
flowchart TD
  S(["isSubsequence(s, t)"]) --> G{"len(s) > len(t)?"}
  G -- yes --> X(["return False"])
  G -- no --> I["i = 0 into s, j = 0 into t"]
  I --> W{"i < len(s) and j < len(t)?"}
  W -- no --> E(["return i == len(s) — every char of s was matched"])
  W -- yes --> Q{"s[i] == t[j]?"}
  Q -- yes --> A["i += 1 — consume the match"]
  Q -- no --> B["j += 1 — always advance t"]
  A --> B
  B --> W
```

## Solution

```py
class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        if len(s) > len(t):
            return False

        i, j = 0, 0
        while i < len(s) and j < len(t):
            if s[i] == t[j]:
                i += 1
            j += 1

        return i == len(s)
```