# 392. Is Subsequence

Given a string **s** and a string **t**, check if **s** is subsequence of **t**.

You may assume that there is only lower case English letters in both **s** and **t**. **t** is potentially a very long (length \~= 500,000) string, and **s** is a short string (<=100).

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).

**Example 1:**\
**s** = `"abc"`, **t** = `"ahbgdc"`

Return `true`.

**Example 2:**\
**s** = `"axc"`, **t** = `"ahbgdc"`

Return `false`.

**Follow up:**\
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

**Credits:**\
Special thanks to [@pbrother](https://leetcode.com/pbrother/) for adding this problem and creating all test cases.

```cpp
// Two Pointers
bool isSubsequence(string s, string t) { // time: O(s_len); space: O(1)
    int s_idx = 0;
    for (int t_idx = 0; t_idx < t.length() && s_idx < s.length(); ++t_idx) {
        if (s[s_idx] == t[t_idx]) ++s_idx;
    }
    return s_idx == s.length();
}
```

```cpp
// Follow-Up: Pre-Processing + Binary Search
bool isSubsequence(string s, string t) {
    vector<vector<int> > idx(26);
    for (int i = 0; i < t.length(); ++i) idx[t[i] - 'a'].push_back(i);
    int prev = -1;
    for (char c : s) {
        vector<int> vec = idx[c - 'a'];
        if (vec.empty()) return false;
        int pos = upper_bound(vec.begin(), vec.end(), prev) - vec.begin();
        if (pos == vec.size()) return false;
        prev = vec[pos];
    }
    return true;
}
```
