# 392. Is Subsequence

[Link](https://leetcode.com/problems/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?&#x20;

***Ans: Instead of using indexOf() funciton to find the index in t***&#x20;

* ***Create a hashMap\<Character, List\<Integer>> to store the indexs of each char in t***
* ***write a binary search for finding the index of s.charAt(t) in list .***

***TimeComplexity: O(klogn), where k is the s.length, and n is the t.length***

**Solution**

IndexOf cost O(n), where n is the length of t

Time complexity : O(kn)

```
  public static boolean isSubsequence(String s, String t) {
        int previousIndexInT = 0;
        for(int i = 0; i < s.length(); i++){
            int indexInT = t.indexOf(s.charAt(i), previousIndexInT);
            if(indexInT == -1){
                return false;
            }
            previousIndexInT = indexInT + 1; 
            //REMEMBER 要從下一個開始找，不然會一直在同一個
        }
        return true;
    }
```
