Re: Java binary search using TreeSet got TLE

I think the Map and TreeSet could be simplified by Array and binarySearch. Since we scan T from beginning to the end (index itself is in increasing order), List will be sufficient. Then we can use binarySearch to replace with TreeSet ability which is a little overkill for this problem. Here is my solution.

```
// Follow-up: O(N) time for pre-processing, O(Mlog?) for each S.
// Eg-1. s="abc", t="bahbgdca"
// idx=[a={1,7}, b={0,3}, c={6}]
// i=0 ('a'): prev=1
// i=1 ('b'): prev=3
// i=2 ('c'): prev=6 (return true)
// Eg-2. s="abc", t="bahgdcb"
// idx=[a={1}, b={0,6}, c={5}]
// i=0 ('a'): prev=1
// i=1 ('b'): prev=6
// i=2 ('c'): prev=? (return false)
public boolean isSubsequence(String s, String t) {
List<Integer>[] idx = new List[256]; // Just for clarity
for (int i = 0; i < t.length(); i++) {
if (idx[t.charAt(i)] == null)
idx[t.charAt(i)] = new ArrayList<>();
idx[t.charAt(i)].add(i);
}
int prev = 0;
for (int i = 0; i < s.length(); i++) {
if (idx[s.charAt(i)] == null) return false; // Note: char of S does NOT exist in T causing NPE
int j = Collections.binarySearch(idx[s.charAt(i)], prev);
if (j < 0) j = -j - 1;
if (j == idx[s.charAt(i)].size()) return false;
prev = idx[s.charAt(i)].get(j) + 1;
}
return true;
}
```