Remove invalid character to make string a valid parenthesis. return all possible result string with least removal operations.

这个题的难点在于，看不出来是要用BFS.

一般看到least会想到用DP.

Solution II: BFS. (divide and conquer by cutting index) -> 一刀就是一个cost！

for any possible remove index in str,

[0, i - 1] [i + 1, end]

if we found a match, then this is minimum removal.

Solution I: DFS