Leetcode 301 - remove invalid parentheses - study one of discussions.

Leetcode301_removeInvalidParentheses.java

/*
Leetcode 301: Remove invalide parentheses
Study one sharing from the disucussion panel:
https://leetcode.com/problems/remove-invalid-parentheses/discuss/75032/Share-my-Java-BFS-solution

What I like to do is to review the note and customize it to fit my reading style.
Feb. 7, 2018

Using breadth first search technique

The idea is straightforward, with the input string s, we generate all possible states by removing 
one '(' or ')', check if they are valid, if found valid ones on the current level, put them to the 
final result list and we are done, otherwise, add them to a queue and carry on to the next level.

The good thing of using BFS is that we can guarantee the number of parentheses that need to be 
removed is minimal, also no recursion call is needed in BFS.

Do not need a stack to check valid parentheses.

Time complexity:

In BFS technique we handle the states level by level, in the worst case, we need to handle all the 
levels, we can analyze the time complexity level by level and add them up to get the final complexity.

On the first level, there’s only one string which is the input string s, let’s say the length of it is 
n, to check whether it’s valid, we need O(n) time. 

On the second level, we remove one '(' or ')' from the first level, so there are C(n, n-1) new strings, 
each of them has n-1 characters, and for each string, we need to check whether it’s valid or not, thus 
the total time complexity on this level is (n-1) x C(n, n-1). 

Come to the third level, total time complexity is (n-2) x C(n, n-2), so on and so forth…

Finally we have this formula:

T(n) = n x C(n, n) + (n-1) x C(n, n-1) + … + 1 x C(n, 1) = n x 2^(n-1).
*/
public class Solution {
    public List<String> removeInvalidParentheses(String s) {
      List<String> res = new ArrayList<>();
      
      // sanity check
      if (s == null) return res;
      
      Set<String> visited = new HashSet<>();
      Queue<String> queue = new LinkedList<>();
      
      // initialize
      queue.add(s);
      visited.add(s);
      
      boolean found = false;
      
      while (!queue.isEmpty()) {
        s = queue.poll();
        
        if (isValid(s)) {
          // found an answer, add to the result
          res.add(s);
          found = true;
        }
      
        if (found) continue;
      
        // generate all possible states
        for (int i = 0; i < s.length(); i++) {
          // we only try to remove left or right paren
          if (s.charAt(i) != '(' && s.charAt(i) != ')') continue;
        
          String t = s.substring(0, i) + s.substring(i + 1);
        
          if (!visited.contains(t)) {
            // for each state, if it's not visited, add it to the queue
            queue.add(t);
            visited.add(t);
          }
        }
      }
      
      return res;
    }
    
    // helper function checks if string s contains valid parantheses
    boolean isValid(String s) {
      int count = 0;
    
      for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (c == '(') count++;
        if (c == ')' && count-- == 0) return false;
      }
    
      return count == 0;
    }
}