22. Generate Parentheses

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

class Solution {
    public List<String> generateParenthesis(int n) {
      List<String> ans = new ArrayList<>();
      if( n == 0){
        return ans;
      }
      StringBuilder current = new StringBuilder();
      dfs( ans, n, current, 0, n, n);
      return ans; 
    }
    private void dfs(List<String> ans, int n, StringBuilder current, int index, int left, int right){
      if(index == n * 2 ){
        ans.add(current.toString()); ///// sb.toString()
        return;
      }


      // pick '(' 
      if( left > 0 ){
        current.append('(');
        
        dfs(ans, n, current, index + 1, left - 1, right);
        current.deleteCharAt(current.length() - 1);
      }


      if(right > 0 && left <= right - 1){
        current.append(')');
       
        dfs(ans, n, current, index+1, left, right - 1);
        current.deleteCharAt(current.length() - 1);
      }
    }    
}

 () three pairs 

6 levels 

every level, add '('  or ')'

 

( or ) 

( or )

( or )

( or )

( or )

( or )

 

only add '(' when there is enough  '('   left , and only add ')' when we used more  '('  than   ')'  and there is enough  ')'  left

do this condition check before adding any parentheses . if satisfies the condition, add the parenthese and dfs on it, and don't forget backtracking on the stringbuilder 

 

 

time complexity 2 ^ ( pairs * 2)

 

 

api : 

current is a stringbuilder
current.deleteCharAt(current.length() - 1);