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generate-parentheses.dart (1104B)

      1 //Given a number n return all combinations of valid parenthesis
      2 //with n sets of parentheses. 
      3 //To do this I used recursion to follow each branch until
      4 //they ended and then returned each set of valid parenthesis.
      5 //The time complexity of this code is 2^n where n is the number
      6 //of sets of parenthesis to generate.
      7 //Time: 273ms Beats: 47.50%
      8 //Memory: 143.2MB Beats: 10%
      9 class Solution {
     10   List<String> generateParenthesis(int n) {
     11       return parenthesis(n , n);
     12   }
     13   List<String> parenthesis(int left , int right){
     14       if(left == 0 && right == 0){
     15           return [""];
     16       }
     17       List<String> return_ls = [];
     18       if(left != 0){
     19           List<String> l1 = parenthesis(left - 1 , right);
     20           for(int i = 0 ; i < l1.length ; ++i){
     21             l1[i] = '(' + l1[i];
     22             return_ls.add(l1[i]);
     23           }
     24       }
     25       if(right != 0 && left < right){
     26           List<String> r1 = parenthesis(left , right - 1);
     27           for(int i = 0 ; i < r1.length ; ++i){
     28               r1[i] = ')' + r1[i];
     29               return_ls.add(r1[i]);
     30           }
     31       }
     32       return return_ls;
     33   }
     34 }