Generate Parentheses

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

For example, givenn= 3, a solution set is:

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

Solution

DFS - Only add valid parantheses

``````class Solution {
public List<String> generateParenthesis(int n) {
List<String> ans = new ArrayList();
backtrack(ans, "", 0, 0, n);
return ans;
}

public void backtrack(List<String> ans, String cur, int open, int close, int max){
if (cur.length() == max * 2) {
return;
}

if (open < max)
backtrack(ans, cur+"(", open+1, close, max);
if (close < open)
backtrack(ans, cur+")", open, close+1, max);
}
}
``````

Another implementation @lisali1203

``````public List<String> generateParenthesis(int n) {
List<String> list = new ArrayList<String>();
generateOneByOne("", list, n, n);
return list;
}
public void generateOneByOne(String sublist, List<String> list, int left, int right){
if(left > right){
return;
}
if(left > 0){
generateOneByOne( sublist + "(" , list, left-1, right);
}
if(right > 0){
generateOneByOne( sublist + ")" , list, left, right-1);
}
if(left == 0 && right == 0){
return;
}
}
``````

Or

``````public List<String> generateParenthesis(int n)
{
if (n > 0) generateParenthesisCore("", n, n, result);
return result;
}

private void generateParenthesisCore(String prefix, int left, int right, List<String> result)
{
if (left == 0 && right == 0) result.add(prefix);
// Has left Parenthesis
if (left > 0) generateParenthesisCore(prefix+'(', left-1, right, result);
// has more right Parenthesis
if (left < right) generateParenthesisCore(prefix+')', left, right-1, result);
}
``````

Reference

https://leetcode.com/problems/generate-parentheses/solution/

Wikipedia: Catalan Number