Given n
pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
Example 1:
Input: n = 3 Output: ["((()))","(()())","(())()","()(())","()()()"]
Example 2:
Input: n = 1 Output: ["()"]
Constraints:
1 <= n <= 8
Idea:
Backtracking
Solution:
/**
* @param {number} n
* @return {string[]}
*/
var generateParenthesis = function(n) {
let ans = [];
backtrack(ans, "", 0, 0, n);
return ans;
}
function backtrack(ans, cur, open, close, max){
// exit recursive condition
if (cur.length === max * 2) {
ans.push(cur.concat());
return;
}
// possible solution
if (open < max)
backtrack(ans, cur + "(", open + 1, close, max);
if (close < open)
backtrack(ans, cur + ")", open, close + 1, max);
}