17. Generate Parentheses
mediumAsked at RampGenerate all combinations of well-formed parentheses for n pairs.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. Return the answer as a list of strings.
Constraints
1 <= n <= 8
Examples
Example 1
n = 3["((()))","(()())","(())()","()(())","()()()"]Example 2
n = 1["()"]Approaches
1. Brute force generate-and-validate
Generate all 2^(2n) strings of length 2n, keep balanced ones.
- Time
- O(2^(2n) * n)
- Space
- O(2^(2n) * n)
function generateParenthesis(n) {
const out = [];
function valid(s) { let c = 0; for (const ch of s) { c += ch === '(' ? 1 : -1; if (c < 0) return false; } return c === 0; }
function build(s) { if (s.length === 2 * n) { if (valid(s)) out.push(s); return; } build(s + '('); build(s + ')'); }
build('');
return out;
}Tradeoff:
2. Backtracking with open/close counts
Track remaining opens and closes; only emit '(' if opens left and only emit ')' if closes outnumber opens already placed.
- Time
- O(4^n / sqrt(n))
- Space
- O(n)
function generateParenthesis(n) {
const out = [];
function build(cur, open, close) {
if (cur.length === 2 * n) { out.push(cur); return; }
if (open < n) build(cur + '(', open + 1, close);
if (close < open) build(cur + ')', open, close + 1);
}
build('', 0, 0);
return out;
}Tradeoff:
Ramp-specific tips
Ramp uses this exact constraint-pruning style inside their approval-graph engine to enumerate valid policy paths without generating dead branches.
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