2. Valid Parentheses
easyAsked at ByteDanceVerify a string of brackets is properly balanced — ByteDance leans on this to confirm stack fluency before pivoting to feed-ranking expression parsers.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given a string containing the characters '()', '[]', '{}', determine if the input string is valid. Brackets must close in the correct order and every opener must have a matching closer.
Constraints
1 <= s.length <= 10^4s consists only of bracket characters
Examples
Example 1
s = "()[]{}"trueExample 2
s = "(]"falseApproaches
1. Brute force replace
Repeatedly remove '()', '[]', '{}' until no change; valid if empty.
- Time
- O(n^2)
- Space
- O(n)
let prev;
while (prev !== s) { prev = s; s = s.replace('()','').replace('[]','').replace('{}',''); }
return s.length === 0;Tradeoff:
2. Stack
Push openers, pop on matching closer. Mismatch or non-empty stack at end means invalid.
- Time
- O(n)
- Space
- O(n)
function isValid(s) {
const pair = { ')':'(', ']':'[', '}':'{' };
const stack = [];
for (const c of s) {
if (c in pair) {
if (stack.pop() !== pair[c]) return false;
} else stack.push(c);
}
return stack.length === 0;
}Tradeoff:
ByteDance-specific tips
ByteDance interviewers want you to articulate the O(n) stack invariant cleanly — ambiguous code that 'almost works' fails their feed-template grading bar.
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