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2. Valid Parentheses

easyAsked at ByteDance

Verify 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 <= 10^4
  • s consists only of bracket characters

Examples

Example 1

Input
s = "()[]{}"
Output
true

Example 2

Input
s = "(]"
Output
false

Approaches

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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