The shape of the problem

Every closing bracket must match the most recent unmatched opener. You scan left-to-right once and decide valid or invalid at the end.

Brute force first

Repeatedly scan the string and delete any adjacent `()`, `[]`, or `{}` pair. Valid iff the whole string collapses to empty. Each pass only removes innermost pairs, so worst case is O(n²).

Trace it on ({[]})

(   push         → stack: [(
{   push         → stack: [( {
[   push         → stack: [( { [
]   pop matches  → stack: [( {
}   pop matches  → stack: [(
)   pop matches  → stack: []   ✓ empty → valid

The stack contains exactly the unmatched openers seen so far, in the order they appeared. Nothing else is ever on it.

Shape of the loop

for c in s:
  if c is opener: push c
  elif stack empty or top != match(c): return false
  else: pop
return stack is empty

Pseudocode only — the full worked solution lives in the Solution tab.