Why does a single counter work for one bracket type but not for three?

With only ( and ), a counter that goes up on (, down on ), fails if it ever drops below zero and accepts only at zero is a complete check, because any ) can close any (. With three types, one counter per type still cannot see the order: for "[{]}" every counter returns to zero without going negative, yet the string is invalid. The stack remembers which type opened last, which is exactly what the counters lose.

What if the string contains other characters besides brackets?

Skip them. Keep the opener branch, and only pop when ch in ")]}", so letters, digits and spaces fall through untouched. With that change is_valid("total = f(x[0]) + {1: 2}[1]") returns True and is_valid("f(x[0)]") returns False.

How do I return the position of the first error instead of True or False?

Push (closer, index) tuples instead of bare closers. Return the current index when a closer finds an empty stack or the wrong top, and after the loop return the index of the oldest unclosed opener, stack[0][1], or -1 if the stack is empty. That gives 2 for "[{]}", 0 for "(()[]", 0 for ")(" and -1 for "{[()()]}".

Is an empty string valid?

Yes. It has no bracket left unmatched, so both solutions return True for "": the pair-removal loop sees no change on its first pass and finds the string empty, and the stack loop never runs and ends with an empty stack.

Why check that the stack is empty at the end?

Because the loop only catches closers that are wrong. An opener that is never closed produces no error while scanning, so "(()[]" passes every comparison and is rejected only by return not expected. Returning True after the loop would accept it.

Valid Parentheses in Python: Pair Removal and Stack Solutions

is_valid.py

Python 3.14

Not run yet

+ 7 hidden

Input

s = "{[()()]}"

Expected

true

Each pair closes inside the one that opened before it, and nothing is left open.

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers

1 · Brute force: delete adjacent pairs until nothing changes

2 · One pass with a stack of expected closers