How to Explain Valid Parentheses in English

The problem

Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. Brackets must close in the correct order, and every closer must match the same type of opener.

Narrate a stack. Interviewers want to hear “push opener, pop on match,” not a regex.

pairs = {")": "(", "]": "[", "}": "{"}
stack = []
for ch in s:
    if ch in "([{":
        stack.append(ch)
    elif not stack or stack.pop() != pairs[ch]:
        return False
return not stack

How to explain it

1. 1. Restate

   Say the problem in your own words.

   One or two sentences. Show you understood the input, the output, and the goal — not that you memorised the prompt.

2. 2. Approach

   Name the method before you code.

   Brute force first if you need it, then the structure you will use: hash map, two pointers, stack, binary search.

3. 3. Example

   Walk one concrete input.

   Pick small numbers. Say what you store, what you compare, and what you return. Interviewers follow an example more easily than abstract talk.

4. 4. Time and space

   One sentence each.

   After the example, before you claim you are done. “Time is O(n) because we scan once. Space is O(n) for the map.”

5. 5. Edge cases

   Name at least one unusual input.

   Empty input, duplicates, already sorted, overflow. Invite a follow-up: “I would also check …”

• “I will use a stack.”
• “I push openers and pop when a closer matches.”
• “If I pop from an empty stack, it is invalid.”
• “At the end the stack must be empty.”

Practise out loud

Model spoken script

I will restate the problem. I have a string of brackets, and I need to say whether every closer matches the right opener in the right order. I will use a stack. I push opening brackets. When I see a closer, I pop and check it matches. If the stack is empty when I try to pop, the string is invalid. For example, '{[]}' — I push curly, push square, then the square closer matches, then the curly closer matches, and the stack is empty, so it is valid. Time is O(n). Space is O(n) in the worst case, all openers. I would also check an empty string — that is valid — and a leftover opener at the end, like '('.