Free AI LeetCode Communication Coach — Valid Parentheses

The problem

Given a string of '(', ')', '{', '}', '[' and ']', determine if brackets close in the correct order.

Example: s = "{[]}" → true

Push, pop, empty check.

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.”
• “At the end the stack must be empty.”

Practise out loud

1. 1. Restate

   I have a string of brackets and I need every closer to match the right opener in order.

2. 2. Approach

   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 pop, it is invalid.

3. 3. Example

   On '{[]}' I push curly, push square, the square closer matches, the curly closer matches, and the stack is empty — valid.

4. 4. Time and space

   Time is O(n). Space is O(n) in the worst case of all openers.

5. 5. Edge cases

   Empty string is valid. A leftover opener at the end is invalid.

Common mistakes when explaining Valid Parentheses

• Forgetting the leftover-opener case.
• Not saying what happens when you pop an empty stack.