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Design a Stack That Supports getMin() in O(1) (Min Stack)

Q: Design a Stack That Supports getMin() in O(1) (Min Stack)

**Target Audience:** Engineers preparing for practical JavaScript coding interviews. **Difficulty:** Easy > **How to read this doc:** the real, defining "revert on pop" behavior this question tests — that `getMin` correctly returns to the PREVIOUS minimum after the current one is popped off — was verified directly: after pushing `5, 2, 7, 1` and popping once, `getMin()` correctly reverted from `1` back to `2`, not staying stuck at the already-removed `1`. ## 1. The problem, restated Build a stack data structure where `push`, `pop`, `top`, AND `getMin` all run in genuine O(1) time — the real, hard part being `getMin`, since a naive approach (scanning the entire stack on every call) would only achieve O(n). ## 2. Clarifying questions a strong candidate asks | Question | Why it matters | | :--- | :--- | | O(1) getMin is the real hard constraint? | Yes — a naive full-scan approach is genuinely O(n) per call; the real design challenge is TRACKING the minimum incrementally instead. | |

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The problem

Problem, as an interviewer would state it: "Design a stack supporting push, pop, top, and getMin — all in O(1) time, including getMin, which must return the current minimum value in the stack at any point."

Examples:

Code

const s = new MinStack();
s.push(5); s.push(2); s.push(7);
s.getMin(); // 2

Clarifying questions expected:

Is a genuine O(1) time requirement for getMin specifically the real, hard constraint here — would scanning the whole stack on each call be acceptable?
After popping the CURRENT minimum off the stack, does getMin need to correctly revert to the PREVIOUS minimum?
Should duplicate values (the same number pushed twice) be handled correctly if one of them happens to be the current minimum?

Code / implementation expected: Yes — real, direct proof that getMin runs in O(1) and correctly reverts to the previous minimum after the current minimum is popped off.

js-coding
utility
algorithm

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Solution

Target Audience: Engineers preparing for practical JavaScript coding interviews. Difficulty: Easy

How to read this doc: the real, defining "revert on pop" behavior this question tests — that getMin correctly returns to the PREVIOUS minimum after the current one is popped off — was verified directly: after pushing 5, 2, 7, 1 and popping once, getMin() correctly reverted from 1 back to 2, not staying stuck at the already-removed 1.

1. The problem, restated

Build a stack data structure where push, pop, top, AND getMin all run in genuine O(1) time — the real, hard part being getMin, since a naive approach (scanning the entire stack on every call) would only achieve O(n).

2. Clarifying questions a strong candidate asks

Question	Why it matters
O(1) getMin is the real hard constraint?	Yes — a naive full-scan approach is ge

The solution is hidden so you can try first

6 min read · 9 sections

Run the code

Edit the code and run it here. Each console line with an expected value becomes a test.

JavaScriptReal, direct proof: MinStack's getMin() runs in O(1) and correctly reverts to the previous minimum after the current minimum is popped off

Practice

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