Min Stack

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Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

  • push(x) — Push element x onto stack.
  • pop() — Removes the element on top of the stack.
  • top() — Get the top element.
  • getMin() — Retrieve the minimum element in the stack.

Example 1:

["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]]



MinStack minStack = new MinStack();




minStack.getMin(); // return -3


minStack.top(); // return 0

minStack.getMin(); // return -2


Methods pop, top and getMin operations will always be called on non-empty stacks.


The solution for this problem is a tricky one. As per the given constraints, we need to solve this problem in a constant time, so basically we can't do any thing with the iterations here. Our main part to solve this problem in a constant time lies in a solution to find the minimum value in a stack because this is not possible if we simply implement stack and solve this.

To solve this problem we will apply two logics here:

  • We will always push the second minimum number on to the stack before we push the upcoming number. This will help us to find the next minimum number in a constant time whenever we pop the top of the stack.

  • If the current minimum number is the top of the stack, we will pop twice and change the current minimum number to be the last minimum number we popped.



class MinStack {
    private Stack<Integer> stack;
    private int min;

    public MinStack() {
        stack = new Stack();
        min = Integer.MAX_VALUE;

    public void push(int x) {

        if (x <= min) {
            min = x;

    public void pop() {

        if (min == stack.pop()) {
            min = stack.pop();

    public int top() {
        return stack.peek();

    public int getMin() {
        return min;


Complexity Analysis:

  • Time Complexity: O(1).
  • Space Complexity: O(n).
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