Move from brute-force thinking to an efficient approach using stack strategy.
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the MinStack class:
MinStack() initializes the stack object.void push(int val) pushes the element val onto the stack.void pop() removes the element on the top of the stack.int top() gets the top element of the stack.int getMin() retrieves the minimum element in the stack.You must implement a solution with O(1) time complexity for each function.
Example 1:
Input ["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]] Output [null,null,null,null,-3,null,0,-2] Explanation MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); // return -3 minStack.pop(); minStack.top(); // return 0 minStack.getMin(); // return -2
Constraints:
-231 <= val <= 231 - 1pop, top and getMin operations will always be called on non-empty stacks.3 * 104 calls will be made to push, pop, top, and getMin.Problem summary: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. Implement the MinStack class: MinStack() initializes the stack object. void push(int val) pushes the element val onto the stack. void pop() removes the element on the top of the stack. int top() gets the top element of the stack. int getMin() retrieves the minimum element in the stack. You must implement a solution with O(1) time complexity for each function.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack · Design
["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]]
sliding-window-maximum)max-stack)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #155: Min Stack
class MinStack {
private Deque<Integer> stk1 = new ArrayDeque<>();
private Deque<Integer> stk2 = new ArrayDeque<>();
public MinStack() {
stk2.push(Integer.MAX_VALUE);
}
public void push(int val) {
stk1.push(val);
stk2.push(Math.min(val, stk2.peek()));
}
public void pop() {
stk1.pop();
stk2.pop();
}
public int top() {
return stk1.peek();
}
public int getMin() {
return stk2.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(val);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
// Accepted solution for LeetCode #155: Min Stack
type MinStack struct {
stk1 []int
stk2 []int
}
func Constructor() MinStack {
return MinStack{[]int{}, []int{math.MaxInt32}}
}
func (this *MinStack) Push(val int) {
this.stk1 = append(this.stk1, val)
this.stk2 = append(this.stk2, min(val, this.stk2[len(this.stk2)-1]))
}
func (this *MinStack) Pop() {
this.stk1 = this.stk1[:len(this.stk1)-1]
this.stk2 = this.stk2[:len(this.stk2)-1]
}
func (this *MinStack) Top() int {
return this.stk1[len(this.stk1)-1]
}
func (this *MinStack) GetMin() int {
return this.stk2[len(this.stk2)-1]
}
/**
* Your MinStack object will be instantiated and called as such:
* obj := Constructor();
* obj.Push(val);
* obj.Pop();
* param_3 := obj.Top();
* param_4 := obj.GetMin();
*/
# Accepted solution for LeetCode #155: Min Stack
class MinStack:
def __init__(self):
self.stk1 = []
self.stk2 = [inf]
def push(self, val: int) -> None:
self.stk1.append(val)
self.stk2.append(min(val, self.stk2[-1]))
def pop(self) -> None:
self.stk1.pop()
self.stk2.pop()
def top(self) -> int:
return self.stk1[-1]
def getMin(self) -> int:
return self.stk2[-1]
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()
// Accepted solution for LeetCode #155: Min Stack
use std::collections::VecDeque;
struct MinStack {
stk1: VecDeque<i32>,
stk2: VecDeque<i32>,
}
/**
* `&self` means the method takes an immutable reference.
* If you need a mutable reference, change it to `&mut self` instead.
*/
impl MinStack {
fn new() -> Self {
Self {
stk1: VecDeque::new(),
stk2: VecDeque::new(),
}
}
fn push(&mut self, x: i32) {
self.stk1.push_back(x);
if self.stk2.is_empty() || *self.stk2.back().unwrap() >= x {
self.stk2.push_back(x);
}
}
fn pop(&mut self) {
let val = self.stk1.pop_back().unwrap();
if *self.stk2.back().unwrap() == val {
self.stk2.pop_back();
}
}
fn top(&self) -> i32 {
*self.stk1.back().unwrap()
}
fn get_min(&self) -> i32 {
*self.stk2.back().unwrap()
}
}
// Accepted solution for LeetCode #155: Min Stack
class MinStack {
stk1: number[];
stk2: number[];
constructor() {
this.stk1 = [];
this.stk2 = [Infinity];
}
push(val: number): void {
this.stk1.push(val);
this.stk2.push(Math.min(val, this.stk2[this.stk2.length - 1]));
}
pop(): void {
this.stk1.pop();
this.stk2.pop();
}
top(): number {
return this.stk1[this.stk1.length - 1];
}
getMin(): number {
return this.stk2[this.stk2.length - 1];
}
}
/**
* Your MinStack object will be instantiated and called as such:
* var obj = new MinStack()
* obj.push(x)
* obj.pop()
* var param_3 = obj.top()
* var param_4 = obj.getMin()
*/
Use this to step through a reusable interview workflow for this problem.
For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.
Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.