https://github.com/emahtab/min-stack
Min Stack
https://github.com/emahtab/min-stack
constant-time leetcode problem-solving stack
Last synced: 3 months ago
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Min Stack
- Host: GitHub
- URL: https://github.com/emahtab/min-stack
- Owner: eMahtab
- Created: 2020-02-21T05:04:56.000Z (over 5 years ago)
- Default Branch: master
- Last Pushed: 2024-10-16T16:24:34.000Z (9 months ago)
- Last Synced: 2025-02-02T03:19:28.898Z (5 months ago)
- Topics: constant-time, leetcode, problem-solving, stack
- Homepage:
- Size: 25.4 KB
- Stars: 1
- Watchers: 0
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# Min Stack
## https://leetcode.com/problems/min-stack
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
1. push(x) -- Push element x onto stack.
2. pop() -- Removes the element on top of the stack.
3. top() -- Get the top element.
4. getMin() -- Retrieve the minimum element in the stack.
```
Example:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
```
## Approach :
The question asks us to implement 4 functions and all of them should have constant runtime, which means no matter how large the input is, those 4 functions should always run in constant time.
Now we already know that, a Stack gives us `push(x)` , `pop()` and `peek()` functions, all of these 3 functions run in constant time.
So If we use Stack, the only additional thing we would have to implement is the `getMin()` function.
We will use one additional stack called the `minStack` to keep track of the minimum element in the stack at any point of time, this will allow us to get the minimum element from the stack in constant time.
# Implementation 1 : getMin() = O(n) (Note, getMin() doesn't run in constant time) 😉
```java
class MinStack {
Stack stack = new Stack<>();
public void push(int x) {
stack.push(x);
}
public void pop() {
stack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
Stack s = new Stack<>();
int min = Integer.MAX_VALUE;
while(!stack.isEmpty()) {
int x = stack.pop();
min = Math.min(min, x);
s.push(x);
}
while(!s.isEmpty()) {
stack.push(s.pop());
}
return min;
}
}
```
# Implementation 2 : getMin() = O(1)
```java
class MinStack {
private Stack stack = new Stack<>();
private Stack minStack = new Stack<>();
public void push(int x) {
stack.push(x);
if (minStack.isEmpty() || x <= minStack.peek()) {
minStack.push(x);
}
}
public void pop() {
if (stack.peek().equals(minStack.peek())) { // Don't use == , stack.peek() == minStack.peek()
minStack.pop();
}
stack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}
```
# Gotcha : Use equals when working with Integer object (don't use == operator)
```java
class Main {
public static void main(String[] args) {
Integer a = new Integer(19);
Integer b = new Integer(19);
System.out.println(a == b); // false
System.out.println(a.equals(b)); // true
}
}
```
### Key Points :
1. Don't forget to update the minStack on both push and pop operations on the main stack
2. Note : We push an element to minStack, only if the element is less than **or equal to** `min_stack.peek()`
As a side note, in Java calling `push()`, `pop()` and `peek()` returns the element, but in this question for `push()` and `pop()` the return type is void. And also calling `pop()` and `peek()` on an empty stack results in `EmptyStackException`
# References :
1. https://www.youtube.com/watch?v=nGwn8_-6e7w
2. https://www.youtube.com/watch?v=WxCuL3jleUA
3. https://leetcode.com/problems/min-stack/solution/