Retrieving the Min element in a stack in O(1) Time

14,887

Solution 1

This wouldn't work if you popped numbers off the stack.

Ex. 2,4,5,3,1. After you pop 1 off, what is your minimum?

The solution is to keep a stack of minimums, not just a single value. If you encounter a value that is less than equal to the current minimum, you need to push it onto the min-stack.

Ex.

Push(4):
Stack: 4
Min-stack: 4

Push(2):
Stack: 4 2
Min-stack: 4 2

Push(2):
Stack: 4 2 2
Min-stack: 4 2 2

Push(5):
Stack: 4 2 2 5
Min-stack: 4 2 2

Push(3):
Stack: 4 2 2 5 3
Min-stack: 4 2 2

Push(1):
Stack: 4 2 2 5 3 1
Min-stack: 4 2 2 1

Pop():
Stack: 4 2 2 5 3
Min-stack: 4 2 2

Pop():
Stack: 4 2 2 5
Min-stack: 4 2 2

Pop():
Stack: 4 2 2
Min-stack: 4 2 2

Pop():
Stack: 4 2
Min-stack: 4 2

Pop():
Stack: 4
Min-stack: 4

Solution 2

Use a linked list to keep track of the minimum value which is gonna be the head.

Note that linkedlist.app= append ( we put the value in the tail ). linkedlist.pre =prepend ( we put the value as the head of the linkedlist)

public class Stack {

int[] elements;
int top;
Linkedlists min;

public Stack(int n) {
    elements = new int[n];
    top = 0;
    min = new Linkedlists();
}

public void realloc(int n) {
    int[] tab = new int[n];
    for (int i = 0; i < top; i++) {
        tab[i] = elements[i];
    }

    elements = tab;
}

public void push(int x) {
    if (top == elements.length) {
        realloc(elements.length * 2);
    }
    if (top == 0) {
        min.pre(x);
    } else if (x < min.head.data) {
        min.pre(x);
    } else {
        min.app(x);
    }
    elements[top++] = x;
}

public int pop() {

    int x = elements[--top];
    if (top == 0) {

    }
    if (this.getMin() == x) {
        min.head = min.head.next;
    }
    elements[top] = 0;
    if (4 * top < elements.length) {
        realloc((elements.length + 1) / 2);
    }

    return x;
}

public void display() {
    for (Object x : elements) {
        System.out.print(x + " ");
    }

}

public int getMin() {
    if (top == 0) {
        return 0;
    }
    return this.min.head.data;
}

public static void main(String[] args) {
    Stack stack = new Stack(4);
    stack.push(2);
    stack.push(3);
    stack.push(1);
    stack.push(4);
    stack.push(5);
    stack.pop();
    stack.pop();
    stack.pop();
    stack.push(1);
    stack.pop();
    stack.pop();
    stack.pop();
    stack.push(2);
    System.out.println(stack.getMin());
    stack.display();

}

}

Share:
14,887
Ali
Author by

Ali

Don't we all like to move fast and break things?

Updated on July 07, 2022

Comments

  • Ali
    Ali almost 2 years

    The reason I'm asking this question is because I cannot see why the way I think cannot be applied to this particular question

    "How would you design a stack which, in addition to push and pop, also has a function min which returns the minimum element? Push, pop and min should all operate in O(1) time"

    My basic solution: Wouldn't it be possible if we had a variable in stack class, that whenever we were pushing an item to stack we would check if it is smaller than our min variable. If it is assign the value to the min, if not ignore.

    You would still get the O(1) as the min function would be;

    int getMinimum(){
      return min;
    }
    

    Why this solution is never mentioned, or what is the fault with the way I think?