Design a dynamic stack using arrays that supports getMin() in O(1) time and O(1) extra space

Design a special dynamic Stack using an array that supports all the stack operations such as push(), pop(), peek(), isEmpty(), and getMin() operations in constant Time and Space complexities.

Examples:

Assuming the right to left orientation as the top to bottom orientation and performing the operations:

1. Push(10): 10 is added to the top of the stack. Thereafter, the stack modifies to {10}.
2. Push(4): 4 is added to the top of the stack. Thereafter, the stack modifies to {10, 4}.
3. Push(9): 9 is added to the top of the stack. Thereafter, the stack modifies to {10, 4, 9}.
4. Push(6): 6 is added to the top of the stack. Thereafter, the stack modifies to {10, 4, 9, 6}.
5. Push(5): 5 is added to the top of the stack. Thereafter, the stack modifies to {10, 4, 9, 6, 5}.
6. Peek(): Prints the top element of the stack 5.
7. getMin(): Prints the minimum element of the stack 4.
8. Pop(): Deletes the top most element, 5 from the stack. Thereafter, the stack modifies to {10, 4, 9, 6}.
9. Pop(): Deletes the top most element, 6 from the stack. Thereafter, the stack modifies to {10, 4, 9}.
10. Pop(): Deletes the top most element, 9 from the stack. Thereafter, the stack modifies to {10, 4}.
11. Pop(): Deletes the top most element, 4 from the stack. Thereafter, the stack modifies to {10}.
12. Peek(): Prints the top element of the stack 10.
13. getMin(): Prints the minimum element of the stack 10.

Approach: To implement a dynamic stack using an array the idea is to double the size of the array every time the array gets full. Follow the steps below to solve the problem:

• Initialize an array, say arr[] with an initial size 5, to implement the stack.
• Also, initialize two variables, say top and minEle to store the index of the top element of the stack and minimum element of the stack.
• Now, perform the following stack operations: 
  • isEmpty(): Checks if the stack is empty or not.
    • Return true if the top is less or equal to 0. Otherwise, return false.
  • Push(x): Inserts x at the top of the stack.
    • If the stack is empty, insert x into the stack and make minEle equal to x.
    • If the stack is not empty, compare x with minEle. Two cases arise:
      • If x is greater than or equal to minEle, simply insert x.
      • If x is less than minEle, insert (2*x – minEle) into the stack and make minEle equal to x.
    • If the array used is full then, double the size of the array and then copy all the elements of the previous array to the new array and then assign the address of the new array to the original array. Thereafter, perform the push operation as discussed above.
  • Pop(): Removes an element from the top of the stack.
    • Let the removed element be y. Two cases arise
    • If y is greater than or equal to minEle, the minimum element in the stack is still minEle.
    • If y is less than minEle, the minimum element now becomes (2*minEle – y), so update minEle as minEle = 2*minEle-y.
  • getMin(): Finds the minimum value of the stack.
    • If the stack is not empty then return the value of minEle. Otherwise, return “-1” and print “Underflow“.

Illustration:

Push(x)

• Number to be Inserted: 3, Stack is empty, so insert 3 into stack and minEle = 3.
• Number to be Inserted: 5, Stack is not empty, 5> minEle, insert 5 into stack and minEle = 3.
• Number to be Inserted: 2, Stack is not empty, 2< minEle, insert (2*2-3 = 1) into stack and minEle = 2.
• Number to be Inserted: 1, Stack is not empty, 1< minEle, insert (2*1-2 = 0) into stack and minEle = 1.
• Number to be Inserted: 1, Stack is not empty, 1 = minEle, insert 1 into stack and minEle = 1.
• Number to be Inserted: -1, Stack is not empty, -1 < minEle, insert (2*-1 – 1 = -3) into stack and minEle = -1.

Pop()

• Initially the minimum element minEle in the stack is -1.
• Number removed: -3, Since -3 is less than the minimum element the original number being removed is minEle which is -1, and the new minEle = 2*-1 – (-3) = 1
• Number removed: 1, 1 == minEle, so number removed is 1 and minEle is still equal to 1.
• Number removed: 0, 0< minEle, original number is minEle which is 1 and new minEle = 2*1 – 0 = 2.
• Number removed: 1, 1< minEle, original number is minEle which is 2 and new minEle = 2*2 – 1 = 3.
• Number removed: 5, 5> minEle, original number is 5 and minEle is still 3

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// A class to create
// our special stack
class Stack {
private:

    // Initial size of
    // the Array
    int Max = 5;

    // Array for the stack
    // implementation
    int* arr = new int(Max);

    // Stores the minimum
    // Element of the stack
    int minEle = 0;

    // Stores the top element
    // of the stack
    int top = 0;

public:
    // Method to check whether
    // stack is empty or not
    bool empty()
    {
        if (top <= 0) {
            return true;
        }
        else {
            return false;
        }
    }
    // Method to push elements
    // to the Special Stack
    void push(int x)
    {
        // If stack is empty
        if (empty()) {

            // Assign x to minEle
            minEle = x;

            // Assign x to arr[top]
            arr[top] = x;

            // Increment top by 1
            top++;
        }
        // If array is full
        else if (top == Max) {

            // Update the Max size
            Max = 2 * Max;

            int* temp = new int(Max);

            // Traverse the array arr[]
            for (int i = 0; i < top; i++) {
                temp[i] = arr[i];
            }

            // If x is less than minEle
            if (x < minEle) {

                // Push 2*x-minEle
                temp[top] = 2 * x - minEle;

                // Assign x to minEle
                minEle = x;

                top++;
            }
            // Else
            else {

                // Push x to stack
                temp[top] = x;
                top++;
            }
            // Assign address of the
            // temp to arr
            arr = temp;
        }
        else {
            // If x is less
            // than minEle
            if (x < minEle) {

                // Push 2*x-minEle
                arr[top] = 2 * x - minEle;
                top++;

                // Update minEle
                minEle = x;
            }
            else {
                // Push x to the
                // stack
                arr[top] = x;
                top++;
            }
        }
    }
    // Method to pop the elements
    // from the stack.
    void pop()
    {
        // If stack is empty
        if (empty()) {
            cout << "Underflow" << endl;
            return;
        }
        // Stores the top element
        // of the stack
        int t = arr[top - 1];

        // If t is less than
        // the minEle
        if (t < minEle) {
            // Pop the minEle
            cout << "Popped element : " << minEle << endl;

            // Update minEle
            minEle = 2 * minEle - t;
        }
        // Else
        else {
            // Pop the topmost element
            cout << "Popped element : " << t << endl;
        }
        top--;
        return;
    }

    // Method to find the topmost
    // element of the stack
    int peek()
    {
        // If stack is empty
        if (empty()) {
            cout << "Underflow" << endl;
            return -1;
        }

        // Stores the top element
        // of the stack
        int t = arr[top - 1];

        // If t is less than
        // the minEle
        if (t < minEle) {
            return minEle;
        }
        // Else
        else {
            return t;
        }
    }
    // Method to find the Minimum
    // element of the Special stack
    int getMin()
    {
        // If stack is empty
        if (empty()) {
            cout << "Underflow" << endl;
            return -1;
        }
        // Else
        else {
            return minEle;
        }
    }
};
// Driver Code
int main()
{
    Stack S;

    S.push(10);
    S.push(4);
    S.push(9);
    S.push(6);
    S.push(5);

    cout << "Top Element : " << S.peek() << endl;

    cout << "Minimum Element : " << S.getMin() << endl;

    S.pop();
    S.pop();
    S.pop();
    S.pop();

    cout << "Top Element : " << S.peek() << endl;
    cout << "Minimum Element : " << S.getMin() << endl;

    return 0;
}

Java

// Java program for the above approach
public class Main
{
    // Initial size of
    // the Array
    static int Max = 5;

    // Array for the stack
    // implementation
    static int[] arr = new int[Max];

    // Stores the minimum
    // Element of the stack
    static int minEle = 0;

    // Stores the top element
    // of the stack
    static int Top = 0;

    // Method to check whether
    // stack is empty or not
    static boolean empty()
    {
        if (Top <= 0) {
            return true;
        }
        else {
            return false;
        }
    }
    // Method to push elements
    // to the Special Stack
    static void push(int x)
    {
        // If stack is empty
        if (empty()) {

            // Assign x to minEle
            minEle = x;

            // Assign x to arr[top]
            arr[Top] = x;

            // Increment top by 1
            Top++;
        }
        // If array is full
        else if (Top == Max) {

            // Update the Max size
            Max = 2 * Max;

            int[] temp = new int[Max];

            // Traverse the array arr[]
            for (int i = 0; i < Top; i++) {
                temp[i] = arr[i];
            }

            // If x is less than minEle
            if (x < minEle) {

                // Push 2*x-minEle
                temp[Top] = 2 * x - minEle;

                // Assign x to minEle
                minEle = x;

                Top++;
            }
            // Else
            else {

                // Push x to stack
                temp[Top] = x;
                Top++;
            }
            // Assign address of the
            // temp to arr
            arr = temp;
        }
        else {
            // If x is less
            // than minEle
            if (x < minEle) {

                // Push 2*x-minEle
                arr[Top] = 2 * x - minEle;
                Top++;

                // Update minEle
                minEle = x;
            }
            else {
                // Push x to the
                // stack
                arr[Top] = x;
                Top++;
            }
        }
    }
    // Method to pop the elements
    // from the stack.
    static void pop()
    {
        // If stack is empty
        if (empty()) {
            System.out.print("Underflow");
            return;
        }
        // Stores the top element
        // of the stack
        int t = arr[Top - 1];

        // If t is less than
        // the minEle
        if (t < minEle) {
            // Pop the minEle
            System.out.println("Popped element : " + minEle);

            // Update minEle
            minEle = 2 * minEle - t;
        }
        // Else
        else {
            // Pop the topmost element
            System.out.println("Popped element : " + t);
        }
        Top--;
        return;
    }

    // Method to find the topmost
    // element of the stack
    static int peek()
    {
        // If stack is empty
        if (empty()) {
            System.out.println("Underflow");
            return -1;
        }

        // Stores the top element
        // of the stack
        int t = arr[Top - 1];

        // If t is less than
        // the minEle
        if (t < minEle) {
            return minEle;
        }
        // Else
        else {
            return t;
        }
    }
    // Method to find the Minimum
    // element of the Special stack
    static int getMin()
    {
        // If stack is empty
        if (empty()) {
            System.out.println("Underflow");
            return -1;
        }
        // Else
        else {
            return minEle;
        }
    }

  // Driver code
    public static void main(String[] args) {
        push(10);
        push(4);
        push(9);
        push(6);
        push(5);

        System.out.println("Top Element : " + peek());

        System.out.println("Minimum Element : " + getMin());

        pop();
        pop();
        pop();
        pop();

        System.out.println("Top Element : " + peek());
        System.out.println("Minimum Element : " + getMin());
    }
}

// This code is contributed by rameshtravel07.

Python3

# Python3 program for the above approach

# Initial size of
# the Array
Max = 5

# Array for the stack
# implementation
arr = [0]*Max

# Stores the minimum
# Element of the stack
minEle = 0

# Stores the top element
# of the stack
Top = 0

# Method to check whether
# stack is empty or not
def empty():

    if (Top <= 0):
        return True
    else:
        return False

# Method to push elements
# to the Special Stack
def push(x):
    global arr, Top, Max, minEle

    # If stack is empty
    if empty():

        # Assign x to minEle
        minEle = x

        # Assign x to arr[top]
        arr[Top] = x

        # Increment top by 1
        Top+=1
    # If array is full
    elif (Top == Max):

        # Update the Max size
        Max = 2 * Max

        temp = [0]*Max

        # Traverse the array arr[]
        for i in range(Top):
            temp[i] = arr[i]

        # If x is less than minEle
        if (x < minEle):
            # Push 2*x-minEle
            temp[Top] = 2 * x - minEle

            # Assign x to minEle
            minEle = x

            Top+=1
        # Else
        else:
            # Push x to stack
            temp[Top] = x
            Top+=1
        # Assign address of the
        # temp to arr
        arr = temp
    else:
        # If x is less
        # than minEle
        if (x < minEle):
            # Push 2*x-minEle
            arr[Top] = 2 * x - minEle
            Top+=1

            # Update minEle
            minEle = x
        else:
            # Push x to the
            # stack
            arr[Top] = x
            Top+=1

# Method to pop the elements
# from the stack.
def pop():
    global Top, minEle

    # If stack is empty
    if empty():
        print("Underflow")
        return

    # Stores the top element
    # of the stack
    t = arr[Top - 1]

    # If t is less than
    # the minEle
    if (t < minEle) :
        # Pop the minEle
        print("Popped element :", minEle)

        # Update minEle
        minEle = 2 * minEle - t
    # Else
    else:
        # Pop the topmost element
        print("Popped element :", t)
    Top-=1
    return

# Method to find the topmost
# element of the stack
def peek():
    # If stack is empty
    if empty():
        print("Underflow")
        return -1

    # Stores the top element
    # of the stack
    t = arr[Top - 1]

    # If t is less than
    # the minEle
    if (t < minEle):
        return minEle
    # Else
    else:
        return t

# Method to find the Minimum
# element of the Special stack
def getMin():
    # If stack is empty
    if empty():
        print("Underflow")
        return -1

    # Else
    else:
        return minEle

push(10)
push(4)
push(9)
push(6)
push(5)

print("Top Element :", peek())

print("Minimum Element :", getMin())

pop()
pop()
pop()
pop()

print("Top Element :", peek())
print("Minimum Element :", getMin())

# This code is contributed by mukesh07.

C#

// C# program for the above approach
using System;
class GFG {

    // Initial size of
    // the Array
    static int Max = 5;

    // Array for the stack
    // implementation
    static int[] arr = new int[Max];

    // Stores the minimum
    // Element of the stack
    static int minEle = 0;

    // Stores the top element
    // of the stack
    static int Top = 0;

    // Method to check whether
    // stack is empty or not
    static bool empty()
    {
        if (Top <= 0) {
            return true;
        }
        else {
            return false;
        }
    }
    // Method to push elements
    // to the Special Stack
    static void push(int x)
    {
        // If stack is empty
        if (empty()) {

            // Assign x to minEle
            minEle = x;

            // Assign x to arr[top]
            arr[Top] = x;

            // Increment top by 1
            Top++;
        }
        // If array is full
        else if (Top == Max) {

            // Update the Max size
            Max = 2 * Max;

            int[] temp = new int[Max];

            // Traverse the array arr[]
            for (int i = 0; i < Top; i++) {
                temp[i] = arr[i];
            }

            // If x is less than minEle
            if (x < minEle) {

                // Push 2*x-minEle
                temp[Top] = 2 * x - minEle;

                // Assign x to minEle
                minEle = x;

                Top++;
            }
            // Else
            else {

                // Push x to stack
                temp[Top] = x;
                Top++;
            }
            // Assign address of the
            // temp to arr
            arr = temp;
        }
        else {
            // If x is less
            // than minEle
            if (x < minEle) {

                // Push 2*x-minEle
                arr[Top] = 2 * x - minEle;
                Top++;

                // Update minEle
                minEle = x;
            }
            else {
                // Push x to the
                // stack
                arr[Top] = x;
                Top++;
            }
        }
    }
    // Method to pop the elements
    // from the stack.
    static void pop()
    {
        // If stack is empty
        if (empty()) {
            Console.WriteLine("Underflow");
            return;
        }
        // Stores the top element
        // of the stack
        int t = arr[Top - 1];

        // If t is less than
        // the minEle
        if (t < minEle) {
            // Pop the minEle
            Console.WriteLine("Popped element : " + minEle);

            // Update minEle
            minEle = 2 * minEle - t;
        }
        // Else
        else {
            // Pop the topmost element
            Console.WriteLine("Popped element : " + t);
        }
        Top--;
        return;
    }

    // Method to find the topmost
    // element of the stack
    static int peek()
    {
        // If stack is empty
        if (empty()) {
            Console.WriteLine("Underflow");
            return -1;
        }

        // Stores the top element
        // of the stack
        int t = arr[Top - 1];

        // If t is less than
        // the minEle
        if (t < minEle) {
            return minEle;
        }
        // Else
        else {
            return t;
        }
    }
    // Method to find the Minimum
    // element of the Special stack
    static int getMin()
    {
        // If stack is empty
        if (empty()) {
            Console.WriteLine("Underflow");
            return -1;
        }
        // Else
        else {
            return minEle;
        }
    }

  static void Main() {
    push(10);
    push(4);
    push(9);
    push(6);
    push(5);

    Console.WriteLine("Top Element : " + peek());

    Console.WriteLine("Minimum Element : " + getMin());

    pop();
    pop();
    pop();
    pop();

    Console.WriteLine("Top Element : " + peek());
    Console.WriteLine("Minimum Element : " + getMin());
  }
}

// This code is contributed by suresh07.

Javascript

<script>
    // Javascript program for the above approach

    // Initial size of
    // the Array
    let Max = 5;

    // Array for the stack
    // implementation
    let arr = new Array(Max);

    // Stores the minimum
    // Element of the stack
    let minEle = 0;

    // Stores the top element
    // of the stack
    let Top = 0;

    // Method to check whether
    // stack is empty or not
    function empty()
    {
        if (Top <= 0) {
            return true;
        }
        else {
            return false;
        }
    }
    // Method to push elements
    // to the Special Stack
    function push(x)
    {
        // If stack is empty
        if (empty()) {

            // Assign x to minEle
            minEle = x;

            // Assign x to arr[top]
            arr[Top] = x;

            // Increment top by 1
            Top++;
        }
        // If array is full
        else if (Top == Max) {

            // Update the Max size
            Max = 2 * Max;

            let temp = new Array(Max);

            // Traverse the array arr[]
            for (let i = 0; i < Top; i++) {
                temp[i] = arr[i];
            }

            // If x is less than minEle
            if (x < minEle) {

                // Push 2*x-minEle
                temp[Top] = 2 * x - minEle;

                // Assign x to minEle
                minEle = x;

                Top++;
            }
            // Else
            else {

                // Push x to stack
                temp[Top] = x;
                Top++;
            }
            // Assign address of the
            // temp to arr
            arr = temp;
        }
        else {
            // If x is less
            // than minEle
            if (x < minEle) {

                // Push 2*x-minEle
                arr[Top] = 2 * x - minEle;
                Top++;

                // Update minEle
                minEle = x;
            }
            else {
                // Push x to the
                // stack
                arr[Top] = x;
                Top++;
            }
        }
    }
    // Method to pop the elements
    // from the stack.
    function pop()
    {
        // If stack is empty
        if (empty()) {
            document.write("Underflow" + "</br>");
            return;
        }
        // Stores the top element
        // of the stack
        let t = arr[Top - 1];

        // If t is less than
        // the minEle
        if (t < minEle) {
            // Pop the minEle
            document.write("Popped element : " + minEle + "</br>");

            // Update minEle
            minEle = 2 * minEle - t;
        }
        // Else
        else {
            // Pop the topmost element
            document.write("Popped element : " + t + "</br>");
        }
        Top--;
        return;
    }

    // Method to find the topmost
    // element of the stack
    function peek()
    {
        // If stack is empty
        if (empty()) {
            document.write("Underflow" + "</br>");
            return -1;
        }

        // Stores the top element
        // of the stack
        let t = arr[Top - 1];

        // If t is less than
        // the minEle
        if (t < minEle) {
            return minEle;
        }
        // Else
        else {
            return t;
        }
    }
    // Method to find the Minimum
    // element of the Special stack
    function getMin()
    {
        // If stack is empty
        if (empty()) {
            document.write("Underflow" + "</br>");
            return -1;
        }
        // Else
        else {
            return minEle;
        }
    }

    push(10);
    push(4);
    push(9);
    push(6);
    push(5);

    document.write("Top Element : " + peek() + "</br>");

    document.write("Minimum Element : " + getMin() + "</br>");

    pop();
    pop();
    pop();
    pop();

    document.write("Top Element : " + peek() + "</br>");
    document.write("Minimum Element : " + getMin() + "</br>");

    // This code is contributed by divyesh072019.
</script>

Output

Top Element : 5
Minimum Element : 4
Popped element : 5
Popped element : 6
Popped element : 9
Popped element : 4
Top Element : 10
Minimum Element : 10

Time Complexity: O(1) for each operation
Auxiliary Space: O(1)