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:
- Push(10): 10 is added to the top of the stack. Thereafter, the stack modifies to {10}.
- Push(4): 4 is added to the top of the stack. Thereafter, the stack modifies to {10, 4}.
- Push(9): 9 is added to the top of the stack. Thereafter, the stack modifies to {10, 4, 9}.
- Push(6): 6 is added to the top of the stack. Thereafter, the stack modifies to {10, 4, 9, 6}.
- Push(5): 5 is added to the top of the stack. Thereafter, the stack modifies to {10, 4, 9, 6, 5}.
- Peek(): Prints the top element of the stack 5.
- getMin(): Prints the minimum element of the stack 4.
- Pop(): Deletes the top most element, 5 from the stack. Thereafter, the stack modifies to {10, 4, 9, 6}.
- Pop(): Deletes the top most element, 6 from the stack. Thereafter, the stack modifies to {10, 4, 9}.
- Pop(): Deletes the top most element, 9 from the stack. Thereafter, the stack modifies to {10, 4}.
- Pop(): Deletes the top most element, 4 from the stack. Thereafter, the stack modifies to {10}.
- Peek(): Prints the top element of the stack 10.
- 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:Â
- 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“.
- isEmpty(): Checks if the stack is empty or not.
Illustration:
Push(x)Â
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- 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()Â
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- 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;Â
// A class to create// our special stackclass Stack {private:       // Initial size of    // the Array    int Max = 5;Â
    // Array for the stack    // implementation    int* arr = new int(Max);Â
    // Stores the minimum    // Element of the stack    int minEle = 0;Â
    // Stores the top element    // of the stack    int top = 0;Â
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()) {Â
            // Assign x to minEle            minEle = x;Â
            // Assign x to arr[top]            arr[top] = x;Â
            // Increment top by 1            top++;        }        // If array is full        else if (top == Max) {Â
            // Update the Max size            Max = 2 * Max;Â
            int* temp = new int(Max);Â
            // Traverse the array arr[]            for (int i = 0; i < top; i++) {                temp[i] = arr[i];            }Â
            // If x is less than minEle            if (x < minEle) {Â
                // Push 2*x-minEle                temp[top] = 2 * x - minEle;Â
                // Assign x to minEle                minEle = x;Â
                top++;            }            // Else            else {Â
                // 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) {Â
                // Push 2*x-minEle                arr[top] = 2 * x - minEle;                top++;Â
                // 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];Â
        // If t is less than        // the minEle        if (t < minEle) {            // Pop the minEle            cout << "Popped element : " << minEle << endl;Â
            // Update minEle            minEle = 2 * minEle - t;        }        // Else        else {            // Pop the topmost element            cout << "Popped element : " << t << endl;        }        top--;        return;    }Â
    // Method to find the topmost    // element of the stack    int peek()    {        // If stack is empty        if (empty()) {            cout << "Underflow" << endl;            return -1;        }Â
        // Stores the top element        // of the stack        int t = arr[top - 1];Â
        // 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 Codeint main(){    Stack S;Â
    S.push(10);    S.push(4);    S.push(9);    S.push(6);    S.push(5);Â
    cout << "Top Element : " << S.peek() << endl;Â
    cout << "Minimum Element : " << S.getMin() << endl;Â
    S.pop();    S.pop();    S.pop();    S.pop();Â
    cout << "Top Element : " << S.peek() << endl;    cout << "Minimum Element : " << S.getMin() << endl;Â
    return 0;} |
Java
// Java program for the above approachpublic 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());    }}Â
// This code is contributed by rameshtravel07. |
Python3
# Python3 program for the above approach     # Initial size of# the ArrayMax = 5Â
# Array for the stack# implementationarr = [0]*MaxÂ
# Stores the minimum# Element of the stackminEle = 0Â
# Stores the top element# of the stackTop = 0Â
# Method to check whether# stack is empty or notdef empty():Â
    if (Top <= 0):        return True    else:        return FalseÂ
# Method to push elements# to the Special Stackdef push(x):    global arr, Top, Max, minEle         # If stack is empty    if empty():               # Assign x to minEle        minEle = xÂ
        # Assign x to arr[top]        arr[Top] = xÂ
        # Increment top by 1        Top+=1    # If array is full    elif (Top == Max):Â
        # Update the Max size        Max = 2 * MaxÂ
        temp = [0]*MaxÂ
        # Traverse the array arr[]        for i in range(Top):            temp[i] = arr[i]Â
        # If x is less than minEle        if (x < minEle):            # Push 2*x-minEle            temp[Top] = 2 * x - minEleÂ
            # Assign x to minEle            minEle = xÂ
            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+=1Â
            # Update minEle            minEle = x        else:            # Push x to the            # stack            arr[Top] = x            Top+=1Â
# Method to pop the elements# from the stack.def pop():Â Â Â Â global Top, minEleÂ
    # If stack is empty    if empty():        print("Underflow")        return         # Stores the top element    # of the stack    t = arr[Top - 1]Â
    # If t is less than    # the minEle    if (t < minEle) :        # Pop the minEle        print("Popped element :", minEle)Â
        # Update minEle        minEle = 2 * minEle - t    # Else    else:        # Pop the topmost element        print("Popped element :", t)    Top-=1    returnÂ
# Method to find the topmost# element of the stackdef peek():    # If stack is empty    if empty():        print("Underflow")        return -1Â
    # Stores the top element    # of the stack    t = arr[Top - 1]Â
    # If t is less than    # the minEle    if (t < minEle):        return minEle    # Else    else:        return tÂ
# Method to find the Minimum# element of the Special stackdef getMin():    # If stack is empty    if empty():        print("Underflow")        return -1           # Else    else:        return minEleÂ
push(10)push(4)push(9)push(6)push(5)Â
print("Top Element :", peek())Â
print("Minimum Element :", getMin())Â
pop()pop()pop()pop()Â
print("Top Element :", peek())print("Minimum Element :", getMin())Â
# This code is contributed by mukesh07. |
C#
// C# program for the above approachusing 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());  }}Â
// 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> |
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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
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Time Complexity: O(1) for each operation
Auxiliary Space: O(1)
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