Stack Complexity Analysis

When working with stacks, it's important to understand the time and space complexity of the fundamental operations: push, pop, and peek. Analyzing the complexity of stack operations helps in determining the efficiency and performance characteristics of a stack-based solution.

Let's break down the complexity of each operation:

• Push: The push operation adds an element to the top of the stack. This operation runs in constant time, O(1), as it only requires updating the top pointer and placing the new element at the top.

• Pop: The pop operation removes the top element from the stack. Similar to push, the pop operation also runs in constant time, O(1), as it only involves updating the top pointer and removing the element from the top.

• Peek: The peek operation retrieves the top element from the stack without modifying the stack. Like push and pop, the peek operation has a time complexity of O(1) since it only involves accessing the top element.

The space complexity of a stack depends on the number of elements stored in it. If a stack holds n elements, the space complexity would be O(n) as it requires memory to store each element.

Understanding the complexity of stack operations helps in selecting the appropriate data structure for a given problem and optimizing its performance. By leveraging the constant time complexity of stack operations, you can design efficient solutions for a wide range of problems.

Now that we've analyzed the complexity of stack operations, let's put our knowledge into practice with some exercises in the next screen.

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}

class Stack {

    private int maxSize;

    private int[] stackArray;

    private int top;

​

    public Stack(int size) {

        maxSize = size;

        stackArray = new int[maxSize];

        top = -1;

    }

​

    public void push(int value) {

        if (top == maxSize - 1) {

            System.out.println("Stack is full. Unable to push.");

            return;

        }

        stackArray[++top] = value;

    }

​

    public int pop() {

        if (top == -1) {

            System.out.println("Stack is empty. Unable to pop.");

            return -1;

        }

        return stackArray[top--];

    }

​

    public int peek() {

        if (top == -1) {