Java program to remove duplicates from a given stack

In this article, we’ll explore two methods to remove duplicate elements from a stack in Java. We’ll compare a straightforward approach with nested loops and a more efficient method using a HashSet. The goal is to demonstrate how to optimize duplicate removal and to evaluate the performance of each approach.

Problem statement

Write a Java program to remove the duplicate element from the stack.

Input

Stack<Integer> data = initData(10L);

Output

Unique elements using Naive Approach: [1, 4, 3, 2, 8, 7, 5]
Time spent for Naive Approach: 18200 nanoseconds

Unique elements using Optimized Approach: [1, 4, 3, 2, 8, 7, 5]
Time spent for Optimized Approach: 34800 nanoseconds

To remove duplicates from a given stack, we have 2 approaches −

• Naïve approach: Create two nested loops to see which elements are already present and prevent adding them to the result stack return.
• HashSet approach: Use a Set to store unique elements, and take advantage of its O(1) complexity to check if an element is present or not.

Generating and cloning random stacks

Below is the Java program first builds a random stack and then creates a duplicate of it for further use −

private static Stack initData(Long size) {
    Stack stack = new Stack < > ();
    Random random = new Random();
    int bound = (int) Math.ceil(size * 0.75);
    for (int i = 0; i < size; ++i) {
        stack.add(random.nextInt(bound) + 1);
    }
    return stack;
}

private static Stack < Integer > manualCloneStack(Stack < Integer > stack) {
    Stack < Integer > newStack = new Stack < > ();
    for (Integer item: stack) {
        newStack.push(item);
    }
    return newStack;
}

initData helps create a Stack with a specified size and random elements ranging from 1 to 100.

manualCloneStack helps generate data by copying data from another stack, using it for performance comparison between the two ideas.

Remove duplicates from a given stack using Naïve approach

Following are the step to remove duplicates from a given stack using Naïve approach −

• Start the timer.
• Create an empty stack to store unique elements.
• Iterate using while loop and continue until the original stack is empty pop the top element.
• Check for duplicates using resultStack.contains(element) to see if the element is already in the result stack.
• If the element is not in the result stack, add it to resultStack.
• Record the end time and calculate the total time spent.
• Return result

Example

Below is the Java program to remove duplicates from a given stack using Naïve approach −

public static Stack idea1(Stack stack) {
  long start = System.nanoTime();
  Stack resultStack = new Stack < > ();

  while (!stack.isEmpty()) {
    int element = stack.pop();
    if (!resultStack.contains(element)) {
      resultStack.add(element);
    }
  }
  System.out.println("Time spent for idea1 is %d nanosecond".formatted(System.nanoTime() - start));
  return resultStack;
}

For Naive approach, we use

<code>while (!stack.isEmpty())</code>

to handle the first loop to travel through all elements in the stack, and the second loop is <pre class="brush:php;toolbar:false">resultStack.contains(element)</pre> to check if the element is already present.

Remove duplicates from a given stack using optimized approach (HashSet)

Following are the step to remove duplicates from a given stack using optimized approach −

• Start the timer
• Create a Set to track seen elements (for O(1) complexity checks).
• Create a temporary stack to store unique elements.
• Iterate using the while loop continue until the stack is empty .
• Remove the top element from the stack.
• To check the duplicates we will use seen.contains(element) to check if the element is already in the set.
• If the element is not in the set, add it to both seen and the temporary stack.
• Record the end time and calculate the total time spent.
• Return the result

Example

Below is the Java program to remove duplicates from a given stack using HashSet −

public static Stack<Integer> idea2(Stack<Integer> stack) {
    long start = System.nanoTime();
    Set<Integer> seen = new HashSet<>();
    Stack<Integer> tempStack = new Stack<>();

    while (!stack.isEmpty()) {
        int element = stack.pop();
        if (!seen.contains(element)) {
            seen.add(element);
            tempStack.push(element);
        }
    }
    System.out.println("Time spent for idea2 is %d nanosecond".formatted(System.nanoTime() - start));
    return tempStack;
}

For optimized approach, we use

<code>Set<Integer> seen</code>

to store unique elements in a Set, take advantage of O(1) complexity when accessing a random element to reduce the calculation cost of checking if an element is present or not.

Using both approaches together

Below are the step to remove duplicates from a given stack using both approaches mentioned above −

• Initialize data we are using the initData(Long size) method to create a stack filled with random integers.
  
• Clone the Stack we are using the cloneStack(Stack stack) method to create an exact copy of the original stack.
• Generate the initial stack with initData.
• Clone this stack using cloneStack.
• Apply the naive approach to remove duplicates from the original stack.
• Apply the optimized approach to remove duplicates from the cloned stack.
• Display the time taken for each method and the unique elements produced by both approaches

Example 

Below is the Java program that removes duplicate elements from a stack using both approaches mentioned above −

import java.util.HashSet;
import java.util.Random;
import java.util.Set;
import java.util.Stack;

public class DuplicateStackElements {
    private static Stack<Integer> initData(Long size) {
        Stack<Integer> stack = new Stack<>();
        Random random = new Random();
        int bound = (int) Math.ceil(size * 0.75);
        for (int i = 0; i < size; ++i) {
            stack.add(random.nextInt(bound) + 1);
        }
        return stack;
    }
    private static Stack<Integer> cloneStack(Stack<Integer> stack) {
        Stack<Integer> newStack = new Stack<>();
        newStack.addAll(stack);
        return newStack;
    }
    public static Stack<Integer> idea1(Stack<Integer> stack) {
        long start = System.nanoTime();
        Stack<Integer> resultStack = new Stack<>();

        while (!stack.isEmpty()) {
            int element = stack.pop();
            if (!resultStack.contains(element)) {
                resultStack.add(element);
            }
        }
        System.out.printf("Time spent for idea1 is %d nanoseconds%n", System.nanoTime() - start);
        return resultStack;
    }
    public static Stack<Integer> idea2(Stack<Integer> stack) {
        long start = System.nanoTime();
        Set<Integer> seen = new HashSet<>();
        Stack<Integer> tempStack = new Stack<>();

        while (!stack.isEmpty()) {
            int element = stack.pop();
            if (!seen.contains(element)) {
                seen.add(element);
                tempStack.push(element);
            }
        }
        System.out.printf("Time spent for idea2 is %d nanoseconds%n", System.nanoTime() - start);
        return tempStack;
    }
    public static void main(String[] args) {
        Stack<Integer> data1 = initData(10L);
        Stack<Integer> data2 = cloneStack(data1);
        System.out.println("Unique elements using idea1: " + idea1(data1));
        System.out.println("Unique elements using idea2: " + idea2(data2));
    }
}

Output

Comparison

* The unit of measurement is nanosecond.

public static void main(String[] args) {
    Stack<Integer> data1 = initData(<number of stack size want to test>);
    Stack<Integer> data2 = manualCloneStack(data1);
    idea1(data1);
    idea2(data2);
}

Method	100 elements	1000 elements	10000 elements	100000 elements	1000000 elements
Idea 1	693100	4051600	19026900	114201800	1157256000
Idea 2	135800	681400	2717800	11489400	36456100

As observed, the time running for Idea 2 is shorter than for Idea 1 because the complexity of Idea 1 is O(n²), while the complexity of Idea 2 is O(n). So, when the number of stacks increases, the time spent on calculations also increases based on it.

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