JavaScript Exercises: Elements from both stacks without duplicates

JavaScript Stack: Exercise-30 with Solution

Union of Two Stacks

Write a JavaScript program that implements a stack and creates a new stack that contains all elements from both stacks without duplicates.

Sample Solution:

JavaScript Code:

class Stack {
  constructor() {
    this.items = [];
  }

  // push element to the stack
  push(element) {
    this.items.push(element);
  }

  // pop element from the stack
  pop() {
    if (this.items.length == 0) {
      return "Underflow";
    }
    return this.items.pop();
  }

  // get the top element of the stack
  peek() {
    return this.items[this.items.length - 1];
  }

  // check if the stack is empty
  isEmpty() {
    return this.items.length == 0;
  }

  // get the size of the stack
  size() {
    return this.items.length;
  }

union_of_two(stack2) {
    const result = new Stack();

    // Add all elements from this stack to the result stack
    for (let i = 0; i < this.items.length; i++) {
      if (!result.items.includes(this.items[i])) {
        result.push(this.items[i]);
      }
    }

    // Add all elements from stack2 to the result stack
    for (let i = 0; i < stack2.items.length; i++) {
      if (!result.items.includes(stack2.items[i])) {
        result.push(stack2.items[i]);
      }
    }
    return result;
  }  
 displayStack(stack) {
  console.log("Stack elements are:");
  let str = "";
  for (let i = 0; i < stack.items.length; i++)
    str += stack.items[i] + " ";
  return str.trim();
 }
}
const stack1 = new Stack();
stack1.push(1);
stack1.push(2);
stack1.push(3);
console.log("Stack1:");
console.log(stack1.displayStack(stack1));
const stack2 = new Stack();
stack2.push(2);
stack2.push(3);
stack2.push(4);
console.log("Stack2:");
console.log(stack2.displayStack(stack2));
const stack3 = new Stack();
stack3.push(1);
stack3.push(2);
stack3.push(3);
console.log("Stack3:");
console.log(stack3.displayStack(stack3));
console.log("New stack that contains all elements from both stacks without duplicates:")
console.log("Union of stack1 and stack2:");
result = stack1.union_of_two(stack2);
console.log(result.displayStack(result));
console.log("Union of stack2 and stack3:");
result = stack2.union_of_two(stack3);
console.log(result.displayStack(result));
console.log("Union of stack3 and stack1:");
result = stack3.union_of_two(stack1);
console.log(result.displayStack(result));

Sample Output:

Stack1:
Stack elements are:
1 2 3
Stack2:
Stack elements are:
2 3 4
Stack3:
Stack elements are:
1 2 3
New stack that contains all elements from both stacks without duplicates:
Union of stack1 and stack2:
Stack elements are:
1 2 3 4
Union of stack2 and stack3:
Stack elements are:
2 3 4 1
Union of stack3 and stack1:
Stack elements are:
1 2 3

Flowchart:

Live Demo:

See the Pen javascript-stack-exercise-30 by w3resource (@w3resource) on CodePen.

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