We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Queues: A Tale of Two Stacks
  2. Discussions

Queues: A Tale of Two Stacks

Sort by

recency

|

422 Discussions

|

  • ygorsansone
     + 1 comment

    Python:

    class MyQueue(object):
    # FIFO
    def __init__(self):
        self.queue = []
    
    def peek(self):
        return self.queue[0]
        
    def pop(self):
        if len(self.queue) != 0:
            return self.queue.pop(0)
        
    def put(self, value):
        self.queue.append(value)

  • vaisakhkolathara
     + 0 comments

    Solution in PHP

    <?php
$_fp = fopen("php://stdin", "r");
/* Enter your code here. Read input from STDIN. Print output to STDOUT */

$q = intval(trim(fgets(STDIN)));
$array = [];
for ($t_itr = 0; $t_itr < $q; $t_itr++) {
    $s = rtrim(fgets(STDIN), "\r\n");
    $arr = array_map('intval', preg_split('/ /', $s, -1, PREG_SPLIT_NO_EMPTY));
    if ($arr[0] == 1 && !empty($arr[1])) {
        array_unshift($array, $arr[1]);
    } elseif ($arr[0] == 2) {
        array_pop($array);
    } elseif ($arr[0] == 3) {
        echo $array[count($array)- 1] . PHP_EOL;
    }
}
fclose($_fp);
?>

  • cherrymeteor
     + 0 comments

    javascript with customized Queue 

    class Queue {
  constructor() {
    this.inbox = []
    this.outbox = []
  }

  put(element) {
    this.inbox.push(element)
  }

  pop() {
    if (this.outbox.length == 0) {
      while (this.inbox.length) this.outbox.push(this.inbox.pop())
    }
    return this.outbox.pop()
  }

  peek() {
    if (this.outbox.length == 0) {
      while (this.inbox.length) this.outbox.push(this.inbox.pop())
    }
    return this.outbox[this.outbox.length - 1]
  }
}

function processData(input) {
  let lines = input.split('\n').slice(1)
  let queue = new Queue()

  for (let i = 0; i < lines.length; i++) {
    let [op, n] = lines[i].split(' ')
    switch (op) {
      case '1': queue.put(n); break
      case '2': queue.pop(); break
      case '3': console.log(queue.peek())
    }
  }
}

  • krystynakruk
     + 0 comments

    JavaScript

    class Queue {
  constructor() {
    this.result = [];
  }
  enque(value) {
    this.result.push(value);
  }

  dequeue() {
    if (this.result.length != 0) {
      this.result.shift();
    }
  }

  print() {
    if (this.result.length != 0) {
      console.log(this.result[0]);
    }
  }
}
function processData(input) {
  let inputArr = input.split("\n");
  let tab = new Queue();
  for (let i = 1; i < inputArr.length; i++) {
    switch (inputArr[i][0]) {
      case "1":
        let value = inputArr[i].slice(1).trim();
        tab.enque(value);
        break;
      case "2":
        tab.dequeue();
        break;
      case "3":
        tab.print();
        break;
    }
  }
}

  • cpjen
     + 1 comment

    With some of these challenges, I have no idea where to start and no idea how anyone else does either, like my brain isn't wired in the necessary way. I suspected this would be another example but on this occasion I'm baffled as to why this is a challenge at all. The simplistic solution below works for all test cases.

    class MyQueue(object):
    def __init__(self):
        self.queue = []
        
    
    def peek(self):
        return self.queue[0]
        
    def pop(self):
        del self.queue[0]
        
        
    def put(self, value):
        self.queue.append(value)