Today I have done Design Circular Queue and leetcode's August LeetCoding Challenge with cpp.

Design Circular Queue

Description

Design your implementation of the circular queue. The circular queue is a linear data structure in which the operations are performed based on FIFO (First In First Out) principle and the last position is connected back to the first position to make a circle. It is also called "Ring Buffer".

One of the benefits of the circular queue is that we can make use of the spaces in front of the queue. In a normal queue, once the queue becomes full, we cannot insert the next element even if there is a space in front of the queue. But using the circular queue, we can use the space to store new values.

Implementation the MyCircularQueue class:

• MyCircularQueue(k) Initializes the object with the size of the queue to be k.
• int Front() Gets the front item from the queue. If the queue is empty, return -1.
• int Rear() Gets the last item from the queue. If the queue is empty, return -1.
• boolean enQueue(int value) Inserts an element into the circular queue. Return true if the operation is successful.
• boolean deQueue() Deletes an element from the circular queue. Return true if the operation is successful.
• boolean isEmpty() Checks whether the circular queue is empty or not.
• boolean isFull() Checks whether the circular queue is full or not.

You must solve the problem without using the built-in queue data structure in your programming language.

Example 1:

Input
["MyCircularQueue", "enQueue", "enQueue", "enQueue", "enQueue", "Rear", "isFull", "deQueue", "enQueue", "Rear"]
[[3], [1], [2], [3], [4], [], [], [], [4], []]
Output
[null, true, true, true, false, 3, true, true, true, 4]

Explanation
MyCircularQueue myCircularQueue = new MyCircularQueue(3);
myCircularQueue.enQueue(1); // return True
myCircularQueue.enQueue(2); // return True
myCircularQueue.enQueue(3); // return True
myCircularQueue.enQueue(4); // return False
myCircularQueue.Rear();     // return 3
myCircularQueue.isFull();   // return True
myCircularQueue.deQueue();  // return True
myCircularQueue.enQueue(4); // return True
myCircularQueue.Rear();     // return 4

Constraints:

• 1 <= k <= 1000
• 0 <= value <= 1000
• At most 3000 calls will be made to enQueue, deQueue, Front, Rear, isEmpty, and isFull.

Solution

class MyCircularQueue {
  vector<int> container;
  int begin;
  int size;
  int capacity;
public:
  MyCircularQueue(int k): capacity(k) {
    container.resize(k);
    begin = 0;
    size = 0;
  }
  
  bool enQueue(int value) {
    if(size == capacity) return false;
    container[(begin + size) % capacity] = value;
    size += 1;
    return true;
  }
  
  bool deQueue() {
    if(!size) return false;
    begin = (begin + 1) % capacity;
    size -= 1;
    return true;
  }
  
  int Front() {
    if(!size) return -1;
    return container[begin];
  }
  
  int Rear() {
    if(!size) return -1;
    return container[(begin + size - 1) % capacity];
  }
  
  bool isEmpty() {
    return size == 0;
  }
  
  bool isFull() {
    return size == capacity; 
  }
};

// Accepted
// 58/58 cases passed (16 ms)
// Your runtime beats 97.99 % of cpp submissions
// Your memory usage beats 40.24 % of cpp submissions (16.9 MB)

August LeetCoding Challenge 26

Description

Verify Preorder Serialization of a Binary Tree

One way to serialize a binary tree is to use preorder traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as '#'.

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where '#' represents a null node.

Given a string of comma-separated values preorder, return true if it is a correct preorder traversal serialization of a binary tree.

It is guaranteed that each comma-separated value in the string must be either an integer or a character '#' representing null pointer.

You may assume that the input format is always valid.

• For example, it could never contain two consecutive commas, such as "1,,3".

Note: You are not allowed to reconstruct the tree.

Example 1:

Input: preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: preorder = "1,#"
Output: false

Example 3:

Input: preorder = "9,#,#,1"
Output: false

Constraints:

• 1 <= preorder.length <= 10^4
• preoder consist of integers in the range [0, 100] and '#' separated by commas ','.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  bool isValidSerialization(string preorder) {
    int count = 1;
    int len = preorder.length();
    for (int i = 0; i < len; ++i) {
      if(preorder[i] == '#') {
        count -= 1;
        i += 1;
        if(count == 0 && i != len) return false;
      } else {
        while(i < len && preorder[i] != ',') {
          i += 1;
        }
        count += 1;
      }
    }
    return !count;
  }
};

// Accepted
// 151/151 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 66.91 % of cpp submissions (6.8 MB)