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Posts tagged as “data structure”

花花酱 LeetCode 1172. Dinner Plate Stacks

You have an infinite number of stacks arranged in a row and numbered (left to right) from 0, each of the stacks has the same maximum capacity.

Implement the DinnerPlates class:

  • DinnerPlates(int capacity) Initializes the object with the maximum capacity of the stacks.
  • void push(int val) pushes the given positive integer val into the leftmost stack with size less than capacity.
  • int pop() returns the value at the top of the rightmost non-empty stack and removes it from that stack, and returns -1 if all stacks are empty.
  • int popAtStack(int index) returns the value at the top of the stack with the given index and removes it from that stack, and returns -1 if the stack with that given index is empty.

Example:

Input: 
["DinnerPlates","push","push","push","push","push","popAtStack","push","push","popAtStack","popAtStack","pop","pop","pop","pop","pop"]
[[2],[1],[2],[3],[4],[5],[0],[20],[21],[0],[2],[],[],[],[],[]]
Output: 

[null,null,null,null,null,null,2,null,null,20,21,5,4,3,1,-1]

Explanation: DinnerPlates D = DinnerPlates(2); // Initialize with capacity = 2 D.push(1); D.push(2); D.push(3); D.push(4); D.push(5); // The stacks are now: 2  4   1  3  5 ﹈ ﹈ ﹈ D.popAtStack(0); // Returns 2. The stacks are now:  4   1  3  5 ﹈ ﹈ ﹈ D.push(20); // The stacks are now: 20 4   1  3  5 ﹈ ﹈ ﹈ D.push(21); // The stacks are now: 20 4 21   1  3  5 ﹈ ﹈ ﹈ D.popAtStack(0); // Returns 20. The stacks are now: 4 21   1  3  5 ﹈ ﹈ ﹈ D.popAtStack(2); // Returns 21. The stacks are now: 4   1  3  5 ﹈ ﹈ ﹈ D.pop() // Returns 5. The stacks are now: 4   1  3 ﹈ ﹈ D.pop() // Returns 4. The stacks are now: 1  3 ﹈ ﹈ D.pop() // Returns 3. The stacks are now: 1 ﹈ D.pop() // Returns 1. There are no stacks. D.pop() // Returns -1. There are still no stacks.

Constraints:

  • 1 <= capacity <= 20000
  • 1 <= val <= 20000
  • 0 <= index <= 100000
  • At most 200000 calls will be made to pushpop, and popAtStack.

Solution: Array of stacks + TreeSet

Store all the stacks in an array, and store the indices of all free stacks in a tree set.
1. push(): find the first free stack and push onto it, if it becomes full, remove it from free set.
2. pop(): pop element from the last stack by calling popAtIndex(stacks.size()-1).
3. popAtIndex(): pop element from given index
3.1 add it to free set if it was full
3.2 remove it from free set if it becomes empty
3.2.1 remove it from stack array if it is the last stack

Time complexity:
Push: O(logn)
Pop: O(logn)
PopAtIndex: O(logn)

Space complexity: O(n)

C++

花花酱 LeetCode 622. Design Circular Queue

Problem

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 can not 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.
Your implementation should support following operations:

  • MyCircularQueue(k): Constructor, set the size of the queue to be k.
  • Front: Get the front item from the queue. If the queue is empty, return -1.
  • Rear: Get the last item from the queue. If the queue is empty, return -1.
  • enQueue(value): Insert an element into the circular queue. Return true if the operation is successful.
  • deQueue(): Delete an element from the circular queue. Return true if the operation is successful.
  • isEmpty(): Checks whether the circular queue is empty or not.
  • isFull(): Checks whether the circular queue is full or not.

Example:

Note:

  • All values will be in the range of [1, 1000].
  • The number of operations will be in the range of [1, 1000].
  • Please do not use the built-in Queue library.

Solution: Deque [Cheating]

C++

Solution2: Array

 

 

花花酱 LeetCode 432. All O`one Data Structure

Problem

题目大意:设计一种数据结构,支持inc/dec/getmaxkey/getminkey操作,必须都在O(1)时间内完成。

https://leetcode.com/problems/all-oone-data-structure/description/

Implement a data structure supporting the following operations:

  1. Inc(Key) – Inserts a new key with value 1. Or increments an existing key by 1. Key is guaranteed to be a non-empty string.
  2. Dec(Key) – If Key’s value is 1, remove it from the data structure. Otherwise decrements an existing key by 1. If the key does not exist, this function does nothing. Key is guaranteed to be a non-empty string.
  3. GetMaxKey() – Returns one of the keys with maximal value. If no element exists, return an empty string "".
  4. GetMinKey() – Returns one of the keys with minimal value. If no element exists, return an empty string "".

Challenge: Perform all these in O(1) time complexity.

Solution

Time complexity: O(1)

Space complexity: O(n), n = # of unique keys

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