Posts tagged as “desgin”

花花酱 LeetCode 1381. Design a Stack With Increment Operation

By zxi on March 15, 2020

Design a stack which supports the following operations.

Implement the CustomStack class:

CustomStack(int maxSize) Initializes the object with maxSize which is the maximum number of elements in the stack or do nothing if the stack reached the maxSize.
void push(int x) Adds x to the top of the stack if the stack hasn’t reached the maxSize.
int pop() Pops and returns the top of stack or -1 if the stack is empty.
void inc(int k, int val) Increments the bottom k elements of the stack by val. If there are less than k elements in the stack, just increment all the elements in the stack.

Example 1:

Input
["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]
Output
[null,null,null,2,null,null,null,null,null,103,202,201,-1]
Explanation
CustomStack customStack = new CustomStack(3); // Stack is Empty []
customStack.push(1);                          // stack becomes [1]
customStack.push(2);                          // stack becomes [1, 2]
customStack.pop();                            // return 2 --> Return top of the stack 2, stack becomes [1]
customStack.push(2);                          // stack becomes [1, 2]
customStack.push(3);                          // stack becomes [1, 2, 3]
customStack.push(4);                          // stack still [1, 2, 3], Don't add another elements as size is 4
customStack.increment(5, 100);                // stack becomes [101, 102, 103]
customStack.increment(2, 100);                // stack becomes [201, 202, 103]
customStack.pop();                            // return 103 --> Return top of the stack 103, stack becomes [201, 202]
customStack.pop();                            // return 202 --> Return top of the stack 102, stack becomes [201]
customStack.pop();                            // return 201 --> Return top of the stack 101, stack becomes []
customStack.pop();                            // return -1 --> Stack is empty return -1.

Solution: Simulation

Time complexity:
init: O(1)
pop: O(1)
push: O(1)
inc: O(k)

C++

// Author: Huahua

class CustomStack {

public:

CustomStack(int maxSize): max_size_(maxSize) {}

void push(int x) {

if (data_.size() == max_size_) return;

data_.push_back(x);

}

int pop() {

if (data_.empty()) return -1;

int val = data_.back();

data_.pop_back();

return val;

}

void increment(int k, int val) {

for (int i = 0; i < min(static_cast<size_t>(k),

data_.size()); ++i)

data_[i] += val;

}

private:

int max_size_;

vector<int> data_;

};

/**

* Your CustomStack object will be instantiated and called as such:

* CustomStack* obj = new CustomStack(maxSize);

* obj->push(x);

* int param_2 = obj->pop();

* obj->increment(k,val);

花花酱 LeetCode 707. Design Linked List

By zxi on August 9, 2018

Problem

Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.

Implement these functions in your linked list class:

get(index) : Get the value of the index-th node in the linked list. If the index is invalid, return -1.
addAtHead(val) : Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
addAtTail(val) : Append a node of value val to the last element of the linked list.
addAtIndex(index, val) : Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
deleteAtIndex(index) : Delete the index-th node in the linked list, if the index is valid.

Example:

MyLinkedList linkedList = new MyLinkedList();

linkedList.addAtHead(1);

linkedList.addAtTail(3);

linkedList.addAtIndex(1, 2); // linked list becomes 1->2->3

linkedList.get(1); // returns 2

linkedList.deleteAtIndex(1); // now the linked list is 1->3

linkedList.get(1); // returns 3

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 LinkedList library.

Solution: Single linked list

Keep tracking head and tail of the list.

Time Complexity:

addAtHead, addAtTail O(1)

addAtIndex O(index)

deleteAtIndex O(index)

Space complexity: O(1)

C++

Tracking head/tail and size of the list.

// Author: Huahua

// Running time: 24 ms

class MyLinkedList {

public:

MyLinkedList(): head_(nullptr), tail_(nullptr), size_(0) {}

~MyLinkedList() {

Node* node = head_;

while (node) {

Node* cur = node;

node = node->next;

delete cur;

}

head_ = nullptr;

tail_ = nullptr;

}

int get(int index) {

if (index < 0 || index >= size_) return -1;

auto node = head_;

while (index--)

node = node->next;

return node->val;

}

void addAtHead(int val) {

head_ = new Node(val, head_);

if (size_++ == 0)

tail_ = head_;

}

void addAtTail(int val) {

auto node = new Node(val);

if (size_++ == 0) {

head_ = tail_ = node;

} else {

tail_->next = node;

tail_ = tail_->next;

}

void addAtIndex(int index, int val) {

if (index < 0 || index > size_) return;

if (index == 0) return addAtHead(val);

if (index == size_) return addAtTail(val);

Node dummy(0, head_);

Node* prev = &dummy;

while (index--) prev = prev->next;

prev->next = new Node(val, prev->next);

++size_;

}

void deleteAtIndex(int index) {

if (index < 0 || index >= size_) return;

Node dummy(0, head_);

Node* prev = &dummy;

for (int i = 0; i < index; ++i)

prev = prev->next;

Node* node_to_delete = prev->next;

prev->next = prev->next->next;

if (index == 0) head_ = prev->next;

if (index == size_ - 1) tail_ = prev;

delete node_to_delete;

--size_;

}

private:

struct Node {

int val;

Node* next;

Node(int _val): Node(_val, nullptr) {}

Node(int _val, Node* _next): val(_val), next(_next) {}

};

Node* head_;

Node* tail_;

int size_;

};

// Author: Huahua

// Running time: 28 ms

class MyLinkedList {

public:

MyLinkedList(): head_(nullptr), tail_(nullptr), dummy_(0), size_(0) {}

~MyLinkedList() {

Node* node = head_;

while (node) {

Node* cur = node;

node = node->next;

delete cur;

}

head_ = nullptr;

tail_ = nullptr;

}

int get(int index) {

if (index < 0 || index >= size_) return -1;

return getNode(index)->val;

}

void addAtHead(int val) {

head_ = new Node(val, head_);

if (size_++ == 0)

tail_ = head_;

}

void addAtTail(int val) {

auto node = new Node(val);

if (size_++ == 0) {

head_ = tail_ = node;

} else {

tail_->next = node;

tail_ = tail_->next;

}

void addAtIndex(int index, int val) {

if (index < 0 || index > size_) return;

if (index == 0) return addAtHead(val);

if (index == size_) return addAtTail(val);

Node* prev = getNode(index - 1);

prev->next = new Node(val, prev->next);

++size_;

}

void deleteAtIndex(int index) {

if (index < 0 || index >= size_) return;

Node* prev = getNode(index - 1);

Node* node_to_delete = prev->next;

prev->next = node_to_delete->next;

if (index == 0) head_ = prev->next;

if (index == size_ - 1) tail_ = prev;

delete node_to_delete;

--size_;

}

private:

struct Node {

int val;

Node* next;

Node(int _val): Node(_val, nullptr) {}

Node(int _val, Node* _next): val(_val), next(_next) {}

};

Node* head_; // Does not own

Node* tail_; // Does not own

Node dummy_;

int size_;

Node* getNode(int index) {

dummy_.next = head_;

Node* n = &dummy_;

for (int i = 0; i <= index; ++i)

n = n->next;

return n;

}

};

Python3

"""

Author: Huahua

Running time: 132 ms

"""

class Node:

def __init__(self, val, _next=None):

self.val = val

self.next = _next

class MyLinkedList:

def __init__(self):

self.head = self.tail = None

self.size = 0

def getNode(self, index):

n = Node(0, self.head)

for i in range(index + 1):

n = n.next

return n

def get(self, index):

if index < 0 or index >= self.size: return -1

return self.getNode(index).val

def addAtHead(self, val):

n = Node(val, self.head)

self.head = n

if self.size == 0:

self.tail = n

self.size += 1

def addAtTail(self, val):

n = Node(val)

if self.size == 0:

self.head = self.tail = n

else:

self.tail.next = n

self.tail = n

self.size += 1

def addAtIndex(self, index, val):

if index < 0 or index > self.size: return

if index == 0: return self.addAtHead(val)

if index == self.size: return self.addAtTail(val)

prev = self.getNode(index - 1)

n = Node(val, prev.next)

prev.next = n

self.size += 1

def deleteAtIndex(self, index):

if index < 0 or index >= self.size: return

prev = self.getNode(index - 1)

prev.next = prev.next.next

if index == 0: self.head = prev.next

if index == self.size - 1: self.tail = prev

self.size -= 1

Java

// Author: Huahua

// Running time: 74 ms

class MyLinkedList {

class Node {

public int val;

public Node next;

public Node(int val) { this.val = val; this.next = null; }

public Node(int val, Node next) { this.val = val; this.next = next; }

}

private Node head;

private Node tail;

private int size;

public MyLinkedList() {

this.head = this.tail = null;

this.size = 0;

}

private Node getNode(int index) {

Node n = new Node(0, this.head);

while (index-- >= 0) {

n = n.next;

}

return n;

}

public int get(int index) {

if (index < 0 || index >= size) return -1;

return getNode(index).val;

}

public void addAtHead(int val) {

this.head = new Node(val, this.head);

if (this.size++ == 0)

this.tail = this.head;

}

public void addAtTail(int val) {

Node n = new Node(val);

if (this.size++ == 0)

this.head = this.tail = n;

else

this.tail = this.tail.next = n;

}

public void addAtIndex(int index, int val) {

if (index < 0 || index > this.size) return;

if (index == 0) { this.addAtHead(val); return; }

if (index == size) { this.addAtTail(val); return; }

Node prev = this.getNode(index - 1);

prev.next = new Node(val, prev.next);

++this.size;

}

public void deleteAtIndex(int index) {

if (index < 0 || index >= this.size) return;

Node prev = this.getNode(index - 1);

prev.next = prev.next.next;

if (index == 0) this.head = prev.next;

if (index == this.size - 1) this.tail = prev;

--this.size;

}

花花酱 LeetCode 872. Implement Rand10() Using Rand7()

By zxi on July 16, 2018

Problem

Given a function rand7 which generates a uniform random integer in the range 1 to 7, write a function rand10 which generates a uniform random integer in the range 1 to 10.

Do NOT use system’s Math.random().

Example 1:

Input: 1
Output: [7]

Example 2:

Input: 2
Output: [8,4]

Example 3:

Input: 3
Output: [8,1,10]

Note:

rand7 is predefined.
Each testcase has one argument: n, the number of times that rand10 is called.

Solution: Math

Time complexity: O(49/40) = O(1)

// Author: Huahua

// Running time: 112 ms

class Solution {

public:

int rand10() {

int target = 40;

while (target >= 40)

target = 7 * (rand7() - 1) + (rand7() - 1);

return target % 10 + 1;

}

};

Time complexity: O(7/6 + 7 / 5) = O(1)

// Author: Huahua

// Running time: 112 ms

class Solution {

public:

int rand10() {

int i = INT_MAX;

int j = INT_MAX;

while (i > 6) i = rand7(); // i = [1, 2, 3, 4, 5, 6]

while (j > 5) j = rand7(); // j = [1, 2, 3, 4, 5]

return j + 5 * (i & 1);

}

};

花花酱 LeetCode 641. Design Circular Deque

By zxi on July 14, 2018

Problem

Design your implementation of the circular double-ended queue (deque).
Your implementation should support following operations:

MyCircularDeque(k): Constructor, set the size of the deque to be k.
insertFront(): Adds an item at the front of Deque. Return true if the operation is successful.
insertLast(): Adds an item at the rear of Deque. Return true if the operation is successful.
deleteFront(): Deletes an item from the front of Deque. Return true if the operation is successful.
deleteLast(): Deletes an item from the rear of Deque. Return true if the operation is successful.
getFront(): Gets the front item from the Deque. If the deque is empty, return -1.
getRear(): Gets the last item from Deque. If the deque is empty, return -1.
isEmpty(): Checks whether Deque is empty or not.
isFull(): Checks whether Deque is full or not.

Example:

MyCircularDeque circularDeque = new MycircularDeque(3); // set the size to be 3
circularDeque.insertLast(1);			// return true
circularDeque.insertLast(2);			// return true
circularDeque.insertFront(3);			// return true
circularDeque.insertFront(4);			// return false, the queue is full
circularDeque.getRear();  				// return 32
circularDeque.isFull();				// return true
circularDeque.deleteLast();			// return true
circularDeque.insertFront(4);			// return true
circularDeque.getFront();				// return 4

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 Deque library.

Solution

Using head and tail to pointer to the head and the tail in the circular buffer.

// Author: Huahua

// Running time: 20 ms

class MyCircularDeque {

public:

/** Initialize your data structure here. Set the size of the deque to be k. */

MyCircularDeque(int k): k_(k), q_(k), head_(1), tail_(-1), size_(0) {}

/** Adds an item at the front of Deque. Return true if the operation is successful. */

bool insertFront(int value) {

if (isFull()) return false;

head_ = (head_ - 1 + k_) % k_;

q_[head_] = value;

if (size_ == 0) tail_ = head_;

++size_;

return true;

}

/** Adds an item at the rear of Deque. Return true if the operation is successful. */

bool insertLast(int value) {

if (isFull()) return false;

tail_ = (tail_ + 1 + k_) % k_;

q_[tail_] = value;

if (size_ == 0) head_ = tail_;

++size_;

return true;

}

/** Deletes an item from the front of Deque. Return true if the operation is successful. */

bool deleteFront() {

if (isEmpty()) return false;

head_ = (head_ + 1 + k_) % k_;

--size_;

return true;

}

/** Deletes an item from the rear of Deque. Return true if the operation is successful. */

bool deleteLast() {

if (isEmpty()) return false;

tail_ = (tail_ - 1 + k_) % k_;

--size_;

return true;

}

/** Get the front item from the deque. */

int getFront() {

if (isEmpty()) return -1;

return q_[head_];

}

/** Get the last item from the deque. */

int getRear() {

if (isEmpty()) return -1;

return q_[tail_];

}

/** Checks whether the circular deque is empty or not. */

bool isEmpty() {

return size_ == 0;

}

/** Checks whether the circular deque is full or not. */

bool isFull() {

return size_ == k_;

}

private:

const int k_;

int head_;

int tail_;

int size_;

vector<int> q_;

};

Posts tagged as “desgin”

花花酱 LeetCode 1381. Design a Stack With Increment Operation

Solution: Simulation

C++

花花酱 LeetCode 707. Design Linked List

Problem

Solution: Single linked list

花花酱 LeetCode 872. Implement Rand10() Using Rand7()

Problem

Solution: Math

花花酱 LeetCode 641. Design Circular Deque

Problem

Solution

Related Problems