Press "Enter" to skip to content

Posts tagged as “deque”

花花酱 LeetCode 1813. Sentence Similarity III

A sentence is a list of words that are separated by a single space with no leading or trailing spaces. For example, "Hello World""HELLO""hello world hello world" are all sentences. Words consist of only uppercase and lowercase English letters.

Two sentences sentence1 and sentence2 are similar if it is possible to insert an arbitrary sentence (possibly empty) inside one of these sentences such that the two sentences become equal. For example, sentence1 = "Hello my name is Jane" and sentence2 = "Hello Jane" can be made equal by inserting "my name is" between "Hello" and "Jane" in sentence2.

Given two sentences sentence1 and sentence2, return true if sentence1 and sentence2 are similar. Otherwise, return false.

Example 1:

Input: sentence1 = "My name is Haley", sentence2 = "My Haley"
Output: true
Explanation: sentence2 can be turned to sentence1 by inserting "name is" between "My" and "Haley".

Example 2:

Input: sentence1 = "of", sentence2 = "A lot of words"
Output: false
Explanation: No single sentence can be inserted inside one of the sentences to make it equal to the other.

Example 3:

Input: sentence1 = "Eating right now", sentence2 = "Eating"
Output: true
Explanation: sentence2 can be turned to sentence1 by inserting "right now" at the end of the sentence.

Example 4:

Input: sentence1 = "Luky", sentence2 = "Lucccky"
Output: false

Constraints:

  • 1 <= sentence1.length, sentence2.length <= 100
  • sentence1 and sentence2 consist of lowercase and uppercase English letters and spaces.
  • The words in sentence1 and sentence2 are separated by a single space.

Solution: Dequeue / Common Prefix + Suffix

Break sequences to words, store them in two deques. Pop the common prefix and suffix. At least one of the deque should be empty.

Time complexity: O(m+n)
Space complexity: O(m+n)

C++

Python3

花花酱 LeetCode 1585. Check If String Is Transformable With Substring Sort Operations

Given two strings s and t, you want to transform string s into string t using the following operation any number of times:

  • Choose a non-empty substring in s and sort it in-place so the characters are in ascending order.

For example, applying the operation on the underlined substring in "14234" results in "12344".

Return true if it is possible to transform string s into string t. Otherwise, return false.

substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "84532", t = "34852"
Output: true
Explanation: You can transform s into t using the following sort operations:
"84532" (from index 2 to 3) -> "84352"
"84352" (from index 0 to 2) -> "34852"

Example 2:

Input: s = "34521", t = "23415"
Output: true
Explanation: You can transform s into t using the following sort operations:
"34521" -> "23451"
"23451" -> "23415"

Example 3:

Input: s = "12345", t = "12435"
Output: false

Example 4:

Input: s = "1", t = "2"
Output: false

Constraints:

  • s.length == t.length
  • 1 <= s.length <= 105
  • s and t only contain digits from '0' to '9'.

Solution: Queue

We can move a smaller digit from right to left by sorting two adjacent digits.
e.g. 18572 -> 18527 -> 18257 -> 12857, but we can not move a larger to the left of a smaller one.

Thus, for each digit in the target string, we find the first occurrence of it in s, and try to move it to the front by checking if there is any smaller one in front of it.

Time complexity: O(n)
Space complexity: O(n)

C++

Python3

花花酱 LeetCode 919. Complete Binary Tree Inserter

Problem

complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible.

Write a data structure CBTInserter that is initialized with a complete binary tree and supports the following operations:

  • CBTInserter(TreeNode root) initializes the data structure on a given tree with head node root;
  • CBTInserter.insert(int v) will insert a TreeNode into the tree with value node.val = v so that the tree remains complete, and returns the value of the parent of the inserted TreeNode;
  • CBTInserter.get_root() will return the head node of the tree.

Example 1:

Input: inputs = ["CBTInserter","insert","get_root"], inputs = [[[1]],[2],[]]
Output: [null,1,[1,2]]

Example 2:

Input: inputs = ["CBTInserter","insert","insert","get_root"], inputs = [[[1,2,3,4,5,6]],[7],[8],[]]
Output: [null,3,4,[1,2,3,4,5,6,7,8]]

Note:

  1. The initial given tree is complete and contains between 1 and 1000 nodes.
  2. CBTInserter.insert is called at most 10000 times per test case.
  3. Every value of a given or inserted node is between 0 and 5000.

Solution 2: Deque

Using a deck to keep track of insertable nodes (potential parents) in order.

Time complexity: O(1) / O(n) first call

Space complexity: O(n)

C++

花花酱 LeetCode 641. Design Circular Deque

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.

Related Problems

花花酱 LeetCode 622. Design Circular Queue

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.

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:

MyCircularQueue circularQueue = new MyCircularQueue(3); // set the size to be 3
circularQueue.enQueue(1);  // return true
circularQueue.enQueue(2);  // return true
circularQueue.enQueue(3);  // return true
circularQueue.enQueue(4);  // return false, the queue is full
circularQueue.Rear();  // return 3
circularQueue.isFull();  // return true
circularQueue.deQueue();  // return true
circularQueue.enQueue(4);  // return true
circularQueue.Rear();  // return 4

Note:

  • All values will be in the range of [0, 1000].
  • The number of operations will be in the range of [1, 1000].
  • Please do not use the built-in Queue library.

Solution: Simulate with an array

We need a fixed length array, and the head location as well as the size of the current queue.

We can use q[head] to access the front, and q[(head + size – 1) % k] to access the rear.

Time complexity: O(1) for all the operations.
Space complexity: O(k)

C++

Java

Python3