easy – Page 49 – Huahua’s Tech Road

花花酱 LeetCode 1114. Print in Order

By zxi on July 23, 2019

Suppose we have a class:

public class Foo {

  public void first() { print("first"); }

  public void second() { print("second"); }

  public void third() { print("third"); }

}

The same instance of Foo will be passed to three different threads. Thread A will call first(), thread B will call second(), and thread C will call third(). Design a mechanism and modify the program to ensure that second() is executed after first(), and third() is executed after second().

Example 1:

Input: [1,2,3]
Output: "firstsecondthird"
Explanation: There are three threads being fired asynchronously. The input [1,2,3] means thread A calls first(), thread B calls second(), and thread C calls third(). "firstsecondthird" is the correct output.

Example 2:

Input: [1,3,2]
Output: "firstsecondthird"
Explanation: The input [1,3,2] means thread A calls first(), thread B calls third(), and thread C calls second(). "firstsecondthird" is the correct output.

Note:

We do not know how the threads will be scheduled in the operating system, even though the numbers in the input seems to imply the ordering. The input format you see is mainly to ensure our tests’ comprehensiveness.

Solution 1: Mutex

C++

// Author: Huahua, 128 ms / 8.2 MB

class Foo {

public:

Foo() {

first_.lock();

second_.lock();

}

void first(function<void()> printFirst) {

// printFirst() outputs "first". Do not change or remove this line.

printFirst();

first_.unlock();

}

void second(function<void()> printSecond) {

first_.lock();

// printSecond() outputs "second". Do not change or remove this line.

printSecond();

first_.unlock();

second_.unlock();

}

void third(function<void()> printThird) {

second_.lock();

// printThird() outputs "third". Do not change or remove this line.

printThird();

second_.unlock();

}

private:

std::mutex first_;

std::mutex second_;

};

Solution 2: Busy waiting

C++

// Author: Huahua, 176 ms / 9.2 MB

#include <unistd.h>

class Foo {

public:

Foo() {}

void first(function<void()> printFirst) {

// printFirst() outputs "first". Do not change or remove this line.

printFirst();

state_ = 1;

}

void second(function<void()> printSecond) {

while (state_ != 1) usleep(1);

// printSecond() outputs "second". Do not change or remove this line.

printSecond();

state_ = 2;

}

void third(function<void()> printThird) {

while (state_ != 2) usleep(1);

// printThird() outputs "third". Do not change or remove this line.

printThird();

state_ = 3;

}

private:

int state_ = 0;

};

花花酱 LeetCode 217. Contains Duplicate

By zxi on July 22, 2019

Given an array of integers, find if the array contains any duplicates.

Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.

Example 1:

Input: [1,2,3,1]
Output: true

Example 2:

Input: [1,2,3,4]
Output: false

Example 3:

Input: [1,1,1,3,3,4,3,2,4,2]
Output: true

Solution 1: HashTable

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

C++

// Author: Huahua, 48ms/16.5MB

class Solution {

public:

bool containsDuplicate(vector<int>& nums) {

unordered_set<int> s;

for (int num : nums)

if (!s.insert(num).second) return true;

return false;

}

};

Solution 2: Sorting

Time complexity: O(nlogn)
Space complexity: O(1)

C++

// Author: Huahua, 28ms/11.4MB

class Solution {

public:

bool containsDuplicate(vector<int>& nums) {

sort(begin(nums), end(nums));

for (int i = 1; i < nums.size(); ++i)

if (nums[i] == nums[i - 1]) return true;

return false;

}

};

花花酱 LeetCode 226. Invert Binary Tree

By zxi on July 22, 2019

Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

Solution: Recursion

Recursive invert the left and right subtrees and swap them.

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

C++

// Author: Huahua

class Solution {

public:

TreeNode* invertTree(TreeNode* root) {

if (!root) return nullptr;

TreeNode* tmp = root->right;

root->right = invertTree(root->left);

root->left = invertTree(tmp);

return root;

}

};

Python3

# Author: Huahua

class Solution:

def invertTree(self, root: TreeNode) -> TreeNode:

if not root: return None

root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)

return root

花花酱 LeetCode 1128. Number of Equivalent Domino Pairs

By zxi on July 20, 2019

Given a list of dominoes, dominoes[i] = [a, b] is equivalent to dominoes[j] = [c, d] if and only if either (a==c and b==d), or (a==d and b==c) – that is, one domino can be rotated to be equal to another domino.

Return the number of pairs (i, j) for which 0 <= i < j < dominoes.length, and dominoes[i] is equivalent to dominoes[j].

Example 1:

Input: dominoes = [[1,2],[2,1],[3,4],[5,6]]
Output: 1

Constraints:

1 <= dominoes.length <= 40000
1 <= dominoes[i][j] <= 9

Solution: HashTable

Count how many times each key occurred so far.

Time complexity: O(n)
Space complexity: O(100)

C++

class Solution {

public:

int numEquivDominoPairs(vector<vector<int>>& dominoes) {

vector<int> m(100);

int ans = 0;

for (const auto& d : dominoes) {

int k1 = d[0] * 10 + d[1];

int k2 = d[1] * 10 + d[0];

ans += m[k1];

if (k1 != k2) ans += m[k2];

++m[k1];

}

return ans;

}

};

花花酱 LeetCode 1108. Defanging an IP Address

By zxi on July 11, 2019

Given a valid (IPv4) IP address, return a defanged version of that IP address.

A defanged IP address replaces every period "." with "[.]".

Example 1:

Input: address = "1.1.1.1"
Output: "1[.]1[.]1[.]1"

Example 2:

Input: address = "255.100.50.0"
Output: "255[.]100[.]50[.]0"

Constraints:

The given address is a valid IPv4 address.

Solution: String

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

C++

class Solution {

public:

string defangIPaddr(string address) {

string output;

for (const char c : address) {

if (c == '.') output += "[.]";

else output += c;

}

return output;

}

};

Posts tagged as “easy”

花花酱 LeetCode 1114. Print in Order

Solution 1: Mutex

C++

Solution 2: Busy waiting

C++

花花酱 LeetCode 217. Contains Duplicate

Solution 1: HashTable

C++

Solution 2: Sorting

C++

花花酱 LeetCode 226. Invert Binary Tree

Solution: Recursion

C++

Python3

花花酱 LeetCode 1128. Number of Equivalent Domino Pairs

Solution: HashTable

C++

花花酱 LeetCode 1108. Defanging an IP Address

Solution: String

C++