Posts tagged as “queue”

花花酱 LeetCode 2073. Time Needed to Buy Tickets

By zxi on November 15, 2021

There are n people in a line queuing to buy tickets, where the 0^th person is at the front of the line and the (n - 1)^th person is at the back of the line.

You are given a 0-indexed integer array tickets of length n where the number of tickets that the i^th person would like to buy is tickets[i].

Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously) in order to buy more tickets. If a person does not have any tickets left to buy, the person will leave the line.

Return the time taken for the person at position k(0-indexed) to finish buying tickets.

Example 1:

Input: tickets = [2,3,2], k = 2
Output: 6
Explanation: 
- In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1].
- In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0].
The person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.

Example 2:

Input: tickets = [5,1,1,1], k = 0
Output: 8
Explanation:
- In the first pass, everyone in the line buys a ticket and the line becomes [4, 0, 0, 0].
- In the next 4 passes, only the person in position 0 is buying tickets.
The person at position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.

Constraints:

n == tickets.length
1 <= n <= 100
1 <= tickets[i] <= 100
0 <= k < n

Solution 1: Simulation

Time complexity: O(n * tickets[k])
Space complexity: O(n) / O(1)

C++

// Author: huahua

class Solution {

public:

int timeRequiredToBuy(vector<int>& tickets, int k) {

const int n = tickets.size();

queue<pair<int, int>> q;

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

q.emplace(i, tickets[i]);

int ans = 0;

while (!q.empty()) {

++ans;

auto [i, t] = q.front(); q.pop();

if (--t == 0 && k == i) return ans;

if (t) q.emplace(i, t);

}

return -1;

}

};

Solution 2: Math

Each person before k will have tickets[k] rounds, each person after k will have tickets[k] – 1 rounds.

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

C++

// Author: huahua

class Solution {

public:

int timeRequiredToBuy(vector<int>& tickets, int k) {

int ans = 0;

for (int i = 0; i < tickets.size(); ++i)

ans += min(tickets[i], tickets[k] - (i > k));

return ans;

}

};

花花酱 LeetCode 1823. Find the Winner of the Circular Game

By zxi on April 12, 2021

There are n friends that are playing a game. The friends are sitting in a circle and are numbered from 1 to n in clockwise order. More formally, moving clockwise from the i^th friend brings you to the (i+1)^th friend for 1 <= i < n, and moving clockwise from the n^th friend brings you to the 1^st friend.

The rules of the game are as follows:

Start at the 1^st friend.
Count the next k friends in the clockwise direction including the friend you started at. The counting wraps around the circle and may count some friends more than once.
The last friend you counted leaves the circle and loses the game.
If there is still more than one friend in the circle, go back to step 2 starting from the friend immediately clockwise of the friend who just lost and repeat.
Else, the last friend in the circle wins the game.

Given the number of friends, n, and an integer k, return the winner of the game.

Example 1:

Input: n = 5, k = 2
Output: 3
Explanation: Here are the steps of the game:
1) Start at friend 1.
2) Count 2 friends clockwise, which are friends 1 and 2.
3) Friend 2 leaves the circle. Next start is friend 3.
4) Count 2 friends clockwise, which are friends 3 and 4.
5) Friend 4 leaves the circle. Next start is friend 5.
6) Count 2 friends clockwise, which are friends 5 and 1.
7) Friend 1 leaves the circle. Next start is friend 3.
8) Count 2 friends clockwise, which are friends 3 and 5.
9) Friend 5 leaves the circle. Only friend 3 is left, so they are the winner.

Example 2:

Input: n = 6, k = 5
Output: 1
Explanation: The friends leave in this order: 5, 4, 6, 2, 3. The winner is friend 1.

Constraints:

1 <= k <= n <= 500

Solution 1: Simulation w/ Queue / List

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

C++/Queue

// Author: Huahua, 56 ms, 24.3 MB

class Solution {

public:

int findTheWinner(int n, int k) {

queue<int> q;

for (int i = 1; i <= n; ++i)

q.push(i);

while (q.size() != 1) {

for (int i = 1; i < k; ++i) {

int x = q.front();

q.pop();

q.push(x);

}

q.pop();

}

return q.front();

}

};

C++/List

// Author: Huahua, 12 ms, 6.8 MB

class Solution {

public:

int findTheWinner(int n, int k) {

list<int> l;

for (int i = 1; i <= n; ++i)

l.push_back(i);

auto it = l.begin();

while (l.size() != 1) {

for (int i = 1; i < k; ++i)

if (++it == l.end())

it = l.begin();

it = l.erase(it);

if (it == l.end())

it = l.begin();

}

return l.front();

}

};

花花酱 LeetCode 1701. Average Waiting Time

By zxi on December 26, 2020

There is a restaurant with a single chef. You are given an array customers, where customers[i] = [arrival_i, time_i]:

arrival_i is the arrival time of the i^th customer. The arrival times are sorted in non-decreasing order.
time_i is the time needed to prepare the order of the i^th customer.

When a customer arrives, he gives the chef his order, and the chef starts preparing it once he is idle. The customer waits till the chef finishes preparing his order. The chef does not prepare food for more than one customer at a time. The chef prepares food for customers in the order they were given in the input.

Return the average waiting time of all customers. Solutions within 10^-5 from the actual answer are considered accepted.

Example 1:

Input: customers = [[1,2],[2,5],[4,3]]
Output: 5.00000
Explanation:
1) The first customer arrives at time 1, the chef takes his order and starts preparing it immediately at time 1, and finishes at time 3, so the waiting time of the first customer is 3 - 1 = 2.
2) The second customer arrives at time 2, the chef takes his order and starts preparing it at time 3, and finishes at time 8, so the waiting time of the second customer is 8 - 2 = 6.
3) The third customer arrives at time 4, the chef takes his order and starts preparing it at time 8, and finishes at time 11, so the waiting time of the third customer is 11 - 4 = 7.
So the average waiting time = (2 + 6 + 7) / 3 = 5.

Example 2:

Input: customers = [[5,2],[5,4],[10,3],[20,1]]
Output: 3.25000
Explanation:
1) The first customer arrives at time 5, the chef takes his order and starts preparing it immediately at time 5, and finishes at time 7, so the waiting time of the first customer is 7 - 5 = 2.
2) The second customer arrives at time 5, the chef takes his order and starts preparing it at time 7, and finishes at time 11, so the waiting time of the second customer is 11 - 5 = 6.
3) The third customer arrives at time 10, the chef takes his order and starts preparing it at time 11, and finishes at time 14, so the waiting time of the third customer is 14 - 10 = 4.
4) The fourth customer arrives at time 20, the chef takes his order and starts preparing it immediately at time 20, and finishes at time 21, so the waiting time of the fourth customer is 21 - 20 = 1.
So the average waiting time = (2 + 6 + 4 + 1) / 4 = 3.25.

Constraints:

1 <= customers.length <= 10⁵
1 <= arrival_i, time_i <= 10⁴
arrival_i<= arrival_i+1

Solution: Simulation

When a customer arrives, if the arrival time is greater than current, then advance the clock to arrival time. Advance the clock by cooking time. Waiting time = current time – arrival time.

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

C++

// Author: Huahua

class Solution {

public:

double averageWaitingTime(vector<vector<int>>& customers) {

int t = 0;

double w = 0;

for (const auto& c : customers) {

if (c[0] > t) t = c[0];

t += c[1];

w += t - c[0];

}

return w / customers.size();

}

};

Python3

# Author: Huahua

class Solution:

def averageWaitingTime(self, customers: List[List[int]]) -> float:

cur = 0

waiting = 0

for arrival, time in customers:

cur = max(cur, arrival) + time

waiting += cur - arrival

return waiting / len(customers)

花花酱 LeetCode 1700. Number of Students Unable to Eat Lunch

By zxi on December 26, 2020

The school cafeteria offers circular and square sandwiches at lunch break, referred to by numbers 0 and 1 respectively. All students stand in a queue. Each student either prefers square or circular sandwiches.

The number of sandwiches in the cafeteria is equal to the number of students. The sandwiches are placed in a stack. At each step:

If the student at the front of the queue prefers the sandwich on the top of the stack, they will take it and leave the queue.
Otherwise, they will leave it and go to the queue’s end.

This continues until none of the queue students want to take the top sandwich and are thus unable to eat.

You are given two integer arrays students and sandwiches where sandwiches[i] is the type of the i^th sandwich in the stack (i = 0 is the top of the stack) and students[j] is the preference of the j^th student in the initial queue (j = 0 is the front of the queue). Return the number of students that are unable to eat.

Example 1:

Input: students = [1,1,0,0], sandwiches = [0,1,0,1]
Output: 0 
Explanation:
- Front student leaves the top sandwich and returns to the end of the line making students = [1,0,0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [0,0,1,1].
- Front student takes the top sandwich and leaves the line making students = [0,1,1] and sandwiches = [1,0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [1,1,0].
- Front student takes the top sandwich and leaves the line making students = [1,0] and sandwiches = [0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [0,1].
- Front student takes the top sandwich and leaves the line making students = [1] and sandwiches = [1].
- Front student takes the top sandwich and leaves the line making students = [] and sandwiches = [].
Hence all students are able to eat.

Example 2:

Input: students = [1,1,1,0,0,1], sandwiches = [1,0,0,0,1,1]
Output: 3

Constraints:

1 <= students.length, sandwiches.length <= 100
students.length == sandwiches.length
sandwiches[i] is 0 or 1.
students[i] is 0 or 1.

Solution 1: Simulation

Time complexity: O(n^2)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int countStudents(vector<int>& students, vector<int>& sandwiches) {

queue<int> q;

for (int s : students) q.push(s);

int c = 0;

int i = 0;

while (!q.empty()) {

int s = q.front(); q.pop();

if (s == sandwiches[i]) {

++i;

c = 0;

} else {

q.push(s);

}

if (++c > q.size()) break;

}

return q.size();

}

};

Solution 2: Counting

Count student’s preferences. Then process students from 1 to n, if there is no sandwich for current student then we can stop, since he/she will block all the students behind him/her.

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

C++

// Author: Huahua

class Solution {

public:

int countStudents(vector<int>& students, vector<int>& sandwiches) {

const int n = students.size();

vector<int> c(2);

for (int p : students) ++c[p];

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

if (--c[sandwiches[i]] < 0) return n - i;

return 0;

}

};

花花酱 LeetCode 1417. Reformat The String

By zxi on April 20, 2020

Given alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits).

You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.

Return the reformatted string or return an empty string if it is impossible to reformat the string.

Example 1:

Input: s = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.

Example 2:

Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.

Example 3:

Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.

Example 4:

Input: s = "covid2019"
Output: "c2o0v1i9d"

Example 5:

Input: s = "ab123"
Output: "1a2b3"

Constraints:

1 <= s.length <= 500
s consists of only lowercase English letters and/or digits.

Solution: Two streams

Create two stacks, one for alphas, another for numbers. If the larger stack has more than one element than the other one then no solution, return “”. Otherwise, interleave two stacks, start with the larger one.

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

C++

// Author: Huahua

class Solution {

public:

string reformat(string s) {

string q1;

string q2;

for (char c : s) {

if (isalpha(c))

q1 += c;

else

q2 += c;

}

if (q1.length() < q2.length())

swap(q1, q2);

if (q1.length() > q2.length() + 1)

return "";

string ans;

while (!q1.empty()) {

ans += q1.back();

q1.pop_back();

if (q2.empty()) break;

ans += q2.back();

q2.pop_back();

}

return ans;

}

};