events Archives - Huahua's Tech Road

There is a party where n friends numbered from 0 to n - 1 are attending. There is an infinite number of chairs in this party that are numbered from 0 to infinity. When a friend arrives at the party, they sit on the unoccupied chair with the smallest number.

For example, if chairs 0, 1, and 5 are occupied when a friend comes, they will sit on chair number 2.

When a friend leaves the party, their chair becomes unoccupied at the moment they leave. If another friend arrives at that same moment, they can sit in that chair.

You are given a 0-indexed 2D integer array times where times[i] = [arrival_i, leaving_i], indicating the arrival and leaving times of the i^th friend respectively, and an integer targetFriend. All arrival times are distinct.

Return the chair number that the friend numbered targetFriend will sit on.

Example 1:

Input: times = [[1,4],[2,3],[4,6]], targetFriend = 1
Output: 1
Explanation: 
- Friend 0 arrives at time 1 and sits on chair 0.
- Friend 1 arrives at time 2 and sits on chair 1.
- Friend 1 leaves at time 3 and chair 1 becomes empty.
- Friend 0 leaves at time 4 and chair 0 becomes empty.
- Friend 2 arrives at time 4 and sits on chair 0.
Since friend 1 sat on chair 1, we return 1.

Example 2:

Input: times = [[3,10],[1,5],[2,6]], targetFriend = 0
Output: 2
Explanation: 
- Friend 1 arrives at time 1 and sits on chair 0.
- Friend 2 arrives at time 2 and sits on chair 1.
- Friend 0 arrives at time 3 and sits on chair 2.
- Friend 1 leaves at time 5 and chair 0 becomes empty.
- Friend 2 leaves at time 6 and chair 1 becomes empty.
- Friend 0 leaves at time 10 and chair 2 becomes empty.
Since friend 0 sat on chair 2, we return 2.

Constraints:

n == times.length
2 <= n <= 10⁴
times[i].length == 2
1 <= arrival_i < leaving_i <= 10⁵
0 <= targetFriend <= n - 1
Each arrival_i time is distinct.

Solution: Treeset + Simulation

Use a treeset to track available chairs, sort events by time.
note: process leaving events first.

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

C++

// Author: Huahua

class Solution {

public:

int smallestChair(vector<vector<int>>& times, int targetFriend) {

const int n = times.size();

vector<pair<int, int>> arrival(n);

vector<pair<int, int>> leaving(n);

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

arrival[i] = {times[i][0], i};

leaving[i] = {times[i][1], i};

}

vector<int> pos(n);

iota(begin(pos), end(pos), 0); // pos = [0, 1, ..., n -1]

set<int> aval(begin(pos), end(pos));

sort(begin(arrival), end(arrival));

sort(begin(leaving), end(leaving));

for (int i = 0, j = 0; i < n; ++i) {

const auto [t, u] = arrival[i];

while (j < n && leaving[j].first <= t)

aval.insert(pos[leaving[j++].second]);

aval.erase(pos[u] = *begin(aval));

if (u == targetFriend) return pos[u];

}

return -1;

}

};

Solution: Sort + Heap

Sort events by start time, process them from left to right.

Use a min heap to store the events processed so far, a variable cur to track the max value of a non-overlapping event.

For a given event, pop all non-overlapping events whose end time is smaller than its start time and update cur.

ans = max(val + cur)

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

C++

// Author: Huahua

class Solution {

public:

int maxTwoEvents(vector<vector<int>>& events) {

using E = pair<int,int>;

sort(begin(events), end(events));

priority_queue<E, vector<E>, greater<E>> q; // (end_time, val)

int ans = 0;

int cur = 0;

for (const auto& e : events) {

while (!q.empty() && q.top().first < e[0]) {

cur = max(cur, q.top().second);

q.pop();

}

ans = max(ans, cur + e[2]);

q.emplace(e[1], e[2]);

}

return ans;

}

};

Posts tagged as “events”

花花酱 LeetCode 1942. The Number of the Smallest Unoccupied Chair

Solution: Treeset + Simulation

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花花酱 LeetCode 2054. Two Best Non-Overlapping Events

Solution: Sort + Heap

C++