##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/855.html

# 855. Exam Room (Medium)

In an exam room, there are N seats in a single row, numbered 0, 1, 2, ..., N-1.

When a student enters the room, they must sit in the seat that maximizes the distance to the closest person.  If there are multiple such seats, they sit in the seat with the lowest number.  (Also, if no one is in the room, then the student sits at seat number 0.)

Return a class ExamRoom(int N) that exposes two functions: ExamRoom.seat() returning an int representing what seat the student sat in, and ExamRoom.leave(int p) representing that the student in seat number p now leaves the room.  It is guaranteed that any calls to ExamRoom.leave(p) have a student sitting in seat p.

Example 1:

Input: ["ExamRoom","seat","seat","seat","seat","leave","seat"], [[10],[],[],[],[],[4],[]]
Output: [null,0,9,4,2,null,5]
Explanation:
ExamRoom(10) -> null
seat() -> 0, no one is in the room, then the student sits at seat number 0.
seat() -> 9, the student sits at the last seat number 9.
seat() -> 4, the student sits at the last seat number 4.
seat() -> 2, the student sits at the last seat number 2.
leave(4) -> null
seat() -> 5, the student sits at the last seat number 5.


Note:

1. 1 <= N <= 10^9
2. ExamRoom.seat() and ExamRoom.leave() will be called at most 10^4 times across all test cases.
3. Calls to ExamRoom.leave(p) are guaranteed to have a student currently sitting in seat number p.

Related Topics:
Ordered Map

Similar Questions:

## Solution 1.

• class ExamRoom {
private TreeSet<int[]> ts = new TreeSet<>((a, b) -> {
int d1 = dist(a), d2 = dist(b);
return d1 == d2 ? a[0] - b[0] : d2 - d1;
});
private Map<Integer, Integer> left = new HashMap<>();
private Map<Integer, Integer> right = new HashMap<>();
private int n;

public ExamRoom(int n) {
this.n = n;
}

public int seat() {
int[] s = ts.first();
int p = (s[0] + s[1]) >> 1;
if (s[0] == -1) {
p = 0;
} else if (s[1] == n) {
p = n - 1;
}
del(s);
return p;
}

public void leave(int p) {
int l = left.get(p), r = right.get(p);
del(new int[] {l, p});
del(new int[] {p, r});
}

private int dist(int[] s) {
int l = s[0], r = s[1];
return l == -1 || r == n ? r - l - 1 : (r - l) >> 1;
}

left.put(s[1], s[0]);
right.put(s[0], s[1]);
}

private void del(int[] s) {
ts.remove(s);
left.remove(s[1]);
right.remove(s[0]);
}
}

/**
* Your ExamRoom object will be instantiated and called as such:
* ExamRoom obj = new ExamRoom(n);
* int param_1 = obj.seat();
* obj.leave(p);
*/

• int N;

int dist(const pair<int, int>& p) {
auto [l, r] = p;
if (l == -1 || r == N) return r - l - 1;
return (r - l) >> 1;
}

struct cmp {
bool operator()(const pair<int, int>& a, const pair<int, int>& b) const {
int d1 = dist(a), d2 = dist(b);
return d1 == d2 ? a.first < b.first : d1 > d2;
};
};

class ExamRoom {
public:
ExamRoom(int n) {
N = n;
this->n = n;
}

int seat() {
auto s = *ts.begin();
int p = (s.first + s.second) >> 1;
if (s.first == -1) {
p = 0;
} else if (s.second == n) {
p = n - 1;
}
del(s);
return p;
}

void leave(int p) {
int l = left[p], r = right[p];
del({l, p});
del({p, r});
}

private:
set<pair<int, int>, cmp> ts;
unordered_map<int, int> left;
unordered_map<int, int> right;
int n;

ts.insert(s);
left[s.second] = s.first;
right[s.first] = s.second;
}

void del(pair<int, int> s) {
ts.erase(s);
left.erase(s.second);
right.erase(s.first);
}
};

/**
* Your ExamRoom object will be instantiated and called as such:
* ExamRoom* obj = new ExamRoom(n);
* int param_1 = obj->seat();
* obj->leave(p);
*/

• from sortedcontainers import SortedList

class ExamRoom:
def __init__(self, n: int):
def dist(x):
l, r = x
return r - l - 1 if l == -1 or r == n else (r - l) >> 1

self.n = n
self.ts = SortedList(key=lambda x: (-dist(x), x[0]))
self.left = {}
self.right = {}

def seat(self) -> int:
s = self.ts[0]
p = (s[0] + s[1]) >> 1
if s[0] == -1:
p = 0
elif s[1] == self.n:
p = self.n - 1
self.delete(s)
return p

def leave(self, p: int) -> None:
l, r = self.left[p], self.right[p]
self.delete((l, p))
self.delete((p, r))

self.left[s[1]] = s[0]
self.right[s[0]] = s[1]

def delete(self, s):
self.ts.remove(s)
self.left.pop(s[1])
self.right.pop(s[0])

# Your ExamRoom object will be instantiated and called as such:
# obj = ExamRoom(n)
# param_1 = obj.seat()
# obj.leave(p)


• type ExamRoom struct {
rbt   *redblacktree.Tree
left  map[int]int
right map[int]int
n     int
}

func Constructor(n int) ExamRoom {
dist := func(s []int) int {
if s[0] == -1 || s[1] == n {
return s[1] - s[0] - 1
}
return (s[1] - s[0]) >> 1
}
cmp := func(a, b interface{}) int {
x, y := a.([]int), b.([]int)
d1, d2 := dist(x), dist(y)
if d1 == d2 {
return x[0] - y[0]
}
return d2 - d1
}
this := ExamRoom{redblacktree.NewWith(cmp), map[int]int{}, map[int]int{}, n}
return this
}

func (this *ExamRoom) Seat() int {
s := this.rbt.Left().Key.([]int)
p := (s[0] + s[1]) >> 1
if s[0] == -1 {
p = 0
} else if s[1] == this.n {
p = this.n - 1
}
this.del(s)
return p
}

func (this *ExamRoom) Leave(p int) {
l, _ := this.left[p]
r, _ := this.right[p]
this.del([]int{l, p})
this.del([]int{p, r})
}

func (this *ExamRoom) add(s []int) {
this.rbt.Put(s, struct{}{})
this.left[s[1]] = s[0]
this.right[s[0]] = s[1]
}

func (this *ExamRoom) del(s []int) {
this.rbt.Remove(s)
delete(this.left, s[1])
delete(this.right, s[0])
}

/**
* Your ExamRoom object will be instantiated and called as such:
* obj := Constructor(n);
* param_1 := obj.Seat();
* obj.Leave(p);
*/