# Program to find closest room from queries in Python

Suppose there is an array called rooms. where rooms[i] contains a pair [roomId_i, size_i] denotes a room whose id is roomId_i and size is size_i. All room numbers are distinct. We also have another array queries, where queries[j] contains a pair [preferred_j, minSize_j]. The answer to the jth query is the room number id of a room such that −

• The room has size of at least minSize_j, and

• |id - preferred_j| is minimized.

Now, if there is a tie in the absolute difference, then use the room with the smallest id. If there is no such room, return -1. So we have to find an array called answer whose length is same as queries, that contains the answer to the jth query.

So, if the input is like rooms = [[2,2],[1,2],[3,2]] queries = [[3,1],[3,3],[5,2]], then the output will be [3, -1, 3] because

• For query [3,1]: The room 3 is the closest because |3 - 3| = 0, and its size of 2 is at least 1, so the answer is 3.

• For query [3,3]: There are no rooms whose size is at least 3, so the answer is -1.

• For query [5,2]: The room 3 is the closest because |3 - 5| = 2, and its size of 2 is at least 2, so the answer is 3.

To solve this, we will follow these steps −

• sort rooms based on size, when size are same then based on room id

• queries = a list of pairs (qid,size,i) for index i, and pair (qid, size) in queries

• sort the queries in reverse order based on size, if sizes are same, then based on preferred, when both are same then based on index

• ans := an array of size same as size of queries and fill with -1

• X := a new list

• for each (qid, size, i) in queries, do

• while rooms and size of last item of rooms >= size, do

• (idr, p) := deleted last element from rooms

• sort X after inserting idr

• if X is not empty, then

• j := index where to insert qid to remain X sorted

• if j is same as size of X , then

• ans[i] := last element of X

• otherwise when j is same as 0, then

• ans[i] := X[0]

• otherwise,

• if X[j] - qid < qid - X[j-1], then

• ans[i] := X[j]

• otherwise,

• ans[i] := X[j-1]

• return ans

## Example

Let us see the following implementation to get better understanding

import bisect
def solve(rooms, queries):
rooms.sort(key = lambda x: (x[1], x[0]))
queries = [(qid,size,i) for i, (qid, size) in enumerate(queries)]
queries.sort(key = lambda x: (x[1], x[0], x[2]), reverse = True)
ans = [-1] * len(queries)
X = []
for qid, size, i in queries:
while rooms and rooms[-1][1] >= size:
idr, _ = rooms.pop()
bisect.insort(X, idr)
if X:
j = bisect.bisect(X, qid)
if j == len(X):
ans[i] = X[-1]
elif j == 0:
ans[i] = X[0]
else:
if X[j] - qid < qid - X[j-1]:
ans[i] = X[j]
else:
ans[i] = X[j-1]
return ans

rooms = [[2,2],[1,2],[3,2]]
queries = [[3,1],[3,3],[5,2]]
print(solve(rooms, queries))

## Input

[[2,2],[1,2],[3,2]], [[3,1],[3,3],[5,2]]


## Output

[3, -1, 3]

Updated on: 08-Oct-2021

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