TAOCP 2.1 Exercise 5

We are asked to find, for each query, a hotel room whose size meets a minimum requirement and whose room number is closest to a preferred room number.

taocpmathematicsalgorithmsvolume-1medium

Section 2.1: Introduction

Exercise 5. [21] Give an algorithm that essentially undoes the effect of exercise 4: Assuming that the pile is not empty and that its bottom card is face down, your algorithm should

remove the bottom card and make NEWCARD link to it. (This algorithm is sometimes called "cheating" in solitaire games.)

Verified: no
Solve time: -

Solution

Problem Understanding

We are asked to find, for each query, a hotel room whose size meets a minimum requirement and whose room number is closest to a preferred room number. Formally, each room is described by a pair [roomIdi, sizei], and each query by [preferredj, minSizej]. The task is to return a list of room numbers, one per query, that minimizes abs(id - preferredj) among all rooms with sizei >= minSizej. In case of a tie, the smallest room number is chosen. If no room meets the minimum size requirement, the answer is -1.

The input constraints indicate large-scale data: up to 10^5 rooms and 10^4 queries. Room IDs can be as large as 10^7, so direct indexing by room ID is impractical. The uniqueness of room IDs allows us to store them in ordered structures without duplicates.

Edge cases include queries where no rooms meet the minimum size, multiple rooms equally close to the preferred ID, and rooms with the same size but different IDs.

Approaches

Brute Force

The naive approach examines each query individually. For a query [preferredj, minSizej], one iterates over all rooms, checks if sizei >= minSizej, and maintains the room with the smallest abs(id - preferredj) (choosing the smaller ID on a tie). This produces correct results because it evaluates all possibilities.

However, the time complexity is O(n * k), which in the worst case is 10^9 operations. This is far too slow for the problem constraints.

Optimal Approach

The key insight is that queries can be processed efficiently if we organize the rooms by size in descending order. For each query, we only need rooms with size greater than or equal to the query's minSize. By sorting rooms and queries by size, we can insert eligible rooms into an ordered data structure (like a balanced BST or SortedList in Python) that allows for logarithmic time retrieval of the closest room ID. The ordered set maintains all room IDs with sizes meeting or exceeding the current query’s minSize. For a preferred room ID p, we can efficiently find the closest room ID using binary search to locate the floor and ceiling in the set.

This approach reduces the time complexity to O(n log n + k log n), dominated by sorting and balanced set operations.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * k) O(1) Check all rooms for each query
Optimal O((n + k) log n) O(n) Sort rooms and queries, use ordered set for closest lookup

Algorithm Walkthrough

  1. Sort Rooms by Size Descending: First, we sort all rooms in descending order of their size, so we can process the largest rooms first.
  2. Attach Indices to Queries and Sort: For each query [preferredj, minSizej], attach its index in the query list and sort all queries in descending order of minSizej.
  3. Initialize an Ordered Set: This set will contain room IDs eligible for the current query, ordered by their numeric value for efficient closest-room lookup.
  4. Process Queries: Initialize a pointer i = 0 over the sorted rooms. For each query (starting with the largest minSize):
  • While the current room's size sizei >= minSizej, insert roomIdi into the ordered set and increment i.
  • The set now contains all rooms that satisfy the size requirement.
  1. Find Closest Room: Use binary search in the ordered set to locate the closest room ID to preferredj. Check both the predecessor and successor to resolve ties according to the smallest absolute difference and smallest ID.
  2. Store Answer: Save the result in the answer array at the query’s original index.
  3. Return Result: Once all queries are processed, return the answer array.

This approach guarantees that for each query, the set contains exactly the rooms that satisfy the minimum size, and the binary search ensures that the closest ID is found efficiently.

Python Solution

from typing import List
from sortedcontainers import SortedList

class Solution:
    def closestRoom(self, rooms: List[List[int]], queries: List[List[int]]) -> List[int]:
        # Step 1: Sort rooms by size descending
        rooms.sort(key=lambda x: -x[1])
        
        # Step 2: Attach original index and sort queries by minSize descending
        queries_with_index = [(minSize, preferred, idx) for idx, (preferred, minSize) in enumerate(queries)]
        queries_with_index.sort(reverse=True)
        
        result = [-1] * len(queries)
        sorted_ids = SortedList()
        room_idx = 0
        
        for minSize, preferred, q_idx in queries_with_index:
            # Step 3: Add all rooms that meet minSize to the ordered set
            while room_idx < len(rooms) and rooms[room_idx][1] >= minSize:
                sorted_ids.add(rooms[room_idx][0])
                room_idx += 1
            
            # Step 4: If no rooms meet minSize, answer is -1
            if not sorted_ids:
                continue
            
            # Step 5: Binary search for closest ID
            pos = sorted_ids.bisect_left(preferred)
            candidates = []
            if pos < len(sorted_ids):
                candidates.append(sorted_ids[pos])
            if pos > 0:
                candidates.append(sorted_ids[pos - 1])
            
            # Step 6: Choose closest by abs diff, break tie with smaller ID
            closest = min(candidates, key=lambda x: (abs(x - preferred), x))
            result[q_idx] = closest
        
        return result

In the Python implementation, we use SortedList from the sortedcontainers library as an efficient ordered set. Rooms are inserted in decreasing order of size, and each query considers only eligible rooms. The binary search in the ordered set ensures O(log n) lookup for each query.

Go Solution

package main

import (
	"sort"
)

func closestRoom(rooms [][]int, queries [][]int) []int {
	type room struct {
		id, size int
	}
	type query struct {
		preferred, minSize, index int
	}
	
	n, k := len(rooms), len(queries)
	roomList := make([]room, n)
	for i, r := range rooms {
		roomList[i] = room{r[0], r[1]}
	}
	sort.Slice(roomList, func(i, j int) bool {
		return roomList[i].size > roomList[j].size
	})
	
	queryList := make([]query, k)
	for i, q := range queries {
		queryList[i] = query{q[0], q[1], i}
	}
	sort.Slice(queryList, func(i, j int) bool {
		return queryList[i].minSize > queryList[j].minSize
	})
	
	res := make([]int, k)
	for i := range res {
		res[i] = -1
	}
	
	var sortedIds []int
	roomIdx := 0
	
	for _, q := range queryList {
		for roomIdx < n && roomList[roomIdx].size >= q.minSize {
			id := roomList[roomIdx].id
			pos := sort.Search(len(sortedIds), func(i int) bool { return sortedIds[i] >= id })
			sortedIds = append(sortedIds, 0)
			copy(sortedIds[pos+1:], sortedIds[pos:])
			sortedIds[pos] = id
			roomIdx++
		}
		if len(sortedIds) == 0 {
			continue
		}
		pos := sort.Search(len(sortedIds), func(i int) bool { return sortedIds[i] >= q.preferred })
		candidates := []int{}
		if pos < len(sortedIds) {
			candidates = append(candidates, sortedIds[pos])
		}
		if pos > 0 {
			candidates = append(candidates, sortedIds[pos-1])
		}
		closest := candidates[0]
		for _, c := range candidates[1:] {
			if abs(c-q.preferred) < abs(closest-q.preferred) || (abs(c-q.preferred) == abs(closest-q.preferred) && c < closest) {
				closest = c
			}
		}
		res[q.index] = closest
	}
	return res
}

func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}

In Go, sortedIds is maintained as a sorted slice. We use sort.Search for binary search and manually insert elements to maintain order. This preserves the same algorithmic logic as the Python solution, but requires explicit slice management.

Worked Examples

Example 1: rooms = [[2,2],[1,2],[3,2]], queries = [[3,1],[3,3],[5,2]]

Rooms sorted descending by size: [[2,2],[1,2],[3,2]] (all same size, order stable).

Queries sorted descending by minSize: [[3,3,1],[5,2,2],[3,1,0]]

  • Query [3,3]: no room >= 3 → answer = -1
  • Query [5,2]: rooms >=2 → add 2,1,3 → sortedIds = [1,2,3], closest to 5 → 3 →