Meeting Rooms II (Minimum Conference Rooms)

🎯

Meeting Rooms II (Minimum Conference Rooms)

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Imagine you are managing conference rooms in a busy office. You need to find the minimum number of rooms required so that all meetings can happen without overlap.

💡 This problem involves scheduling intervals and understanding how overlapping intervals affect resource allocation. Beginners often struggle to efficiently track overlaps and optimize room usage.

📋

Problem Statement

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms required to hold all the meetings without conflicts.

1 ≤ n ≤ 10^50 ≤ start_i < end_i ≤ 10^9Intervals may overlap arbitrarily

💡

Example

Input"[[0,30],[5,10],[15,20]]"

Output2

The first meeting overlaps with the second, so two rooms are needed. The third meeting starts after the second ends, so two rooms suffice.

• All meetings are non-overlapping → output is 1
• All meetings overlap completely → output is n
• Meetings with same start and end times → count as overlapping
• Single meeting → output is 1

⚠️

Common Mistakes

Confusing overlapping condition (e.g., using <= instead of <)

Incorrect room count due to counting meetings that just touch but do not overlap

✅ Use strict inequality for overlap: start_j < end_i and end_j > start_i

Not sorting intervals before processing

Heap or two-pointer logic fails, leading to wrong room counts

✅ Always sort intervals by start time before applying optimized approaches

Using a max-heap instead of min-heap for end times

Cannot efficiently find earliest ending meeting, causing incorrect room reuse

✅ Use a min-heap to track earliest end time for room freeing decisions

Not updating max_rooms after freeing a room in two-pointer approach

Underestimates the number of rooms needed

✅ Update max_rooms after each increment or decrement of used_rooms

🧠

Brute Force (Check Overlaps for Each Meeting)

💡 This approach exists to build intuition by directly comparing every meeting with every other to count overlaps, illustrating the problem's complexity before optimization.

Intuition

For each meeting, count how many other meetings overlap with it. The maximum overlap count across all meetings is the minimum number of rooms needed.

Algorithm

1. For each meeting, iterate over all other meetings.
2. Check if the current pair of meetings overlap.
3. Count the number of overlaps for each meeting.
4. Return the maximum overlap count found.

💡 This nested iteration is straightforward but inefficient, helping beginners grasp the overlap concept before moving to better methods.

</>

Code

def minMeetingRooms(intervals):
    max_rooms = 0
    n = len(intervals)
    for i in range(n):
        count = 0
        for j in range(n):
            if intervals[j][0] < intervals[i][1] and intervals[j][1] > intervals[i][0]:
                count += 1
        max_rooms = max(max_rooms, count)
    return max_rooms

# Driver code
intervals = [[0,30],[5,10],[15,20]]
print(minMeetingRooms(intervals))  # Output: 2

Line Notes

max_rooms = 0Initialize maximum rooms needed to zero before counting overlaps.

for i in range(n):Outer loop selects each meeting to compare against others.

if intervals[j][0] < intervals[i][1] and intervals[j][1] > intervals[i][0]:Check if meetings i and j overlap by comparing start and end times. Strict inequalities ensure meetings that only touch are counted as overlapping, matching problem definition.

max_rooms = max(max_rooms, count)Update the maximum number of overlapping meetings found so far.

import java.util.*;
public class Solution {
    public static int minMeetingRooms(int[][] intervals) {
        int maxRooms = 0;
        int n = intervals.length;
        for (int i = 0; i < n; i++) {
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (intervals[j][0] < intervals[i][1] && intervals[j][1] > intervals[i][0]) {
                    count++;
                }
            }
            maxRooms = Math.max(maxRooms, count);
        }
        return maxRooms;
    }

    public static void main(String[] args) {
        int[][] intervals = {{0,30},{5,10},{15,20}};
        System.out.println(minMeetingRooms(intervals)); // Output: 2
    }
}

Line Notes

int maxRooms = 0;Initialize variable to track maximum rooms needed.

for (int i = 0; i < n; i++) {Outer loop iterates over each meeting.

if (intervals[j][0] < intervals[i][1] && intervals[j][1] > intervals[i][0]) {Check if two meetings overlap by comparing intervals. Strict inequalities ensure touching intervals count as overlapping.

maxRooms = Math.max(maxRooms, count);Update maximum overlap count after checking all pairs.

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int minMeetingRooms(vector<vector<int>>& intervals) {
    int maxRooms = 0;
    int n = intervals.size();
    for (int i = 0; i < n; i++) {
        int count = 0;
        for (int j = 0; j < n; j++) {
            if (intervals[j][0] < intervals[i][1] && intervals[j][1] > intervals[i][0]) {
                count++;
            }
        }
        maxRooms = max(maxRooms, count);
    }
    return maxRooms;
}

int main() {
    vector<vector<int>> intervals = {{0,30},{5,10},{15,20}};
    cout << minMeetingRooms(intervals) << endl; // Output: 2
    return 0;
}

Line Notes

int maxRooms = 0;Initialize maximum rooms counter before processing intervals.

for (int i = 0; i < n; i++) {Outer loop selects each interval for comparison.

if (intervals[j][0] < intervals[i][1] && intervals[j][1] > intervals[i][0]) {Overlap condition between two intervals. Strict inequalities ensure touching intervals count as overlapping.

maxRooms = max(maxRooms, count);Keep track of the highest overlap count found.

function minMeetingRooms(intervals) {
    let maxRooms = 0;
    const n = intervals.length;
    for (let i = 0; i < n; i++) {
        let count = 0;
        for (let j = 0; j < n; j++) {
            if (intervals[j][0] < intervals[i][1] && intervals[j][1] > intervals[i][0]) {
                count++;
            }
        }
        maxRooms = Math.max(maxRooms, count);
    }
    return maxRooms;
}

// Test
console.log(minMeetingRooms([[0,30],[5,10],[15,20]])); // Output: 2

Line Notes

let maxRooms = 0;Initialize variable to track maximum concurrent meetings.

for (let i = 0; i < n; i++) {Outer loop picks each meeting to compare.

if (intervals[j][0] < intervals[i][1] && intervals[j][1] > intervals[i][0]) {Check if two meetings overlap by comparing start and end times. Strict inequalities ensure touching intervals count as overlapping.

maxRooms = Math.max(maxRooms, count);Update maximum rooms needed after counting overlaps.

⏱

Complexity

TimeO(n^2)

SpaceO(1)

We compare each meeting with every other meeting, resulting in n*n comparisons.

💡 For n=20 meetings, this means 400 comparisons, which grows quickly and becomes impractical for large n.

Interview Verdict: TLE for large inputs; useful only to introduce the problem and brute force concept.

This approach is too slow for big inputs but helps understand the problem's nature and overlap checking.

🧠

Sorting + Two Pointers (Sweep Line with Separate Start and End Arrays)

💡 This approach introduces sorting and two-pointer technique to efficiently count overlaps by processing start and end times separately.

Intuition

Sort all start times and end times. Use two pointers to track how many meetings are ongoing by comparing next start and end times.

Algorithm

1. Extract and sort all start times and end times separately.
2. Initialize two pointers for start and end arrays.
3. Iterate through start times; if a meeting starts before the earliest ending meeting ends, increment room count.
4. Otherwise, move the end pointer forward to free a room.
5. Track the maximum number of rooms needed during iteration.

💡 This approach cleverly uses sorted arrays and pointers to simulate meeting overlaps without nested loops.

</>

Code

def minMeetingRooms(intervals):
    starts = sorted(i[0] for i in intervals)
    ends = sorted(i[1] for i in intervals)
    s_ptr = e_ptr = 0
    used_rooms = 0
    max_rooms = 0
    while s_ptr < len(intervals):
        if starts[s_ptr] < ends[e_ptr]:
            used_rooms += 1
            s_ptr += 1
        else:
            used_rooms -= 1
            e_ptr += 1
        max_rooms = max(max_rooms, used_rooms)
    return max_rooms

# Driver code
intervals = [[0,30],[5,10],[15,20]]
print(minMeetingRooms(intervals))  # Output: 2

Line Notes

starts = sorted(i[0] for i in intervals)Sort all meeting start times to process in chronological order.

ends = sorted(i[1] for i in intervals)Sort all meeting end times to know when rooms free up.

if starts[s_ptr] < ends[e_ptr]:If next meeting starts before earliest meeting ends, need a new room.

used_rooms -= 1If a meeting ended before next starts, free a room by moving end pointer.

import java.util.*;
public class Solution {
    public static int minMeetingRooms(int[][] intervals) {
        int n = intervals.length;
        int[] starts = new int[n];
        int[] ends = new int[n];
        for (int i = 0; i < n; i++) {
            starts[i] = intervals[i][0];
            ends[i] = intervals[i][1];
        }
        Arrays.sort(starts);
        Arrays.sort(ends);
        int s_ptr = 0, e_ptr = 0, usedRooms = 0, maxRooms = 0;
        while (s_ptr < n) {
            if (starts[s_ptr] < ends[e_ptr]) {
                usedRooms++;
                s_ptr++;
            } else {
                usedRooms--;
                e_ptr++;
            }
            maxRooms = Math.max(maxRooms, usedRooms);
        }
        return maxRooms;
    }

    public static void main(String[] args) {
        int[][] intervals = {{0,30},{5,10},{15,20}};
        System.out.println(minMeetingRooms(intervals)); // Output: 2
    }
}

Line Notes

int[] starts = new int[n];Create array to hold all start times separately.

Arrays.sort(starts);Sort start times to process meetings in chronological order.

if (starts[s_ptr] < ends[e_ptr]) {Check if next meeting starts before earliest meeting ends to allocate room.

usedRooms--Free a room when a meeting ends before next starts by moving end pointer.

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int minMeetingRooms(vector<vector<int>>& intervals) {
    int n = intervals.size();
    vector<int> starts(n), ends(n);
    for (int i = 0; i < n; i++) {
        starts[i] = intervals[i][0];
        ends[i] = intervals[i][1];
    }
    sort(starts.begin(), starts.end());
    sort(ends.begin(), ends.end());
    int s_ptr = 0, e_ptr = 0, usedRooms = 0, maxRooms = 0;
    while (s_ptr < n) {
        if (starts[s_ptr] < ends[e_ptr]) {
            usedRooms++;
            s_ptr++;
        } else {
            usedRooms--;
            e_ptr++;
        }
        maxRooms = max(maxRooms, usedRooms);
    }
    return maxRooms;
}

int main() {
    vector<vector<int>> intervals = {{0,30},{5,10},{15,20}};
    cout << minMeetingRooms(intervals) << endl; // Output: 2
    return 0;
}

Line Notes

vector<int> starts(n), ends(n);Separate arrays for start and end times to process events.

sort(starts.begin(), starts.end());Sort start times to simulate meeting start events in order.

if (starts[s_ptr] < ends[e_ptr]) {If next meeting starts before earliest ends, allocate a room.

usedRooms--Free a room when a meeting ends before next starts by advancing end pointer.

function minMeetingRooms(intervals) {
    const starts = intervals.map(i => i[0]).sort((a,b) => a - b);
    const ends = intervals.map(i => i[1]).sort((a,b) => a - b);
    let s_ptr = 0, e_ptr = 0, usedRooms = 0, maxRooms = 0;
    while (s_ptr < intervals.length) {
        if (starts[s_ptr] < ends[e_ptr]) {
            usedRooms++;
            s_ptr++;
        } else {
            usedRooms--;
            e_ptr++;
        }
        maxRooms = Math.max(maxRooms, usedRooms);
    }
    return maxRooms;
}

// Test
console.log(minMeetingRooms([[0,30],[5,10],[15,20]])); // Output: 2

Line Notes

const starts = intervals.map(i => i[0]).sort((a,b) => a - b);Extract and sort start times to process chronologically.

if (starts[s_ptr] < ends[e_ptr]) {Compare next start with earliest end to decide if new room needed.

usedRooms--Free a room when a meeting ends before next starts by moving end pointer.

maxRooms = Math.max(maxRooms, usedRooms);Track maximum concurrent meetings to find minimum rooms.

⏱

Complexity

TimeO(n log n)

SpaceO(n)

Sorting start and end arrays takes O(n log n), and the two-pointer sweep is O(n).

💡 For n=20, sorting takes about 20*log2(20) ≈ 86 operations, much faster than brute force.

Interview Verdict: Accepted and efficient for large inputs; a common interview solution.

This approach balances clarity and efficiency, making it a strong candidate solution.

🧠

Min-Heap (Priority Queue) to Track End Times

💡 This approach uses a min-heap to dynamically track the earliest ending meeting, allowing efficient room reuse decisions.

Intuition

Sort meetings by start time. Use a min-heap to keep track of meeting end times currently occupying rooms. If the earliest ending meeting ends before the next starts, reuse that room.

Algorithm

1. Sort intervals by start time.
2. Initialize a min-heap and add the end time of the first meeting.
3. For each subsequent meeting, compare its start time with the earliest end time in the heap.
4. If the meeting starts after or when the earliest meeting ends, pop from heap to reuse room.
5. Push the current meeting's end time into the heap.
6. The heap size at the end is the minimum number of rooms required.

💡 This approach simulates room allocation dynamically, using the heap to efficiently find the earliest room that frees up.

</>

Code

import heapq

def minMeetingRooms(intervals):
    if not intervals:
        return 0
    intervals.sort(key=lambda x: x[0])
    heap = []
    heapq.heappush(heap, intervals[0][1])
    for i in range(1, len(intervals)):
        if intervals[i][0] >= heap[0]:
            heapq.heappop(heap)
        heapq.heappush(heap, intervals[i][1])
    return len(heap)

# Driver code
intervals = [[0,30],[5,10],[15,20]]
print(minMeetingRooms(intervals))  # Output: 2

Line Notes

intervals.sort(key=lambda x: x[0])Sort meetings by start time to process chronologically.

heapq.heappush(heap, intervals[0][1])Add first meeting's end time to heap representing occupied rooms.

if intervals[i][0] >= heap[0]:If current meeting starts after earliest meeting ends, free that room.

return len(heap)Heap size represents number of rooms currently occupied, i.e., minimum rooms needed.

import java.util.*;
public class Solution {
    public static int minMeetingRooms(int[][] intervals) {
        if (intervals.length == 0) return 0;
        Arrays.sort(intervals, (a,b) -> a[0] - b[0]);
        PriorityQueue<Integer> heap = new PriorityQueue<>();
        heap.add(intervals[0][1]);
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >= heap.peek()) {
                heap.poll();
            }
            heap.add(intervals[i][1]);
        }
        return heap.size();
    }

    public static void main(String[] args) {
        int[][] intervals = {{0,30},{5,10},{15,20}};
        System.out.println(minMeetingRooms(intervals)); // Output: 2
    }
}

Line Notes

Arrays.sort(intervals, (a,b) -> a[0] - b[0]);Sort intervals by start time for chronological processing.

PriorityQueue<Integer> heap = new PriorityQueue<>();Min-heap to track earliest ending meeting times.

if (intervals[i][0] >= heap.peek()) {Check if current meeting can reuse room freed by earliest ending meeting.

return heap.size();Heap size equals minimum rooms needed after processing all meetings.

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;

int minMeetingRooms(vector<vector<int>>& intervals) {
    if (intervals.empty()) return 0;
    sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a[0] < b[0]; });
    priority_queue<int, vector<int>, greater<int>> minHeap;
    minHeap.push(intervals[0][1]);
    for (int i = 1; i < intervals.size(); i++) {
        if (intervals[i][0] >= minHeap.top()) {
            minHeap.pop();
        }
        minHeap.push(intervals[i][1]);
    }
    return minHeap.size();
}

int main() {
    vector<vector<int>> intervals = {{0,30},{5,10},{15,20}};
    cout << minMeetingRooms(intervals) << endl; // Output: 2
    return 0;
}

Line Notes

sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a[0] < b[0]; });Sort intervals by start time to process in order.

priority_queue<int, vector<int>, greater<int>> minHeap;Min-heap to keep track of earliest meeting end times.

if (intervals[i][0] >= minHeap.top()) {If current meeting starts after earliest ends, reuse that room by popping heap.

return minHeap.size();Heap size at end is minimum number of rooms needed.

function minMeetingRooms(intervals) {
    if (intervals.length === 0) return 0;
    intervals.sort((a,b) => a[0] - b[0]);
    const heap = [];
    function heapPush(val) {
        heap.push(val);
        let i = heap.length - 1;
        while (i > 0) {
            let parent = Math.floor((i - 1) / 2);
            if (heap[parent] <= heap[i]) break;
            [heap[parent], heap[i]] = [heap[i], heap[parent]];
            i = parent;
        }
    }
    function heapPop() {
        const top = heap[0];
        const end = heap.pop();
        if (heap.length > 0) {
            heap[0] = end;
            let i = 0;
            while (true) {
                let left = 2 * i + 1, right = 2 * i + 2, smallest = i;
                if (left < heap.length && heap[left] < heap[smallest]) smallest = left;
                if (right < heap.length && heap[right] < heap[smallest]) smallest = right;
                if (smallest === i) break;
                [heap[i], heap[smallest]] = [heap[smallest], heap[i]];
                i = smallest;
            }
        }
        return top;
    }
    function heapPeek() {
        return heap[0];
    }
    heapPush(intervals[0][1]);
    for (let i = 1; i < intervals.length; i++) {
        if (intervals[i][0] >= heapPeek()) {
            heapPop();
        }
        heapPush(intervals[i][1]);
    }
    return heap.length;
}

// Test
console.log(minMeetingRooms([[0,30],[5,10],[15,20]])); // Output: 2

Line Notes

intervals.sort((a,b) => a[0] - b[0]);Sort intervals by start time to process meetings chronologically.

function heapPush(val)Custom min-heap push to maintain earliest end times at top.

if (intervals[i][0] >= heapPeek()) {If current meeting starts after earliest ends, reuse that room by popping heap.

return heap.length;Heap size after processing all meetings is minimum rooms needed.

⏱

Complexity

TimeO(n log n)

SpaceO(n)

Sorting takes O(n log n), and each meeting is pushed and popped from heap at most once, each heap operation O(log n).

💡 For n=20, sorting and heap operations are efficient and scalable for large inputs.

Interview Verdict: Accepted and efficient; preferred in many interviews for clarity and heap usage demonstration.

This approach shows mastery of data structures and is often favored for its clarity and efficiency.

📊

All Approaches - One-Glance Tradeoffs

💡 The two-pointer and min-heap approaches are efficient and practical; brute force is only for initial understanding.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n^2)	O(1)	No	N/A	Mention only - never code due to inefficiency
2. Sorting + Two Pointers	O(n log n)	O(n)	No	No (only counts rooms)	Good for quick implementation and explanation
3. Min-Heap	O(n log n)	O(n)	No	Yes (can track room assignments)	Preferred approach to demonstrate data structure knowledge

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice multiple approaches, and prepare to explain tradeoffs clearly in interviews.

How to Present

Clarify the problem and constraints with the interviewer.Start with the brute force approach to show understanding of overlaps.Explain the sorting + two pointers approach for efficiency.Discuss the min-heap approach as an optimal solution.Write clean code and test with examples and edge cases.

Time Allocation

Clarify: 2min → Approach: 5min → Code: 10min → Test: 3min. Total ~20min

What the Interviewer Tests

Interviewer tests your understanding of interval overlaps, sorting, data structures (heap), and ability to optimize from brute force to efficient solutions.

Common Follow-ups

• What if intervals are very large and cannot be sorted in memory? → Use external sorting or streaming algorithms.
• How to return the actual room assignments? → Use a map from room to intervals and assign rooms during heap operations.

💡 These follow-ups test your ability to handle scalability and extend solutions beyond just counting rooms.

🔍

Pattern Recognition

When to Use

1) Problem involves intervals or time ranges, 2) Need to find overlaps or resource allocation, 3) Asked for minimum number of resources (rooms), 4) Overlaps can be arbitrary

Signature Phrases

minimum number of conference roomsmeeting time intervalsoverlapping intervals

NOT This Pattern When

Problems asking for maximum number of non-overlapping intervals or interval partitioning with different constraints

Similar Problems

Meeting Rooms I - checks if a single room sufficesInterval Scheduling - selects maximum non-overlapping intervalsEmployee Free Time - finds gaps between intervals

Practice

1. You are given a list of meeting time intervals represented as pairs of start and end times. You need to determine if a single person can attend all meetings without any overlaps. Which approach guarantees an optimal and efficient solution?

easy

A. Use dynamic programming to find the maximum number of non-overlapping intervals.

B. Sort intervals by start time and check for any overlap by comparing each interval's start with the previous interval's end.

C. Use a brute force nested loop to compare every pair of intervals for overlap.

D. Sort intervals by end time and greedily select intervals that finish earliest to maximize the number of meetings attended.

Solution

1. Step 1: Understand the problem requirement

   The problem asks if a single person can attend all meetings without overlap, so we must check if any intervals overlap.

2. Step 2: Identify the correct approach

   Sorting intervals by start time and checking if any interval starts before the previous one ends guarantees detection of overlaps efficiently.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sorting by start time and checking overlaps is the standard approach [OK]

Hint: Sort by start time and check overlaps [OK]

Common Mistakes:

• Confusing sorting by end time with start time for overlap detection

2. Consider the following code snippet implementing the sweep line algorithm to find the minimum number of platforms required. What is the value of max_platforms returned for the input arrivals = [900, 940, 950] and departures = [910, 1200, 1120]?

easy

A. 4

B. 1

C. 3

D. 2

Solution

1. Step 1: Create and sort events

   Events: (900,1), (940,1), (950,1), (910,-1), (1200,-1), (1120,-1). Sorted: (900,1), (910,-1), (940,1), (950,1), (1120,-1), (1200,-1).

2. Step 2: Sweep through events updating platforms_needed and max_platforms

   At 900 arrival: platforms_needed=1, max=1; at 910 departure: platforms_needed=0; at 940 arrival: platforms_needed=1; at 950 arrival: platforms_needed=2, max=2; at 1120 departure: platforms_needed=1; at 1200 departure: platforms_needed=0.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Maximum concurrent trains is 2 [OK]

Hint: Sort events and track increments/decrements [OK]

Common Mistakes:

• Misordering events when arrival and departure times are close
• Off-by-one errors in counting platforms
• Ignoring departure before arrival at same time

3. Given the following code and input intervals = [[1,4],[3,6],[2,8]], what is the value of count after the loop finishes?

easy

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Sort intervals by start ascending, end descending

   Sorted intervals: [[1,4],[2,8],[3,6]]

2. Step 2: Iterate and update count and max_end

   Iteration 1: end=4 > max_end=0 -> count=1, max_end=4 Iteration 2: end=8 > max_end=4 -> count=2, max_end=8 Iteration 3: end=6 ≤ max_end=8 -> count unchanged

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Count increments twice for uncovered intervals [OK]

Hint: Sort then count intervals with strictly greater end [OK]

Common Mistakes:

• Misorder intervals after sorting
• Off-by-one counting in loop

4. What is the time complexity of the optimal in-place merge intervals algorithm that first sorts the intervals and then merges them in one pass?

medium

A. O(n log n) due to the sorting step dominating the runtime

B. O(n) because merging is done in a single pass

C. O(n²) because each interval is compared with all others during merging

D. O(n log n) because merging requires binary searches for overlaps

Solution

1. Step 1: Identify the sorting step complexity.

   Sorting n intervals by start time takes O(n log n) time.

2. Step 2: Analyze the merging step complexity.

   Merging intervals in one pass is O(n), which is dominated by sorting.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sorting dominates overall complexity, merging is linear. [OK]

Hint: Sorting dominates, merging is linear [OK]

Common Mistakes:

• Ignoring sorting cost
• Assuming nested comparisons cause O(n²)

5. Suppose the car pooling problem is modified so that trips can overlap and passengers can be picked up and dropped off multiple times (i.e., trips can reuse the same passengers). Which of the following algorithmic changes correctly adapts the solution to handle this variant efficiently?

hard

A. Use a segment tree or binary indexed tree to efficiently update and query passenger counts over intervals

B. Switch to a priority queue to process events in order and track current passengers dynamically

C. Use the same difference array approach but multiply passenger counts by the number of trips to simulate reuse

D. Modify the difference array to allow negative passenger counts and track net changes carefully

Solution

1. Step 1: Understand reuse complexity

   Reusing passengers means overlapping intervals can increase and decrease counts multiple times, requiring efficient range updates and queries.

2. Step 2: Identify suitable data structure

   Segment trees or binary indexed trees support efficient interval updates and queries, handling overlapping intervals with reuse.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Segment trees handle dynamic interval sums efficiently [OK]

Hint: Reuse -> need data structure for interval updates [OK]

Common Mistakes:

• Multiplying counts breaks logic
• Priority queue alone can't handle reuse efficiently
• Allowing negative counts without structure causes errors