Car Pooling

🎯

Car Pooling

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Imagine you are managing a carpool service where multiple groups of passengers get on and off at different locations along a route. You need to ensure the car's capacity is never exceeded at any point.

💡 This problem involves tracking overlapping intervals and their cumulative effect on a resource (car capacity). Beginners often struggle because it requires combining interval logic with capacity constraints, and naive solutions can be inefficient.

📋

Problem Statement

You are driving a vehicle that can hold a fixed number of passengers (capacity). Given a list of trips, where each trip is represented as [num_passengers, start_location, end_location], determine if it is possible to pick up and drop off all passengers for all the given trips without exceeding the vehicle's capacity at any point. The vehicle only moves in one direction along a route from location 0 to location 1000.

1 ≤ number of trips ≤ 10^50 ≤ start_location < end_location ≤ 10001 ≤ num_passengers ≤ 10001 ≤ capacity ≤ 10^5

💡

Example

Input"trips = [[2,1,5],[3,3,7]], capacity = 4"

Outputfalse

Between locations 3 and 5, total passengers would be 2 + 3 = 5, which exceeds capacity 4.

Input"trips = [[2,1,5],[3,3,7]], capacity = 5"

Outputtrue

At any point, the total passengers do not exceed capacity 5.

• Single trip with passengers equal to capacity → should return true
• Trips with no overlapping intervals → should return true
• Trips where passengers get off exactly when others get on → should return true
• Trips with maximum number of passengers and locations → test performance and correctness

⚠️

Common Mistakes

Not sorting events before processing

Passenger counts are updated out of order, leading to incorrect capacity checks

✅ Always sort events by location before processing

Incrementing passenger count at end location instead of decrementing

Capacity calculations become incorrect, possibly allowing overcapacity

✅ Subtract passengers at the end location to mark dropoff

Using brute force approach for large inputs

Solution times out or runs too slowly in interviews

✅ Use event sorting or difference array approaches for efficiency

Not handling edge cases where trips start and end at the same location

Incorrect passenger counts or unnecessary capacity checks

✅ Trips with start == end do not affect capacity and can be ignored

Forgetting to check capacity after each event update

Missed overcapacity situations leading to wrong answers

✅ Check capacity immediately after updating current passengers

🧠

Brute Force (Simulate Each Location)

💡 This approach exists to build intuition by simulating the problem literally, even though it is inefficient. It helps beginners understand the problem constraints and why optimization is needed.

Intuition

We simulate the entire route from start to end, tracking how many passengers are in the car at each location by iterating over all trips for every location.

Algorithm

1. Initialize an array to represent passenger count at each location from 0 to max location.
2. For each trip, increment passenger count for all locations from start to end-1.
3. Iterate through the array to check if passenger count exceeds capacity at any location.
4. Return false if capacity exceeded, else true.

💡 This approach is straightforward but inefficient because it checks every location for every trip, leading to a large number of operations.

</>

Code

def carPooling(trips, capacity):
    max_location = 0
    for _, start, end in trips:
        max_location = max(max_location, end)
    passenger_count = [0] * (max_location + 1)
    for num, start, end in trips:
        for loc in range(start, end):
            passenger_count[loc] += num
            if passenger_count[loc] > capacity:
                return False
    return True

# Example usage
if __name__ == '__main__':
    print(carPooling([[2,1,5],[3,3,7]], 4))  # False
    print(carPooling([[2,1,5],[3,3,7]], 5))  # True

Line Notes

max_location = 0Initialize max_location to find the furthest drop-off point

for _, start, end in trips:Find the maximum end location to size the passenger_count array

passenger_count = [0] * (max_location + 1)Create an array to track passengers at each location

for loc in range(start, end):Increment passenger count for every location the passengers occupy

if passenger_count[loc] > capacity:Check if capacity is exceeded at any location

return TrueIf no location exceeds capacity, return True

import java.util.*;
public class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        int maxLocation = 0;
        for (int[] trip : trips) {
            maxLocation = Math.max(maxLocation, trip[2]);
        }
        int[] passengerCount = new int[maxLocation + 1];
        for (int[] trip : trips) {
            int num = trip[0], start = trip[1], end = trip[2];
            for (int loc = start; loc < end; loc++) {
                passengerCount[loc] += num;
                if (passengerCount[loc] > capacity) {
                    return false;
                }
            }
        }
        return true;
    }
    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.carPooling(new int[][]{{2,1,5},{3,3,7}}, 4)); // false
        System.out.println(sol.carPooling(new int[][]{{2,1,5},{3,3,7}}, 5)); // true
    }
}

Line Notes

int maxLocation = 0;Initialize maxLocation to find the furthest drop-off point

for (int[] trip : trips)Iterate over trips to find max end location

int[] passengerCount = new int[maxLocation + 1];Array to track passengers at each location

for (int loc = start; loc < end; loc++)Increment passenger count for each location in trip interval

if (passengerCount[loc] > capacity)Check if capacity exceeded at any location

return true;Return true if capacity never exceeded

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

bool carPooling(vector<vector<int>>& trips, int capacity) {
    int maxLocation = 0;
    for (auto& trip : trips) {
        maxLocation = max(maxLocation, trip[2]);
    }
    vector<int> passengerCount(maxLocation + 1, 0);
    for (auto& trip : trips) {
        int num = trip[0], start = trip[1], end = trip[2];
        for (int loc = start; loc < end; loc++) {
            passengerCount[loc] += num;
            if (passengerCount[loc] > capacity) {
                return false;
            }
        }
    }
    return true;
}

int main() {
    vector<vector<int>> trips1 = {{2,1,5},{3,3,7}};
    cout << (carPooling(trips1, 4) ? "true" : "false") << endl; // false
    cout << (carPooling(trips1, 5) ? "true" : "false") << endl; // true
    return 0;
}

Line Notes

int maxLocation = 0;Find the maximum end location to size the vector

vector<int> passengerCount(maxLocation + 1, 0);Initialize vector to track passengers at each location

for (int loc = start; loc < end; loc++)Increment passenger count for each location in the trip interval

if (passengerCount[loc] > capacity)Check if capacity is exceeded at any location

return true;Return true if capacity never exceeded

function carPooling(trips, capacity) {
    let maxLocation = 0;
    for (const [, start, end] of trips) {
        maxLocation = Math.max(maxLocation, end);
    }
    const passengerCount = new Array(maxLocation + 1).fill(0);
    for (const [num, start, end] of trips) {
        for (let loc = start; loc < end; loc++) {
            passengerCount[loc] += num;
            if (passengerCount[loc] > capacity) {
                return false;
            }
        }
    }
    return true;
}

// Example usage
console.log(carPooling([[2,1,5],[3,3,7]], 4)); // false
console.log(carPooling([[2,1,5],[3,3,7]], 5)); // true

Line Notes

let maxLocation = 0;Initialize maxLocation to find the furthest drop-off point

const passengerCount = new Array(maxLocation + 1).fill(0);Create array to track passengers at each location

for (let loc = start; loc < end; loc++)Increment passenger count for each location in trip interval

if (passengerCount[loc] > capacity)Check if capacity exceeded at any location

return true;Return true if capacity never exceeded

⏱

Complexity

TimeO(n * L) where n is number of trips and L is max location range

SpaceO(L) for passenger count array

For each trip, we update passenger counts for each location in its interval, which can be up to max location length. This leads to potentially large time complexity.

💡 If n=100,000 and max location=1000, this approach could do up to 100 million operations, which is too slow for interviews.

Interview Verdict: TLE / Inefficient for large inputs

This approach is too slow for large inputs but is useful to understand the problem and why optimization is needed.

🧠

Sorting Events and Sweep Line

💡 This approach improves efficiency by transforming trips into events and processing them in order, which is a common pattern for interval problems.

Intuition

Convert each trip into two events: passengers getting on and passengers getting off. Sort these events by location and simulate passenger changes in order.

Algorithm

1. Create a list of events: (location, passenger change), +num for pickup, -num for dropoff.
2. Sort events by location.
3. Iterate through events, updating current passengers by event passenger change.
4. If current passengers exceed capacity at any point, return false.
5. Return true if capacity never exceeded.

💡 This approach reduces complexity by only processing points where passenger count changes, avoiding unnecessary checks.

</>

Code

def carPooling(trips, capacity):
    events = []
    for num, start, end in trips:
        events.append((start, num))
        events.append((end, -num))
    events.sort()
    current_passengers = 0
    for _, change in events:
        current_passengers += change
        if current_passengers > capacity:
            return False
    return True

# Example usage
if __name__ == '__main__':
    print(carPooling([[2,1,5],[3,3,7]], 4))  # False
    print(carPooling([[2,1,5],[3,3,7]], 5))  # True

Line Notes

events = []Initialize list to hold all pickup and dropoff events

events.append((start, num))Add pickup event with positive passenger count

events.append((end, -num))Add dropoff event with negative passenger count

events.sort()Sort events by location to process in order

current_passengers += changeUpdate current passengers based on event

if current_passengers > capacity:Check if capacity exceeded after event

import java.util.*;
public class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        List<int[]> events = new ArrayList<>();
        for (int[] trip : trips) {
            events.add(new int[]{trip[1], trip[0]}); // pickup
            events.add(new int[]{trip[2], -trip[0]}); // dropoff
        }
        events.sort((a, b) -> Integer.compare(a[0], b[0]));
        int currentPassengers = 0;
        for (int[] event : events) {
            currentPassengers += event[1];
            if (currentPassengers > capacity) {
                return false;
            }
        }
        return true;
    }
    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.carPooling(new int[][]{{2,1,5},{3,3,7}}, 4)); // false
        System.out.println(sol.carPooling(new int[][]{{2,1,5},{3,3,7}}, 5)); // true
    }
}

Line Notes

List<int[]> events = new ArrayList<>();Create list to hold pickup and dropoff events

events.add(new int[]{trip[1], trip[0]});Add pickup event with positive passenger count

events.add(new int[]{trip[2], -trip[0]});Add dropoff event with negative passenger count

events.sort((a, b) -> Integer.compare(a[0], b[0]));Sort events by location ascending

currentPassengers += event[1];Update current passengers after event

if (currentPassengers > capacity)Check if capacity exceeded

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

bool carPooling(vector<vector<int>>& trips, int capacity) {
    vector<pair<int,int>> events;
    for (auto& trip : trips) {
        events.emplace_back(trip[1], trip[0]); // pickup
        events.emplace_back(trip[2], -trip[0]); // dropoff
    }
    sort(events.begin(), events.end());
    int currentPassengers = 0;
    for (auto& event : events) {
        currentPassengers += event.second;
        if (currentPassengers > capacity) {
            return false;
        }
    }
    return true;
}

int main() {
    vector<vector<int>> trips1 = {{2,1,5},{3,3,7}};
    cout << (carPooling(trips1, 4) ? "true" : "false") << endl; // false
    cout << (carPooling(trips1, 5) ? "true" : "false") << endl; // true
    return 0;
}

Line Notes

vector<pair<int,int>> events;Create vector to hold pickup and dropoff events

events.emplace_back(trip[1], trip[0]);Add pickup event with positive passenger count

events.emplace_back(trip[2], -trip[0]);Add dropoff event with negative passenger count

sort(events.begin(), events.end());Sort events by location ascending

currentPassengers += event.second;Update current passengers after event

if (currentPassengers > capacity)Check if capacity exceeded

function carPooling(trips, capacity) {
    const events = [];
    for (const [num, start, end] of trips) {
        events.push([start, num]); // pickup
        events.push([end, -num]); // dropoff
    }
    events.sort((a, b) => a[0] - b[0]);
    let currentPassengers = 0;
    for (const [, change] of events) {
        currentPassengers += change;
        if (currentPassengers > capacity) {
            return false;
        }
    }
    return true;
}

// Example usage
console.log(carPooling([[2,1,5],[3,3,7]], 4)); // false
console.log(carPooling([[2,1,5],[3,3,7]], 5)); // true

Line Notes

const events = [];Initialize array to hold pickup and dropoff events

events.push([start, num]);Add pickup event with positive passenger count

events.push([end, -num]);Add dropoff event with negative passenger count

events.sort((a, b) => a[0] - b[0]);Sort events by location ascending

currentPassengers += change;Update current passengers after event

if (currentPassengers > capacity)Check if capacity exceeded

⏱

Complexity

TimeO(n log n) due to sorting events

SpaceO(2n) for events array

We create two events per trip and sort them, then iterate once to check capacity, which is efficient for large inputs.

💡 For n=100,000 trips, sorting 200,000 events is feasible within typical interview time constraints.

Interview Verdict: Accepted / Efficient

This approach is efficient and commonly accepted in interviews for interval capacity tracking problems.

🧠

Difference Array / Prefix Sum Optimization

💡 This approach uses a difference array to track passenger changes at start and end points, then uses prefix sums to find the passenger count at each location efficiently.

Intuition

Instead of processing every location for every trip, mark only the changes at start and end locations, then accumulate these changes to get passenger counts.

Algorithm

1. Initialize an array of zeros for all locations up to max location.
2. For each trip, add passengers at start location and subtract passengers at end location.
3. Compute prefix sums over this array to get current passengers at each location.
4. Check if at any location passengers exceed capacity.
5. Return true if capacity never exceeded.

💡 This approach avoids sorting and processes the entire route in linear time, leveraging prefix sums for efficiency.

</>

Code

def carPooling(trips, capacity):
    max_location = 0
    for _, start, end in trips:
        max_location = max(max_location, end)
    diff = [0] * (max_location + 1)
    for num, start, end in trips:
        diff[start] += num
        diff[end] -= num
    current_passengers = 0
    for i in range(max_location + 1):
        current_passengers += diff[i]
        if current_passengers > capacity:
            return False
    return True

# Example usage
if __name__ == '__main__':
    print(carPooling([[2,1,5],[3,3,7]], 4))  # False
    print(carPooling([[2,1,5],[3,3,7]], 5))  # True

Line Notes

diff = [0] * (max_location + 1)Initialize difference array to track passenger changes

diff[start] += numAdd passengers at start location

diff[end] -= numSubtract passengers at end location

current_passengers += diff[i]Accumulate passenger changes to get current count

if current_passengers > capacity:Check if capacity exceeded at any location

import java.util.*;
public class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        int maxLocation = 0;
        for (int[] trip : trips) {
            maxLocation = Math.max(maxLocation, trip[2]);
        }
        int[] diff = new int[maxLocation + 1];
        for (int[] trip : trips) {
            diff[trip[1]] += trip[0];
            diff[trip[2]] -= trip[0];
        }
        int currentPassengers = 0;
        for (int i = 0; i <= maxLocation; i++) {
            currentPassengers += diff[i];
            if (currentPassengers > capacity) {
                return false;
            }
        }
        return true;
    }
    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.carPooling(new int[][]{{2,1,5},{3,3,7}}, 4)); // false
        System.out.println(sol.carPooling(new int[][]{{2,1,5},{3,3,7}}, 5)); // true
    }
}

Line Notes

int[] diff = new int[maxLocation + 1];Initialize difference array for passenger changes

diff[trip[1]] += trip[0];Add passengers at start location

diff[trip[2]] -= trip[0];Subtract passengers at end location

currentPassengers += diff[i];Accumulate passenger changes to get current count

if (currentPassengers > capacity)Check if capacity exceeded

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

bool carPooling(vector<vector<int>>& trips, int capacity) {
    int maxLocation = 0;
    for (auto& trip : trips) {
        maxLocation = max(maxLocation, trip[2]);
    }
    vector<int> diff(maxLocation + 1, 0);
    for (auto& trip : trips) {
        diff[trip[1]] += trip[0];
        diff[trip[2]] -= trip[0];
    }
    int currentPassengers = 0;
    for (int i = 0; i <= maxLocation; i++) {
        currentPassengers += diff[i];
        if (currentPassengers > capacity) {
            return false;
        }
    }
    return true;
}

int main() {
    vector<vector<int>> trips1 = {{2,1,5},{3,3,7}};
    cout << (carPooling(trips1, 4) ? "true" : "false") << endl; // false
    cout << (carPooling(trips1, 5) ? "true" : "false") << endl; // true
    return 0;
}

Line Notes

vector<int> diff(maxLocation + 1, 0);Initialize difference array for passenger changes

diff[trip[1]] += trip[0];Add passengers at start location

diff[trip[2]] -= trip[0];Subtract passengers at end location

currentPassengers += diff[i];Accumulate passenger changes to get current count

if (currentPassengers > capacity)Check if capacity exceeded

function carPooling(trips, capacity) {
    let maxLocation = 0;
    for (const [, start, end] of trips) {
        maxLocation = Math.max(maxLocation, end);
    }
    const diff = new Array(maxLocation + 1).fill(0);
    for (const [num, start, end] of trips) {
        diff[start] += num;
        diff[end] -= num;
    }
    let currentPassengers = 0;
    for (let i = 0; i <= maxLocation; i++) {
        currentPassengers += diff[i];
        if (currentPassengers > capacity) {
            return false;
        }
    }
    return true;
}

// Example usage
console.log(carPooling([[2,1,5],[3,3,7]], 4)); // false
console.log(carPooling([[2,1,5],[3,3,7]], 5)); // true

Line Notes

const diff = new Array(maxLocation + 1).fill(0);Initialize difference array for passenger changes

diff[start] += num;Add passengers at start location

diff[end] -= num;Subtract passengers at end location

currentPassengers += diff[i];Accumulate passenger changes to get current count

if (currentPassengers > capacity)Check if capacity exceeded

⏱

Complexity

TimeO(n + L) where n is number of trips and L is max location

SpaceO(L) for difference array

We mark changes for each trip and then do a single pass prefix sum over the locations, which is efficient for large inputs.

💡 For n=100,000 and max location=1000, this approach is very fast because it avoids sorting and nested loops.

Interview Verdict: Accepted / Most optimal for this problem

This is the best approach in terms of time and space for this problem and is recommended in interviews.

📊

All Approaches - One-Glance Tradeoffs

💡 In most interviews, the difference array or sweep line approach is best to implement due to efficiency and clarity.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n * L)	O(L)	No	N/A	Mention only - never code due to inefficiency
2. Sorting Events and Sweep Line	O(n log n)	O(n)	No	N/A	Good to code if input size is large and sorting is acceptable
3. Difference Array / Prefix Sum	O(n + L)	O(L)	No	N/A	Best approach to code for optimal performance

💼

Interview Strategy

💡 Use this guide to understand the problem deeply before interviews. Start by clarifying the problem, then explain the brute force approach to show understanding. Progress to optimized solutions and discuss tradeoffs.

How to Present

Clarify the problem constraints and input/output format.Describe the brute force approach and its inefficiency.Introduce the sweep line approach with event sorting.Explain the difference array optimization as the best solution.Write clean, tested code and discuss complexity.

Time Allocation

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

What the Interviewer Tests

The interviewer tests your ability to handle interval events efficiently, your understanding of sorting and prefix sums, and your ability to optimize naive solutions.

Common Follow-ups

• What if locations are very large numbers? → Use coordinate compression.
• Can you handle trips that start and end at the same location? → They don't affect capacity.
• How to extend if vehicle can move back and forth? → More complex event handling needed.
• What if capacity changes over time? → Use segment trees or binary indexed trees.

💡 These follow-ups test your ability to adapt the solution to real-world complexities and edge cases.

🔍

Pattern Recognition

When to Use

Use this pattern when you need to track cumulative effects over intervals and check constraints like capacity or overlap.

Signature Phrases

'Given trips with start and end locations''Check if capacity is exceeded at any point''Passengers get on and off at different locations'

NOT This Pattern When

Problems that require interval merging or maximum interval coverage without capacity constraints

Similar Problems

Meeting Rooms II - also involves interval overlap and resource allocationMy Calendar I - event booking with overlap checksMinimum Number of Platforms - scheduling with capacity constraints

Practice

1. Consider the following Python code that merges intervals in-place after sorting. Given the input intervals = [[1,4],[2,5],[7,9]], what is the returned list after the function completes?

easy

A. [[1,5],[7,9]]

B. [[1,4],[2,5],[7,9]]

C. [[1,5],[2,5],[7,9]]

D. [[1,4],[7,9]]

Solution

1. Step 1: Sort intervals by start time.

   Input is already sorted: [[1,4],[2,5],[7,9]].

2. Step 2: Merge intervals in one pass.

   Compare [2,5] with [1,4]: overlap since 2 <= 4, merge to [1,5]. Then compare [7,9] with [1,5]: no overlap, move index forward and assign [7,9]. Final intervals: [[1,5],[7,9]].

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Overlapping intervals merged correctly, non-overlapping preserved. [OK]

Hint: Check if current start ≤ last merged end to merge [OK]

Common Mistakes:

• Returning original intervals without merging
• Off-by-one slicing errors

2. Consider the following Python function that returns the maximum number of non-overlapping intervals. What is the value of the variable removals after the second iteration of the loop when the input is [[1,3],[2,4],[3,5]]?

from typing import List

def max_non_overlapping_intervals(intervals: List[List[int]]) -> int:
    intervals.sort(key=lambda x: x[0])
    removals = 0
    last_end = intervals[0][1]
    for i in range(1, len(intervals)):
        if intervals[i][0] < last_end:
            removals += 1
            last_end = min(last_end, intervals[i][1])
        else:
            last_end = intervals[i][1]
    return len(intervals) - removals

intervals = [[1,3],[2,4],[3,5]]
print(max_non_overlapping_intervals(intervals))

easy

A. 1

B. 3

C. 2

D. 0

Solution

1. Step 1: Trace first iteration (i=1)

   Intervals sorted by start: [[1,3],[2,4],[3,5]]. last_end=3. At i=1, intervals[1][0]=2 < 3, so removals=1, last_end=min(3,4)=3.

2. Step 2: Trace second iteration (i=2)

   At i=2, intervals[2][0]=3 <= last_end=3 is false, so last_end=5, removals remains 1.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   removals incremented once at i=1 only [OK]

Hint: Check overlap condition carefully at each iteration [OK]

Common Mistakes:

• Off-by-one error in loop iteration
• Misunderstanding when to increment removals
• Confusing last_end update logic

3. What is the time complexity of the optimal solution that sorts intervals by start ascending and end descending, then scans once to count uncovered intervals?

medium

A. O(n log n) due to sorting plus O(n) scanning

B. O(n log n) due to sorting, but scanning is O(n²) in worst case

C. O(n) because scanning is linear and sorting is negligible

D. O(n²) due to nested coverage checks

Solution

1. Step 1: Identify sorting cost

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

2. Step 2: Identify scanning cost

   Single pass scanning to count uncovered intervals is O(n).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sorting dominates, scanning is linear [OK]

Hint: Sorting dominates complexity, scanning is linear [OK]

Common Mistakes:

• Confuse scanning as nested loops O(n²)
• Ignore sorting cost and say O(n)

4. Suppose the problem changes: intervals can be reused infinitely many times, and you want to count how many intervals contain each point considering unlimited reuse (e.g., intervals repeat periodically). Which modification to the line sweep approach correctly handles this scenario?

hard

A. No change needed; line sweep already counts intervals correctly regardless of reuse.

B. Preprocess intervals to merge overlapping intervals into one, then run line sweep once.

C. Use a segment tree or binary indexed tree to handle repeated intervals efficiently instead of line sweep.

D. Modify events to include multiple copies of intervals explicitly and run line sweep on expanded events.

Solution

1. Step 1: Understand reuse impact

   Infinite reuse means intervals repeat periodically, so naive line sweep with explicit events cannot handle infinite copies.

2. Step 2: Recognize data structure need

   Segment trees or binary indexed trees can efficiently query counts over ranges and handle repeated intervals without enumerating all copies.

3. Step 3: Why other options fail

   No change (A) ignores infinite reuse; merging (B) loses count detail; explicit expansion (D) is infeasible for infinite repeats.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Segment trees handle range queries with repeated intervals efficiently [OK]

Hint: Infinite reuse requires data structures, not event enumeration [OK]

Common Mistakes:

• Assuming line sweep handles infinite repeats
• Trying to expand events explicitly
• Merging intervals loses count accuracy

5. Suppose balloons can be reused multiple times after bursting (i.e., an arrow can burst a balloon multiple times if shot at different points). How does this affect the algorithm to find the minimum number of arrows needed?

hard

A. The original greedy algorithm still applies without changes.

B. We must use dynamic programming to track multiple bursts per balloon.

C. The problem reduces to counting unique balloons since reuse removes overlap constraints.

D. We can shoot arrows only at balloon start points to minimize arrows.

Solution

1. Step 1: Understand reuse effect

   If balloons can be burst multiple times independently, overlapping intervals no longer reduce arrow count.

2. Step 2: Simplify problem

   Each balloon requires at least one arrow, so minimum arrows equal number of unique balloons.

3. Step 3: Algorithm implication

   Overlap and interval sorting become irrelevant; counting balloons suffices.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Reuse breaks overlap optimization, so count balloons directly [OK]

Hint: Reuse removes overlap benefit; count balloons directly [OK]

Common Mistakes:

• Assuming greedy still works unchanged
• Trying to track multiple bursts with DP unnecessarily
• Ignoring that reuse breaks interval overlap logic