Minimum Number of Arrows to Burst Balloons

Medium

Minimum Number of Arrows to Burst Balloons

Medium

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the start and end points of each balloon, and what data type are they (e.g., integers)?
Can the input array of balloon coordinates be empty or null?
If two balloons share an exact point (e.g., [1, 5] and [5, 8]), do they count as overlapping and can be burst by a single arrow?
If no balloons are present, should I return 0?
Are the balloons guaranteed to be sorted by their starting x coordinate?

Brute Force Solution

Approach

We want to find the fewest arrows to burst a set of balloons. The brute force way involves trying every single combination of arrow placements to see which one uses the least arrows while bursting all the balloons.

Here's how the algorithm would work step-by-step:

Imagine we have a set of balloons. We'll start by trying to burst all balloons with just one arrow.
We pick a spot and check if all balloons can be burst from that spot. If so, we found our answer!
If not, we then try every combination of two arrow spots. We check each pair of spots to see if those two arrows can burst all the balloons.
If not, we try every combination of three arrow spots, and so on.
We keep increasing the number of arrows we are using, testing every possible location for each arrow, until we find a set of arrow locations that bursts all the balloons.
Of all the arrow sets that burst all balloons, the solution is the set that uses the fewest number of arrows.

Code Implementation

def minimum_arrows_brute_force(balloons):
    number_of_balloons = len(balloons)
    
    for number_of_arrows in range(1, number_of_balloons + 1):
        
        arrow_placements = find_all_combinations(balloons, number_of_arrows)
        
        for current_arrow_placement in arrow_placements:

            balloons_burst = True
            for balloon_start, balloon_end in balloons:
                balloon_burst_by_any_arrow = False
                for arrow_position in current_arrow_placement:

                    if balloon_start <= arrow_position <= balloon_end:
                        balloon_burst_by_any_arrow = True
                        break
                if not balloon_burst_by_any_arrow:
                    balloons_burst = False
                    break
            # If all balloons are burst, return the number of arrows
            if balloons_burst:
                return number_of_arrows

    return number_of_balloons

def find_all_combinations(balloons, number_of_arrows):
    
    all_possible_arrow_locations = []
    for balloon_start, balloon_end in balloons:
        for position in range(balloon_start, balloon_end + 1):
            all_possible_arrow_locations.append(position)
    
    all_possible_arrow_locations = sorted(list(set(all_possible_arrow_locations)))

    combinations = find_combinations_recursive(all_possible_arrow_locations, number_of_arrows, 0, [], [])
    return combinations

def find_combinations_recursive(locations, number_to_choose, current_index, current_combination, all_combinations):
    if len(current_combination) == number_to_choose:
        all_combinations.append(current_combination[:])
        return

    if current_index >= len(locations):
        return

    # Include the current element
    current_combination.append(locations[current_index])

    find_combinations_recursive(locations, number_to_choose, current_index + 1, current_combination, all_combinations)
    current_combination.pop()

    # Exclude the current element
    find_combinations_recursive(locations, number_to_choose, current_index + 1, current_combination, all_combinations)

Big(O) Analysis

Time Complexity

O(n^n) – The described brute force algorithm iterates through every possible combination of arrow placements to burst all balloons. In the worst case, we might need to test up to n arrows, where n is the number of balloons. For each number of arrows (from 1 to n), we have to consider every possible combination of arrow positions. In the absolute worst case, we are checking all possible combinations of arrow locations to see if they burst all the balloons, leading to a complexity that grows factorially. This is approximated as O(n^n) as it considers all positions for all possible arrows (1 to n).

Space Complexity

O(1) – The provided brute-force approach, as described, doesn't utilize any significant auxiliary data structures. It primarily involves iterating through combinations of arrow placements and checking if those placements burst all balloons. The number of arrow placements being tested at a time, and the loop indices, would take up constant space. Therefore, the space complexity is O(1), as the extra space used remains constant regardless of the number of balloons (N).

Optimal Solution

Approach

The goal is to use the fewest arrows possible to burst all balloons. The key insight is to focus on overlapping balloons: if balloons overlap, a single arrow can burst them all. By strategically choosing where to shoot, we maximize the number of balloons burst per arrow.

Here's how the algorithm would work step-by-step:

First, think of each balloon as a segment on a number line, defined by its start and end points.
Sort all the balloons based on where they end on the number line. This is a crucial first step.
Start by shooting an arrow at the endpoint of the first balloon (which is also the smallest endpoint, because we sorted them).
Count how many other balloons this arrow also bursts. A balloon is burst by this arrow if its start point is less than or equal to the arrow's position.
Move to the next balloon that hasn't been burst yet. Shoot an arrow at its endpoint, and repeat the process of bursting as many balloons as possible.
Keep doing this until all the balloons are burst. The total number of arrows you used is the minimum number required.

Code Implementation

def minimum_arrows_to_burst_balloons(balloons):
    if not balloons:
        return 0

    # Sort balloons based on their end points.
    balloons.sort(key=lambda x: x[1])

    number_of_arrows = 0
    arrow_position = float('-inf')

    for balloon in balloons:
        # If the current balloon's start is after
        # the current arrow, we need a new arrow.
        if balloon[0] > arrow_position:
            number_of_arrows += 1
            # Update the arrow position to the end
            # of the current balloon.
            arrow_position = balloon[1]

    return number_of_arrows

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this algorithm is sorting the balloons based on their end points. Sorting algorithms like merge sort or quicksort have a time complexity of O(n log n), where n is the number of balloons. After sorting, we iterate through the balloons once to determine the minimum number of arrows. This iteration takes O(n) time. Therefore, the overall time complexity is O(n log n) + O(n), which simplifies to O(n log n) since sorting dominates.

Space Complexity

O(1) – The algorithm primarily sorts the input balloons in place, which doesn't contribute to auxiliary space complexity, assuming the sorting algorithm used is in-place or its space cost is negligible compared to the input size. It uses a constant number of variables, such as the arrow count and the current arrow position, regardless of the number of balloons (N). No additional data structures like arrays or hash maps are created that scale with the input size. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty input points array	Return 0 since no balloons need to be burst.
Single point in the input array	Return 1 since one arrow is enough to burst the single balloon.
Points array with all identical intervals	Return 1 since all balloons can be burst with one arrow.
Points array with intervals that are greatly overlapping	The greedy approach should efficiently combine these with fewer arrows.
Points array with intervals that are completely disjointed	The algorithm should require one arrow per disjoint interval.
Points array with very large interval values	Ensure that the sorting and comparison operations handle the large values without integer overflow issues.
Points array is already sorted in increasing order of end points	The algorithm should still function correctly and efficiently without needing to be sorted.
Points array where intervals have the same start and end points	The algorithm should treat these as valid balloons and count them correctly.