Eliminate Maximum Number of Monsters

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ArraysGreedy Algorithms

You are playing a video game where you are defending your city from a group of n monsters. You are given a 0-indexed integer array dist of size n, where dist[i] is the initial distance in kilometers of the i^th monster from the city.

The monsters walk toward the city at a constant speed. The speed of each monster is given to you in an integer array speed of size n, where speed[i] is the speed of the i^th monster in kilometers per minute.

You have a weapon that, once fully charged, can eliminate a single monster. However, the weapon takes one minute to charge. The weapon is fully charged at the very start.

You lose when any monster reaches your city. If a monster reaches the city at the exact moment the weapon is fully charged, it counts as a loss, and the game ends before you can use your weapon.

Return the maximum number of monsters that you can eliminate before you lose, or n if you can eliminate all the monsters before they reach the city.

Example 1:

Input: dist = [1,3,4], speed = [1,1,1]
Output: 3
Explanation:
In the beginning, the distances of the monsters are [1,3,4]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,2,3]. You eliminate the second monster.
After a minute, the distances of the monsters are [X,X,2]. You eliminate the third monster.
All 3 monsters can be eliminated.

Example 2:

Input: dist = [1,1,2,3], speed = [1,1,1,1]
Output: 1
Explanation:
In the beginning, the distances of the monsters are [1,1,2,3]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,0,1,2], so you lose.
You can only eliminate 1 monster.

Example 3:

Input: dist = [3,2,4], speed = [5,3,2]
Output: 1
Explanation:
In the beginning, the distances of the monsters are [3,2,4]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,0,2], so you lose.
You can only eliminate 1 monster.

Constraints:

n == dist.length == speed.length
1 <= n <= 10⁵
1 <= dist[i], speed[i] <= 10⁵

Try coding it in LeetCode

Eliminate Maximum Number of Monsters

Medium

Asked by:

7 views

Topics:

ArraysGreedy Algorithms

You are playing a video game where you are defending your city from a group of n monsters. You are given a 0-indexed integer array dist of size n, where dist[i] is the initial distance in kilometers of the i^th monster from the city.

The monsters walk toward the city at a constant speed. The speed of each monster is given to you in an integer array speed of size n, where speed[i] is the speed of the i^th monster in kilometers per minute.

You have a weapon that, once fully charged, can eliminate a single monster. However, the weapon takes one minute to charge. The weapon is fully charged at the very start.

You lose when any monster reaches your city. If a monster reaches the city at the exact moment the weapon is fully charged, it counts as a loss, and the game ends before you can use your weapon.

Return the maximum number of monsters that you can eliminate before you lose, or n if you can eliminate all the monsters before they reach the city.

Example 1:

Input: dist = [1,3,4], speed = [1,1,1]
Output: 3
Explanation:
In the beginning, the distances of the monsters are [1,3,4]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,2,3]. You eliminate the second monster.
After a minute, the distances of the monsters are [X,X,2]. You eliminate the third monster.
All 3 monsters can be eliminated.

Example 2:

Input: dist = [1,1,2,3], speed = [1,1,1,1]
Output: 1
Explanation:
In the beginning, the distances of the monsters are [1,1,2,3]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,0,1,2], so you lose.
You can only eliminate 1 monster.

Example 3:

Input: dist = [3,2,4], speed = [5,3,2]
Output: 1
Explanation:
In the beginning, the distances of the monsters are [3,2,4]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,0,2], so you lose.
You can only eliminate 1 monster.

Constraints:

n == dist.length == speed.length
1 <= n <= 10⁵
1 <= dist[i], speed[i] <= 10⁵

Try coding it in LeetCode

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 are the possible ranges for the distance and speed values, and are they integers or floating-point numbers?
Can the distance or speed values be negative or zero? What happens if a monster has a speed of zero?
If multiple monsters can be eliminated in the same round, which ones should I prioritize eliminating?
Is it possible to have more monsters arriving at the city at the same time than I can eliminate in a round?
What should I return if it's impossible to eliminate all the monsters before they reach the city?

Brute Force Solution

Approach

The brute force approach involves simulating every possible scenario of fighting the monsters. We try different orders of shooting the monsters to figure out the best strategy. This means checking all possible combinations of monster takedowns to see which one lets us defeat the most monsters before they reach the city.

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

Consider all possible orders in which we can shoot the monsters. For example, if there are three monsters, consider shooting monster 1 first, then monster 2, then monster 3; or monster 1 first, then monster 3, then monster 2; and so on for all six possible orders.
For each of these shooting orders, simulate the entire scenario. This means, figure out at which moment each monster would reach the city given its initial position and speed.
As we simulate, keep track of how many monsters we can shoot down before they reach the city for that specific shooting order.
Compare the number of monsters defeated in each scenario (each shooting order).
Select the shooting order that allows us to shoot down the maximum number of monsters.

Code Implementation

import itertools

def eliminate_maximum_number_of_monsters(
    distances_to_city, monster_speeds
):
    number_of_monsters = len(distances_to_city)
    maximum_monsters_eliminated = 0

    # Iterate through all possible monster shooting orders
    for shooting_order in itertools.permutations(range(number_of_monsters)):
        monsters_eliminated = 0
        time = 0
        can_eliminate = True

        # Simulate the game with the current shooting order
        for monster_index in shooting_order:
            arrival_time = (
                distances_to_city[monster_index] /
                monster_speeds[monster_index]
            )

            # Check if monster reaches the city before being shot
            if arrival_time <= time:
                can_eliminate = False
                break

            # Increment time after shooting a monster
            time += 1
            monsters_eliminated += 1

        # Update maximum monsters eliminated if needed
        if can_eliminate:
            maximum_monsters_eliminated = max(
                maximum_monsters_eliminated, monsters_eliminated
            )

    return maximum_monsters_eliminated

Big(O) Analysis

Time Complexity

O(n! * n) – The algorithm considers all possible orders of shooting n monsters, which results in n! (n factorial) permutations. For each permutation (shooting order), we simulate the entire scenario to determine how many monsters can be defeated. This simulation involves iterating through the monsters (up to n monsters) to check if they reach the city before being shot. Therefore, for each of the n! permutations, we perform an O(n) operation. Consequently, the overall time complexity is O(n! * n).

Space Complexity

O(N!) – The brute force approach considers all possible orders of shooting monsters, which involves generating permutations. If there are N monsters, there are N! (N factorial) possible permutations. Storing each permutation, either explicitly in a data structure or implicitly through recursive call stacks during generation, would require space proportional to the number of permutations. Therefore, the auxiliary space complexity is O(N!).

Optimal Solution

Approach

The problem asks us to maximize the number of monsters we eliminate. A key insight is to realize we should prioritize eliminating monsters that will reach the city sooner rather than later. By strategically eliminating the closest monsters first, we can defend the city for as long as possible.

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

Calculate how much time it will take each monster to reach the city.
Organize the monsters based on their arrival times, from earliest to latest.
Start eliminating monsters one by one, according to the sorted order.
For each monster, check if you have enough time to eliminate it before it reaches the city.
If you can eliminate it, do so and proceed to the next monster. Note that you use up one unit of time when eliminating a monster.
If you cannot eliminate it because it will reach the city before you have the opportunity to eliminate it, stop and count how many monsters you eliminated.

Code Implementation

def eliminate_maximum_number_of_monsters(distances, speeds):
    number_of_monsters = len(distances)
    arrival_times = []
    for i in range(number_of_monsters):
        arrival_times.append(float(distances[i]) / speeds[i])

    # Sort monsters based on arrival time to prioritize early threats.
    arrival_times.sort()

    eliminated_count = 0
    time = 0

    for arrival_time in arrival_times:
        # Check if we have time to eliminate the monster before it arrives.
        if arrival_time <= time:
            break

        # Increment eliminated count since we can eliminate the monster.
        eliminated_count += 1
        time += 1

    return eliminated_count

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this algorithm is sorting the monsters based on their arrival times. Sorting n elements typically takes O(n log n) time using efficient algorithms like merge sort or quicksort. The subsequent iteration through the sorted monsters takes O(n) time to simulate the elimination process. Therefore, the overall time complexity is O(n log n) + O(n), which simplifies to O(n log n).

Space Complexity

O(N) – The algorithm calculates the time it takes for each monster to reach the city and stores these times. This requires creating an array of size N, where N is the number of monsters, to hold these arrival times. The space used for sorting this arrival time array is typically O(N) in the worst case (e.g., merge sort). Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty dist or speed arrays	Return 0 immediately since there are no monsters to eliminate.
dist and speed arrays of different lengths	Throw an IllegalArgumentException or return an error code, indicating invalid input.
Zero distance for a monster	The time to reach the city is 0, so eliminate it immediately if possible.
Zero speed for a monster	If distance is also zero, handle as above; otherwise, the monster never reaches the city, so don't count it.
Large dist and speed values causing potential integer overflow	Use long data type to store the time to reach the city to prevent overflow.
Monsters all arrive at the same time	The algorithm should correctly eliminate monsters in the order they appear in the input.
No monsters can be eliminated	Return 0 indicating that no monsters could be eliminated before being overtaken.
Very large n (size of dist/speed arrays)	Ensure the sorting algorithm used to determine arrival times has O(n log n) complexity for acceptable performance.