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This approach involves computing the exact time each monster will take to reach the city and then eliminating them in the order of their arrival times. The goal is to eliminate as many monsters as possible before any monster reaches the city.
Time Complexity: O(n log n) due to sorting the times.
Space Complexity: O(n) to store the arrival times.
1def eliminateMaximum(dist, speed):
2 time = [(d + s - 1) // s for d, s in zip(dist, speed)] # ceiling of (dist[i] / speed[i])
3 time.sort()
4 for i, t in enumerate(time):
5 if t <= i:
6 return i
7 return len(time)
8
9print(eliminateMaximum([1, 3, 4], [1, 1, 1]))The Python code calculates the time taken for each monster to reach the city, sorts these times, and checks at each minute if a monster has reached the city. It's designed to stop when a loss condition is met.
This approach improves the efficiency of eliminating the maximum number of monsters by using a binary search method. It pre-calculates when a monster will reach the city and uses efficient search techniques to make decisions dynamically during the elimination process.
Time Complexity: O(n log n) due to the sort operation.
Space Complexity: O(n) to store arrival times.
1from bisect import bisect_right
2
3def eliminateMaximum
By using a library function like bisect_right in Python, the code finds the largest index to insert a value (in this case to simulate minutes passing) while maintaining order. This efficiently tells us the maximum monsters that can be eliminated as time progresses.