853. Car Fleet — LeetCode(Python)
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Problem:
There are n cars going to the same destination along a one-lane road. The destination is target miles away.
You are given two integer array position and speed, both of length n, where position[i] is the position of the ith car and speed[i] is the speed of the ith car (in miles per hour).
A car can never pass another car ahead of it, but it can catch up to it and drive bumper to bumper at the same speed. The faster car will slow down to match the slower car’s speed. The distance between these two cars is ignored (i.e., they are assumed to have the same position).
A car fleet is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.
If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet.
Return the number of car fleets that will arrive at the destination.
Example 1:
Input: target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]
Output: 3
Explanation:
The cars starting at 10 (speed 2) and 8 (speed 4) become a fleet, meeting each other at 12.
The car starting at 0 does not catch up to any other car, so it is a fleet by itself.
The cars starting at 5 (speed 1) and 3 (speed 3) become a fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.
Note that no other cars meet these fleets before the destination, so the answer is 3.Example 2:
Input: target = 10, position = [3], speed = [3]
Output: 1
Explanation: There is only one car, hence there is only one fleet.Example 3:
Input: target = 100, position = [0,2,4], speed = [4,2,1]
Output: 1
Explanation:
The cars starting at 0 (speed 4) and 2 (speed 2) become a fleet, meeting each other at 4. The fleet moves at speed 2.
Then, the fleet (speed 2) and the car starting at 4 (speed 1) become one fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.Constraints:
n == position.length == speed.length1 <= n <= 1050 < target <= 1060 <= position[i] < target- All the values of
positionare unique. 0 < speed[i] <= 106
Solution:
pairs = [[p,s] for p, s in zip(position, speed)]
stack = []
for p, s in sorted(pairs, reverse=True):
stack.append((target - p) / s)
if len(stack) >= 2 and stack[-1] <= stack[-2]:
stack.pop()
return len(stack)Explanation —
We create a list of pairs.
We then traverse the list of pairs in reverse order. This is important for a few reasons as we are going to see.
When we come across a new pair, we calculate the speed this car takes and push it onto the stack.
If the length of the stack is greater than or equal to 2 and we calculate that the 2nd car will overtake the topmost car, we pop the topmost car, since it is slower and it is not necessary to consider it now.
Since we popped a car, we can say visualize it in such a way that we combined two cars into one fleet.
Finally, we return the length of the stack which tells us how many groups/fleets are present.
Time and Space Complexity —
Iterating over the array of pairs is a linear time operation.
However, sorting of that array is a linear-logarithmic time operation.
Also, we require a linear amount of extra space because we are creating a stack and a pairs array.
Thus, if n is the length of both the position and speed arrays,
Time Complexity: O(n.log(n))
Space Complexity: O(n)
