What if you could instantly know the exact number of platforms needed to keep trains moving without delay?
Why Minimum Number of Platforms in DSA Python?
Imagine a busy train station where many trains arrive and depart at different times. You want to know how many platforms are needed so that no train has to wait for a platform to be free.
Trying to check each train's arrival and departure times manually is slow and confusing. You might miss overlaps or count platforms wrongly, causing delays or wasted space.
The Minimum Number of Platforms concept helps us quickly find the smallest number of platforms needed by comparing all arrival and departure times in order. This way, we know exactly when platforms are busy or free.
for each train: for each other train: if times overlap: count platform needed
sort arrivals and departures
use two pointers to count overlaps
track max platforms neededThis concept lets us efficiently plan platform usage to avoid train delays and optimize station space.
Railway stations use this method to decide how many platforms to build so trains can arrive and leave smoothly without waiting.
Manual checking of train times is slow and error-prone.
Sorting and comparing arrival and departure times simplifies the problem.
We can find the minimum platforms needed to avoid train waiting.
Practice
Solution
Step 1: Understand the problem context
The problem wants to find how many platforms are needed so that no train has to wait for another to leave.Step 2: Identify the key requirement
We must find the minimum number of platforms so that all trains can be accommodated without delay.Final Answer:
The least number of platforms needed so no train waits -> Option AQuick Check:
Minimum platforms = no waiting trains [OK]
- Confusing total trains with platforms needed
- Ignoring overlapping arrival and departure times
- Thinking about train durations instead of platform count
arrivals for the Minimum Number of Platforms problem?Solution
Step 1: Recall Python list sorting method
Python lists have a method calledsort()that sorts the list in place.Step 2: Check each option
arrivals.sort() usesarrivals.sort()which is correct. sort(arrivals) is invalid syntax. arrivals.sorted() uses a non-existent method. arrivals.sort(arrivals) incorrectly passes the list as argument.Final Answer:
arrivals.sort() -> Option CQuick Check:
List.sort() is correct method [OK]
- Using sort() as a standalone function
- Trying to call sorted() as a method
- Passing the list as argument to sort()
arrivals = [900, 940, 950, 1100, 1500, 1800]
departures = [910, 1200, 1120, 1130, 1900, 2000]Solution
Step 1: Sort arrival and departure times
Arrivals sorted: [900, 940, 950, 1100, 1500, 1800]
Departures sorted: [910, 1120, 1130, 1200, 1900, 2000]Step 2: Use two pointers to count platforms
Start with i=0 (arrivals), j=0 (departures), platforms=0, max_platforms=0.
At 900 (arrival), platforms=1.
At 910 (departure), platforms=0.
At 940 (arrival), platforms=1.
At 950 (arrival), platforms=2.
At 1100 (arrival), platforms=2 (since departure at 1120 is after 1100 but 1130 and 1200 are later).
Max platforms needed = 2.Final Answer:
2 -> Option BQuick Check:
Max overlapping trains = 2 [OK]
- Not sorting times before counting
- Mixing arrival and departure pointers
- Counting platforms incorrectly at overlaps
arrivals = [900, 940, 950]
departures = [910, 1200, 1120]
arrivals.sort()
departures.sort()
i = j = 0
platforms = max_platforms = 0
while i < len(arrivals) and j < len(departures):
if arrivals[i] <= departures[j]:
platforms += 1
i += 1
else:
platforms -= 1
j += 1
max_platforms = max(max_platforms, platforms)
print(max_platforms)
Solution
Step 1: Understand the comparison logic
Using <= means if arrival time equals departure time, it counts as needing a new platform, but usually a train leaving at time t frees the platform for a train arriving at time t.Step 2: Correct the comparison
Changing <= to < ensures that if arrival equals departure, platform count decreases first, avoiding unnecessary extra platform.Final Answer:
The comparison should be arrivals[i] < departures[j], not <=. -> Option AQuick Check:
Use < to avoid counting simultaneous arrival and departure as overlap [OK]
- Using <= causes overcounting platforms
- Changing loop condition wrongly to 'or'
- Not sorting departure times
arrivals = [900, 940, 950, 1100, 1500, 1800, 1800]
departures = [910, 1200, 1120, 1130, 1900, 2000, 1800]Choose the best approach.
Solution
Step 1: Sort both arrival and departure lists
Sorting helps to process events in chronological order, which is essential for counting overlapping trains.Step 2: Use two pointers with strict comparison
Use two pointers to traverse arrivals and departures. Usearrivals[i] < departures[j]to decide when to increase or decrease platform count, handling same-time arrivals and departures correctly.Final Answer:
Sort arrivals and departures, then use two pointers with strict less than comparison for arrival vs departure. -> Option DQuick Check:
Sorting + two pointers + < comparison = correct approach [OK]
- Ignoring sorting of departures
- Using <= causing overcount
- Trying to count without sorting or pointers
Ready to prove it?
Pass the test and this pattern moves unknown
