Minimum Number of Platforms in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Minimum Platforms Needed for Given Train Schedule

What is the output of the following code that calculates the minimum number of platforms needed for trains arriving and departing at given times?

DSA Python

def find_min_platforms(arrivals, departures):
    arrivals.sort()
    departures.sort()
    platform_needed = 0
    max_platforms = 0
    i = 0
    j = 0
    n = len(arrivals)
    while i < n and j < n:
        if arrivals[i] < departures[j]:
            platform_needed += 1
            max_platforms = max(max_platforms, platform_needed)
            i += 1
        else:
            platform_needed -= 1
            j += 1
    return max_platforms

arrivals = [900, 940, 950, 1100, 1500, 1800]
departures = [910, 1200, 1120, 1130, 1900, 2000]
print(find_min_platforms(arrivals, departures))

A 3

B 4

C 5

D 2

💡 Hint

Sort arrival and departure times, then count platforms needed by comparing arrivals and departures.

The code sorts arrival and departure times. It uses two pointers to track trains arriving and leaving. When a train arrives before the earliest departure, it increases platform count. When a train departs, it decreases platform count. The maximum platforms needed at any time is 3.

🧠 Conceptual

intermediate

Understanding Platform Calculation Logic

In the minimum number of platforms problem, why do we sort both arrival and departure times before processing?

A To group all arrivals before departures for easier counting

B To process events in chronological order and correctly count overlapping trains

C To match each arrival with its exact departure time

D To ensure departures are always processed before arrivals

💡 Hint

Think about how sorting helps in comparing times step-by-step.

Sorting arrivals and departures allows us to process train events in time order. This helps to count how many trains are at the station simultaneously, which determines the minimum platforms needed.

❓ Predict Output

advanced

Output of Platform Calculation with Overlapping Times

What is the output of this code snippet calculating minimum platforms for overlapping train times?

DSA Python

arrivals = [1000, 1010, 1025, 1030]
departures = [1020, 1035, 1040, 1050]

def min_platforms(arr, dep):
    arr.sort()
    dep.sort()
    i = j = 0
    count = max_count = 0
    n = len(arr)
    while i < n and j < n:
        if arr[i] < dep[j]:
            count += 1
            max_count = max(max_count, count)
            i += 1
        else:
            count -= 1
            j += 1
    return max_count

print(min_platforms(arrivals, departures))

A 1

B 4

C 2

D 3

💡 Hint

Check how many trains overlap at the same time.

The trains arriving at 1000, 1010, and 1025 overlap before the first departure at 1020. So, maximum platforms needed is 3.

🔧 Debug

advanced

Identify the Error in Platform Calculation Code

What error does this code produce when calculating minimum platforms?

DSA Python

def min_platforms(arrivals, departures):
    arrivals.sort()
    departures.sort()
    i = j = 0
    count = max_count = 0
    n = len(arrivals)
    while i < n and j < n:
        if arrivals[i] < departures[j]:
            count += 1
            max_count = max(max_count, count)
            i += 1
        else:
            count -= 1
            j += 1
    return max_count

arrivals = [900, 940, 950]
departures = [910, 1200, 1120]
print(min_platforms(arrivals, departures))

A IndexError: list index out of range

B TypeError: unsupported operand type(s)

C 0

D 3

💡 Hint

Check loop conditions and index usage carefully.

The loop condition uses 'i <= n' which allows i to become equal to n, causing an index out of range error when accessing arrivals[i].

🚀 Application

expert

Calculate Minimum Platforms for Large Schedule

Given the following arrival and departure times, what is the minimum number of platforms needed?

DSA Python

arrivals = [900, 940, 950, 1100, 1500, 1800, 1830, 1845, 1900, 1930]
departures = [910, 1200, 1120, 1130, 1900, 2000, 1850, 1905, 1950, 2000]

def min_platforms(arr, dep):
    arr.sort()
    dep.sort()
    i = j = 0
    count = max_count = 0
    n = len(arr)
    while i < n and j < n:
        if arr[i] < dep[j]:
            count += 1
            max_count = max(max_count, count)
            i += 1
        else:
            count -= 1
            j += 1
    return max_count

print(min_platforms(arrivals, departures))

A 6

B 5

C 4

D 3

💡 Hint

Sort and count overlapping trains carefully.

The maximum overlap occurs around times 1830 to 1850 where 4 trains are at the station simultaneously, so minimum platforms needed is 4.

Practice

1. What does the "Minimum Number of Platforms" problem ask you to find given train arrival and departure times?

easy

A. The least number of platforms needed so no train waits

B. The total number of trains arriving in a day

C. The longest time a train stays on a platform

D. The earliest train arrival time

Solution

1. 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.

2. Step 2: Identify the key requirement

   We must find the minimum number of platforms so that all trains can be accommodated without delay.

3. Final Answer:

   The least number of platforms needed so no train waits -> Option A

4. Quick Check:

   Minimum platforms = no waiting trains [OK]

Hint: Focus on overlapping train times to find platform count [OK]

Common Mistakes:

• Confusing total trains with platforms needed
• Ignoring overlapping arrival and departure times
• Thinking about train durations instead of platform count

2. Which of the following Python code snippets correctly sorts the arrival times list named arrivals for the Minimum Number of Platforms problem?

easy

A. arrivals.sort(arrivals)

B. sort(arrivals)

C. arrivals.sort()

D. arrivals.sorted()

Solution

1. Step 1: Recall Python list sorting method

   Python lists have a method called sort() that sorts the list in place.

2. Step 2: Check each option

   arrivals.sort() uses arrivals.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.

3. Final Answer:

   arrivals.sort() -> Option C

4. Quick Check:

   List.sort() is correct method [OK]

Hint: Use list.sort() to sort lists in place [OK]

Common Mistakes:

• Using sort() as a standalone function
• Trying to call sorted() as a method
• Passing the list as argument to sort()

3. What is the minimum number of platforms needed for the following train schedule?
arrivals = [900, 940, 950, 1100, 1500, 1800]
departures = [910, 1200, 1120, 1130, 1900, 2000]

medium

A. 4

B. 2

C. 3

D. 1

Solution

1. Step 1: Sort arrival and departure times

   Arrivals sorted: [900, 940, 950, 1100, 1500, 1800]
   Departures sorted: [910, 1120, 1130, 1200, 1900, 2000]

2. 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.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Max overlapping trains = 2 [OK]

Hint: Count overlapping trains using sorted arrivals and departures [OK]

Common Mistakes:

• Not sorting times before counting
• Mixing arrival and departure pointers
• Counting platforms incorrectly at overlaps

4. Identify the error in this Python code snippet for finding minimum platforms:

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)

medium

A. The comparison should be arrivals[i] < departures[j], not <=.

B. The loop condition should use 'or' instead of 'and'.

C. The platforms variable should not be incremented.

D. The departures list should not be sorted.

Solution

1. 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.

2. Step 2: Correct the comparison

   Changing <= to < ensures that if arrival equals departure, platform count decreases first, avoiding unnecessary extra platform.

3. Final Answer:

   The comparison should be arrivals[i] < departures[j], not <=. -> Option A

4. Quick Check:

   Use < to avoid counting simultaneous arrival and departure as overlap [OK]

Hint: Use strict less than for arrival vs departure comparison [OK]

Common Mistakes:

• Using <= causes overcounting platforms
• Changing loop condition wrongly to 'or'
• Not sorting departure times

5. Given multiple trains with arrival and departure times, how would you efficiently find the minimum number of platforms needed if the times are in 24-hour format and some trains arrive and depart at the same time? For example:
arrivals = [900, 940, 950, 1100, 1500, 1800, 1800]
departures = [910, 1200, 1120, 1130, 1900, 2000, 1800]
Choose the best approach.

hard

A. Use a dictionary to map arrival times to departure times and count keys.

B. Count trains arriving at each time and sum them directly without sorting.

C. Sort arrivals only and count unique times.

D. Sort arrivals and departures, then use two pointers with strict less than comparison for arrival vs departure.

Solution

1. Step 1: Sort both arrival and departure lists

   Sorting helps to process events in chronological order, which is essential for counting overlapping trains.

2. Step 2: Use two pointers with strict comparison

   Use two pointers to traverse arrivals and departures. Use arrivals[i] < departures[j] to decide when to increase or decrease platform count, handling same-time arrivals and departures correctly.

3. Final Answer:

   Sort arrivals and departures, then use two pointers with strict less than comparison for arrival vs departure. -> Option D

4. Quick Check:

   Sorting + two pointers + < comparison = correct approach [OK]

Hint: Sort both lists and use two pointers with strict comparison [OK]

Common Mistakes:

• Ignoring sorting of departures
• Using <= causing overcount
• Trying to count without sorting or pointers