Minimum Number of Platforms in DSA Javascript - Time & Space Complexity

Time Complexity: Minimum Number of Platforms

O(n log n)

Understanding Time Complexity

We want to find out how the time needed to find the minimum number of platforms changes as the number of trains increases.

Specifically, we ask: How does the running time grow when we have more arrival and departure times?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function findMinPlatforms(arrivals, departures) {
  arrivals.sort((a, b) => a - b);
  departures.sort((a, b) => a - b);

  let platformNeeded = 0, maxPlatforms = 0;
  let i = 0, j = 0;

  while (i < arrivals.length && j < departures.length) {
    if (arrivals[i] <= departures[j]) {
      platformNeeded++;
      i++;
      if (platformNeeded > maxPlatforms) maxPlatforms = platformNeeded;
    } else {
      platformNeeded--;
      j++;
    }
  }

  return maxPlatforms;
}
    

This code finds the minimum number of platforms needed so that no train waits, by sorting arrival and departure times and then comparing them.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the arrival and departure arrays.
• How many times: Each sort runs once over all trains, so twice total.
• Secondary operation: The while loop traverses both arrays once, comparing elements.
• How many times: Up to 2n steps combined, where n is the number of trains.

How Execution Grows With Input

Sorting takes more time as the number of trains grows, and the loop also grows linearly with trains.

Input Size (n)	Approx. Operations
10	About 10 log 10 + 20 steps
100	About 100 log 100 + 200 steps
1000	About 1000 log 1000 + 2000 steps

Pattern observation: Sorting dominates and grows faster than the linear loop, so total time grows a bit faster than just doubling when input doubles.

Final Time Complexity

Time Complexity: O(n log n)

This means as the number of trains grows, the time to find the minimum platforms grows a bit faster than just the number of trains, mainly because of sorting.

Common Mistake

[X] Wrong: "The while loop is the slowest part, so the time complexity is O(n)."

[OK] Correct: Sorting the arrays takes more time than the loop, so the overall time depends on sorting, not just the loop.

Interview Connect

Understanding this time complexity helps you explain how sorting and linear scans work together in real problems, a skill useful in many coding challenges.

Self-Check

"What if the arrival and departure arrays were already sorted? How would the time complexity change?"

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 goal

   The problem asks for the minimum platforms so that no train waits due to platform unavailability.

2. Step 2: Clarify what is being counted

   We count platforms needed simultaneously for overlapping train times, not total trains or times.

3. Final Answer:

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

4. Quick Check:

   Minimum platforms = least needed to avoid waiting [OK]

Hint: Count overlapping trains to find minimum platforms needed [OK]

Common Mistakes:

• Confusing total trains with platforms needed
• Thinking about train durations instead of overlaps
• Ignoring simultaneous arrivals and departures

2. Which of the following JavaScript code snippets correctly sorts an array of train arrival times stored as numbers?

easy

A. arrivals.sort()

B. arrivals.sort((a, b) => a - b)

C. arrivals.sort((a, b) => b - a)

D. arrivals.sort((a, b) => a + b)

Solution

1. Step 1: Understand JavaScript sort behavior

   By default, sort() converts elements to strings and sorts lexicographically, which is incorrect for numbers.

2. Step 2: Use numeric comparison function

   Using (a, b) => a - b sorts numbers in ascending order correctly.

3. Final Answer:

   arrivals.sort((a, b) => a - b) -> Option B

4. Quick Check:

   Numeric sort uses (a, b) => a - b [OK]

Hint: Use numeric compare (a, b) => a - b to sort numbers [OK]

Common Mistakes:

• Using sort() without compare function for numbers
• Using incorrect compare functions like a + b
• Sorting in descending order when ascending needed

3. What is the minimum number of platforms needed for trains with arrival times [900, 940, 950, 1100, 1500, 1800] and departure times [910, 1200, 1120, 1130, 1900, 2000]?

medium

A. 4

B. 2

C. 1

D. 3

Solution

1. Step 1: Sort arrival and departure times

   Arrivals: 900, 940, 950, 1100, 1500, 1800; Departures: 910, 1120, 1130, 1200, 1900, 2000.

2. Step 2: Count platforms needed by comparing arrivals and departures

   At 900 arrival, platform count=1; 940 arrival before 910 departure? No, so platform count=1; 950 arrival before 1120 departure? Yes, platform count=2; 1100 arrival before 1120 departure? Yes, platform count=3; Then departures reduce count as trains leave.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Max overlapping trains = 3 platforms needed [OK]

Hint: Count overlaps by comparing sorted arrivals and departures [OK]

Common Mistakes:

• Not sorting times before counting
• Counting total trains instead of overlaps
• Mixing arrival and departure comparisons

4. Identify the error in this JavaScript snippet for calculating minimum platforms:

let i = 0, j = 0, platforms = 0, maxPlatforms = 0;
arrivals.sort();
departures.sort();
while (i < arrivals.length && j < departures.length) {
  if (arrivals[i] <= departures[j]) {
    platforms++;
    i++;
    if (platforms > maxPlatforms) maxPlatforms = platforms;
  } else {
    platforms--;
    j++;
  }
}
return maxPlatforms;

medium

A. Using <= instead of < in if condition

B. Incrementing i and j in wrong order

C. Sorting arrays without numeric compare function

D. Not initializing maxPlatforms to arrivals.length

Solution

1. Step 1: Check sorting method

   arrivals.sort() and departures.sort() sort lexicographically by default, causing wrong order for numbers.

2. Step 2: Correct sorting for numeric arrays

   Use arrivals.sort((a,b) => a - b) and departures.sort((a,b) => a - b) to sort numbers properly.

3. Final Answer:

   Sorting arrays without numeric compare function -> Option C

4. Quick Check:

   Numeric sort needs compare function [OK]

Hint: Always use numeric compare in sort for numbers [OK]

Common Mistakes:

• Sorting numbers as strings causing wrong order
• Confusing comparison operators in if condition
• Incorrect loop increments

5. Given arrival times [1000, 1010, 1025, 1030, 1040] and departure times [1020, 1035, 1045, 1050, 1100], what is the minimum number of platforms needed if trains can arrive and depart at the same time without conflict?

hard

A. 3

B. 2

C. 4

D. 1

Solution

1. Step 1: Sort arrival and departure times

   Arrivals: 1000, 1010, 1025, 1030, 1040; Departures: 1020, 1035, 1045, 1050, 1100.

2. Step 2: Count platforms considering arrival == departure means no conflict

   Compare arrivals and departures with condition arrivals[i] < departures[j] (strict less than). Counting overlaps: at 1000 arrival platform=1; 1010 arrival platform=2; 1025 arrival platform=3; 1020 departure reduces platform to 2; 1030 arrival platform=3; 1035 departure reduces to 2; 1040 arrival platform=3; then departures reduce count.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Strict less than counts simultaneous arrival/departure correctly [OK]

Hint: Use strict < to handle same-time arrival and departure [OK]

Common Mistakes:

• Using <= causing extra platform count
• Not sorting times before counting
• Ignoring simultaneous arrival and departure rule