Minimum Number of Platforms in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the arrival times array.

DSA Javascript

arrivals.sort((a, b) => a [1] b);

Drag options to blanks, or click blank then click option'

A+

B-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' in the comparator.

Using '*' which does not work for sorting.

The sort function needs a comparator that returns a positive number when a > b, so 'a + b' is incorrect. The correct comparator is 'a - b' to sort ascending.

2fill in blank

medium

Complete the code to increment the platform count when a train arrives before the earliest departure.

DSA Javascript

if (arrivals[i] < departures[j]) {
  platforms_needed [1];
  i++;
}

Drag options to blanks, or click blank then click option'

A++

B-=

C+=

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using '--' which decreases the count.

Using '-=' which subtracts.

When a train arrives before the earliest departure, we need one more platform, so we increment platforms_needed by 1 using '++'.

3fill in blank

hard

Fix the error in updating the maximum platforms needed.

DSA Javascript

max_platforms = Math.max(max_platforms, [1]);

Drag options to blanks, or click blank then click option'

Aplatforms_needed

Bplatforms_needed++

Cplatforms_needed--

Dplatforms_needed + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using platforms_needed++ which changes the value unexpectedly.

Using platforms_needed + 1 which is incorrect here.

We should compare max_platforms with the current platforms_needed, not with platforms_needed++ or other expressions that change the value.

4fill in blank

hard

Fill both blanks to correctly move pointers when a train departs before the next arrival.

DSA Javascript

else {
  platforms_needed [1];
  j [2];
}

Drag options to blanks, or click blank then click option'

A--

B++

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Increasing platforms_needed instead of decreasing.

Moving the wrong pointer.

When a train departs before the next arrival, we reduce platforms_needed by 1 using '--' and move the departure pointer forward using '++'.

5fill in blank

hard

Fill all three blanks to complete the function that calculates minimum platforms.

DSA Javascript

function minPlatforms(arrivals, departures) {
  arrivals.sort((a, b) => a [1] b);
  departures.sort((a, b) => a [2] b);
  let platforms_needed = 0, max_platforms = 0;
  let i = 0, j = 0;
  while (i < arrivals.length && j < departures.length) {
    if (arrivals[i] < departures[j]) {
      platforms_needed [3];
      i++;
    } else {
      platforms_needed--;
      j++;
    }
    if (platforms_needed > max_platforms) max_platforms = platforms_needed;
  }
  return max_platforms;
}

Drag options to blanks, or click blank then click option'

A-

B++

C+

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' or '*' in sort comparator.

Not incrementing platforms_needed correctly.

We sort both arrays ascending using 'a - b'. We increment platforms_needed by 1 when a train arrives before the earliest departure using '++'.

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