Minimum Number of Platforms in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the arrival times array.

DSA Typescript

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

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A+

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' causes incorrect sorting.

Using '*' or '/' in compare function causes runtime errors.

To sort numbers in ascending order, subtract b from a in the compare function.

2fill in blank

medium

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

DSA Typescript

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

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A<

B>=

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' causes wrong platform count.

Using '<=' includes equal times incorrectly.

If arrival time is less than departure time, we need a new platform.

3fill in blank

hard

Fix the error in updating the maximum platforms needed.

DSA Typescript

maxPlatforms = Math.[1](maxPlatforms, platforms);

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Amin

Bmax

Cabs

Dround

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.min causes incorrect minimum platform count.

Using Math.abs or Math.round is unrelated to max calculation.

We use Math.max to keep track of the highest number of platforms needed.

4fill in blank

hard

Fill both blanks to correctly move pointers when a train departs.

DSA Typescript

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

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A--

B++

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Increasing platforms on departure causes errors.

Moving pointers in wrong direction breaks logic.

When a train departs, decrease platforms and move departure pointer forward.

5fill in blank

hard

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

DSA Typescript

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

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

A-

B<

C>

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' in sort causes string concatenation errors.

Using '>' in if condition reverses logic.

Sort both arrays ascending using '-', and compare if arrival < departure to increment platforms.