Minimum Number of Platforms in DSA Javascript - Build from Scratch

Minimum Number of Platforms

📖 Scenario: Imagine a small train station where trains arrive and depart at different times. We want to find out the minimum number of platforms needed so that no train has to wait.

🎯 Goal: You will write a program to find the minimum number of platforms required for the trains based on their arrival and departure times.

📋 What You'll Learn

Create two arrays called arrivals and departures with exact train times

Create a variable called platforms_needed to count platforms

Write a loop to find the minimum number of platforms needed

Print the final number of platforms needed

💡 Why This Matters

🌍 Real World

Train stations and airports use this logic to manage platforms or gates so that vehicles do not wait.

💼 Career

Understanding scheduling and resource allocation is important in software engineering, especially in systems design and operations.

1

Create train arrival and departure arrays

Create two arrays called arrivals and departures with these exact values: arrivals = [900, 940, 950, 1100, 1500, 1800] and departures = [910, 1200, 1120, 1130, 1900, 2000].

DSA Javascript

# Your code here

Hint

Use const to create two arrays named arrivals and departures with the exact times given.

2

Initialize platform counters

Create two variables called platforms_needed and max_platforms and set both to 1. Also create two variables i and j and set both to 1 and 0 respectively.

DSA Javascript

const arrivals = [900, 940, 950, 1100, 1500, 1800];
const departures = [910, 1200, 1120, 1130, 1900, 2000];
# Your code here

Hint

Initialize platforms_needed and max_platforms to 1, and pointers i and j to 1 and 0.

3

Calculate minimum platforms needed

Sort the arrivals and departures arrays. Then use a while loop with condition i < arrivals.length && j < departures.length. Inside the loop, if arrivals[i] <= departures[j], increment platforms_needed and i. Else, decrement platforms_needed and increment j. Update max_platforms to the maximum of itself and platforms_needed inside the loop.

DSA Javascript

const arrivals = [900, 940, 950, 1100, 1500, 1800];
const departures = [910, 1200, 1120, 1130, 1900, 2000];

let platforms_needed = 1;
let max_platforms = 1;
let i = 1;
let j = 0;

// Your code here

Hint

Sort both arrays first. Then use a while loop to compare arrival and departure times. Increase or decrease platforms_needed accordingly and update max_platforms.

4

Print the minimum number of platforms needed

Write console.log(max_platforms) to print the minimum number of platforms needed.

DSA Javascript

const arrivals = [900, 940, 950, 1100, 1500, 1800];
const departures = [910, 1200, 1120, 1130, 1900, 2000];

let platforms_needed = 1;
let max_platforms = 1;
let i = 1;
let j = 0;

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

while (i  max_platforms) {
            max_platforms = platforms_needed;
        }
    } else {
        platforms_needed--;
        j++;
    }
}

// Your code here

Hint

Use console.log(max_platforms) to show the result.

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