What if you could instantly know how many platforms a busy station needs without guessing?
Why Minimum Number of Platforms in DSA Rust?
Imagine a busy train station where many trains arrive and depart at different times. You want to know how many platforms are needed so that no train has to wait for another to leave.
Trying to count platforms manually by checking each train's arrival and departure times is slow and confusing. It's easy to miss overlaps or double count, leading to wrong answers and delays.
The Minimum Number of Platforms concept helps by sorting arrival and departure times and efficiently counting overlaps. This way, you quickly find the smallest number of platforms needed without confusion.
for each train: check all other trains for overlap count platforms needed
sort arrivals and departures
use two pointers to count overlaps
find max platforms neededThis concept enables smooth train scheduling by ensuring enough platforms are available to avoid waiting and delays.
Railway stations use this to plan platform numbers so trains arrive and leave on time without crowding or waiting.
Manual counting is slow and error-prone.
Sorting times and counting overlaps is efficient.
Helps plan resources to avoid delays.
Practice
Solution
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.Step 2: Identify the key requirement
We must find the minimum number of platforms so that all trains can be accommodated without delay.Final Answer:
The least number of platforms needed so no train waits -> Option BQuick Check:
Minimum platforms = no waiting trains [OK]
- Confusing total trains with platforms needed
- Ignoring overlapping arrival and departure times
- Thinking about train speed or travel time
Solution
Step 1: Identify the need for ordering
We need to quickly access the earliest departure time to compare with arrivals.Step 2: Choose a data structure that supports this
A BinaryHeap (priority queue) efficiently gives the smallest or largest element, perfect for earliest departure.Final Answer:
BinaryHeap (priority queue) -> Option AQuick Check:
Priority queue = earliest departure access [OK]
- Using HashMap which doesn't keep order
- Using LinkedList which is inefficient for min extraction
- Using VecDeque which is FIFO, not sorted
let arrivals = vec![900, 940, 950, 1100, 1500, 1800];
let departures = vec![910, 1200, 1120, 1130, 1900, 2000];
// After sorting both vectors
// Calculate minimum platforms needed
// Output the result
// Assume correct implementation of algorithm
println!("{}", min_platforms(arrivals, departures));Solution
Step 1: Sort arrival and departure times
Arrivals sorted: [900, 940, 950, 1100, 1500, 1800] Departures sorted: [910, 1120, 1130, 1200, 1900, 2000]Step 2: Count overlapping trains
At 1100 arrival, trains at station are 3 (940-1200, 950-1120, 1100-1130 overlap). So minimum platforms = 3.Final Answer:
3 -> Option AQuick Check:
Max overlap = 3 platforms [OK]
- Not sorting times before processing
- Counting total trains instead of overlaps
- Mixing arrival and departure times
fn min_platforms(arrivals: Vec<i32>, departures: Vec<i32>) -> i32 {
let mut arr = arrivals.clone();
let mut dep = departures.clone();
arr.sort();
dep.sort();
let mut platform_needed = 1;
let mut max_platforms = 1;
let mut i = 1;
let mut j = 0;
while i < arr.len() && j < dep.len() {
if arr[i] <= dep[j] {
platform_needed += 1;
i += 1;
} else {
platform_needed -= 1;
j += 1;
}
if platform_needed > max_platforms {
max_platforms = platform_needed;
}
}
max_platforms
}Solution
Step 1: Check loop initialization
Variable i starts at 1, skipping the first arrival time comparison which can cause incorrect platform count.Step 2: Understand impact of skipping first arrival
Missing the first arrival means initial platform count may be wrong, leading to incorrect max platforms.Final Answer:
Starting i from 1 skips first arrival comparison -> Option CQuick Check:
Indexing from 1 skips first train [OK]
- Ignoring index start positions
- Thinking sorting is unnecessary
- Confusing clone() with error cause
[100, 140, 150, 200, 215, 400] and departure times [110, 300, 220, 230, 315, 600], what is the minimum number of platforms needed? Explain why this number is required.Solution
Step 1: Sort arrival and departure times
Arrivals sorted: [100, 140, 150, 200, 215, 400] Departures sorted: [110, 220, 230, 300, 315, 600]Step 2: Count overlapping trains step-by-step
At 215 arrival, trains at station: 140-300, 150-220, 200-230, 215-315 (4) overlap. Maximum overlap is 4, so 4 platforms needed.Final Answer:
4 -> Option DQuick Check:
Max overlap = 4 platforms [OK]
- Not sorting times before counting
- Miscounting overlaps at arrival times
- Assuming platforms equal total trains
Ready to prove it?
Pass the test and this pattern moves unknown
