Museum Corridor Sweep

Published atDate not available

Array Easy 0 views

Like0

You can think of this as a small game with a very specific goal. In Museum Corridor Sweep, you are trying to work toward the right number by following one clear idea.

This problem hides a best streak, chain, or longest piece inside a larger list. You are not always taking everything. Instead, you are searching for the strongest run that still follows the rule. A choice that looks good right now may not always lead to the best final result, so careful thinking matters.

For example, if the input is n = 6, corridors = [[0,1],[1,2],[0,2],[3,4]], start = 1, the answer is 3. Rooms 0, 1, and 2 form one connected section; rooms 3 and 4 are separate. Another example is n = 4, corridors = [[0,1],[2,3]], start = 2, which gives 2. The sweep reaches rooms 2 and 3; the other rooms remain unchecked.

This is a friendly practice problem, but it still rewards careful reading. The key is to look for the best hidden streak, not just the first one that seems okay.

Example Input & Output

Example 1

Input

n = 6, corridors = [[0,1],[1,2],[0,2],[3,4]], start = 1

Output

Explanation

Rooms 0, 1, and 2 form one connected section; rooms 3 and 4 are separate.

Example 2

Input

n = 4, corridors = [[0,1],[2,3]], start = 2

Output

Explanation

The sweep reaches rooms 2 and 3; the other rooms remain unchecked.

Example 3

Input

n = 5, corridors = [[0,1],[1,2],[2,3],[3,4]], start = 0

Output

Explanation

The sweep moves through every room in sequence without revisiting any room.

Algorithm Flow

Recommendation Algorithm Flow for Museum Corridor Sweep

Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

java

import java.util.*;
class Solution {
    public int calculate_min_workers(int[][] workers, int length) {
        Arrays.sort(workers, (a, b) -> a[0] - b[0]);
        int res = 0, currentFar = 0, i = 0;
        while (currentFar < length) {
            int nextFar = currentFar;
            while (i < workers.length && workers[i][0] <= currentFar) {
                nextFar = Math.max(nextFar, workers[i][1]);
                i++;
            }
            if (nextFar == currentFar) return -1;
            res++;
            currentFar = nextFar;
        }
        return res;
    }
}

javascript

function calculate_min_workers(workers, length) {
    workers.sort((a, b) => a[0] - b[0]);
    let res = 0, currentFar = 0, i = 0;
    while (currentFar < length) {
        let nextFar = currentFar;
        while (i < workers.length && workers[i][0] <= currentFar) {
            nextFar = Math.max(nextFar, workers[i][1]);
            i++;
        }
        if (nextFar === currentFar) return -1;
        res++;
        currentFar = nextFar;
    }
    return res;
}

php

function calculate_min_workers($workers, $length) {
    usort($workers, function($a, $b) { return $a[0] - $b[0]; });
    $res = 0; $currentFar = 0; $i = 0;
    while ($currentFar < $length) {
        $nextFar = $currentFar;
        while ($i < count($workers) && $workers[$i][0] <= $currentFar) {
            $nextFar = max($nextFar, $workers[$i][1]);
            $i++;
        }
        if ($nextFar == $currentFar) return -1;
        $res++;
        $currentFar = $nextFar;
    }
    return $res;
}

python

def calculate_min_workers(workers: list, length: int) -> int:
    workers.sort()
    res = 0
    current_far = 0
    i = 0
    while current_far < length:
        next_far = current_far
        while i < len(workers) and workers[i][0] <= current_far:
            next_far = max(next_far, workers[i][1])
            i += 1
        if next_far == current_far: return -1
        res += 1
        current_far = next_far
    return res

rust

pub fn calculate_min_workers(workers: Vec<Vec<i32>>, length: i32) -> i32 {
    let mut workers = workers;
    workers.sort_by_key(|w| w[0]);
    let mut res = 0;
    let mut current_far = 0;
    let mut i = 0;
    while current_far < length {
        let mut next_far = current_far;
        while i < workers.len() && workers[i][0] <= current_far {
            next_far = std::cmp::max(next_far, workers[i][1]);
            i += 1;
        }
        if next_far == current_far { return -1; }
        res += 1;
        current_far = next_far;
    }
    res
}

typescript

function calculate_min_workers(workers: number[][], length: number): number {
    workers.sort((a, b) => a[0] - b[0]);
    let res = 0, currentFar = 0, i = 0;
    while (currentFar < length) {
        let nextFar = currentFar;
        while (i < workers.length && workers[i][0] <= currentFar) {
            nextFar = Math.max(nextFar, workers[i][1]);
            i++;
        }
        if (nextFar === currentFar) return -1;
        res++;
        currentFar = nextFar;
    }
    return res;
}

Related Array Challenges

Comments (0)

Join the Discussion

Share your thoughts, ask questions, or help others with this Challenge.

Museum Corridor Sweep

Example Input & Output

Algorithm Flow

Best AnswersThese are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

Solutions Coming Soon

Related Array Challenges

City Signal Span

Celestial Skyline Layout

Aurora Diary Synthesis

Ember Glyph Mosaic

Tidal Melody Chorus

Lantern Pattern Paths

Comments (0)

Join the Discussion

Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.