Skyway Signal Cluster

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This problem feels like a little puzzle you can solve one step at a time. In Skyway Signal Cluster, you are trying to work toward the right number by following one clear idea.

Find skyway signal cluster coverage A good way to think about it is to first understand what goes in, then what rule you must follow, and finally what shape the answer should have.

For example, if the input is n = 4, connectors = [], start = 1, the answer is 1. With no connectors, the broadcast remains at the starting deck. Another example is n = 5, connectors = [[0,1],[1,2],[2,3],[3,4]], start = 2, which gives 5. Every deck is connected through the chain, so the broadcast covers all decks.

This is a friendly practice problem, but it still rewards careful reading. The key is understanding the rule clearly and then applying it carefully.

Example Input & Output

Example 1

Input

n = 4, connectors = [], start = 1

Output

Explanation

With no connectors, the broadcast remains at the starting deck.

Example 2

Input

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

Output

Explanation

Every deck is connected through the chain, so the broadcast covers all decks.

Example 3

Input

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

Output

Explanation

The cluster includes decks 2, 3, and 4; the other decks are unreachable.

Algorithm Flow

Recommendation Algorithm Flow for Skyway Signal Cluster

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 List<List<String>> group_signal_patterns(String[] patterns) {
        Map<String, List<String>> res = new HashMap<>();
        for (String p : patterns) {
            char[] chars = p.toCharArray();
            Arrays.sort(chars);
            String key = new String(chars);
            res.computeIfAbsent(key, x -> new ArrayList<>()).add(p);
        }
        List<List<String>> values = new ArrayList<>(res.values());
        values.sort(Comparator.comparingInt(List::size));
        return values;
    }
}

javascript

function group_signal_patterns(patterns) {
    const res = {};
    for (let p of patterns) {
        const key = p.split('').sort().join('');
        if (!res[key]) res[key] = [];
        res[key].push(p);
    }
    return Object.values(res).sort((a, b) => a.length - b.length);
}

php

function group_signal_patterns($patterns) {
    $res = [];
    foreach ($patterns as $p) {
        $chars = str_split($p);
        sort($chars);
        $key = implode('', $chars);
        $res[$key][] = $p;
    }
    $values = array_values($res);
    usort($values, function($a, $b) { return count($a) - count($b); });
    return $values;
}

python

def group_signal_patterns(patterns: list) -> list:
    res = {}
    for p in patterns:
        key = ''.join(sorted(p))
        if key not in res: res[key] = []
        res[key].append(p)
    return sorted(list(res.values()), key=len)

rust

use std::collections::HashMap;
pub fn group_signal_patterns(patterns: Vec<String>) -> Vec<Vec<String>> {
    let mut res: HashMap<String, Vec<String>> = HashMap::new();
    for p in patterns {
        let mut chars: Vec<char> = p.chars().collect();
        chars.sort();
        let key: String = chars.into_iter().collect();
        res.entry(key).or_insert(Vec::new()).push(p);
    }
    let mut values: Vec<Vec<String>> = res.into_values().collect();
    values.sort_by_key(|v| v.len());
    values
}

typescript

function group_signal_patterns(patterns: string[]): string[][] {
    const res: {[key: string]: string[]} = {};
    for (let p of patterns) {
        const key = p.split('').sort().join('');
        if (!res[key]) res[key] = [];
        res[key].push(p);
    }
    return Object.values(res).sort((a, b) => a.length - b.length);
}

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Skyway Signal Cluster

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.

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