LeetCode 2153: The Number of Passengers in Each Bus II Solution

Master LeetCode problem 2153 (The Number of Passengers in Each Bus II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2153. The Number of Passengers in Each Bus II

Hard

Database

Problem Explanation

Explanation:

To solve this problem, we can use a binary search approach. We can define the search space as the range of possible values for the number of passengers in each bus. Then, we can perform a binary search within this search space to find the minimum number of passengers such that all passengers can be transported within m buses and t time units.

Here are the steps:

Define the search space for the number of passengers in each bus as [1, sum(passenger times)].
Perform a binary search within this search space.
For each mid value, simulate the process of transporting passengers in m buses and check if it can be done within t time units.
Adjust the search space based on the result of the simulation.
Repeat the binary search until we find the minimum number of passengers.

Time complexity: O(n log(sum(passenger times))), where n is the number of passengers. Space complexity: O(1)

: :

Solution Code

class Solution {
    public int numBusesToDestination(int[][] routes, int source, int target) {
        int n = routes.length;
        Map<Integer, Set<Integer>> stopToBuses = new HashMap<>();
        
        for (int bus = 0; bus < n; bus++) {
            for (int stop : routes[bus]) {
                if (!stopToBuses.containsKey(stop)) {
                    stopToBuses.put(stop, new HashSet<>());
                }
                stopToBuses.get(stop).add(bus);
            }
        }
        
        Set<Integer> visitedBuses = new HashSet<>();
        Set<Integer> visitedStops = new HashSet<>();
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(source);
        int level = 0;
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            level++;
            for (int i = 0; i < size; i++) {
                int currStop = queue.poll();
                if (currStop == target) {
                    return level - 1;
                }
                for (int bus : stopToBuses.get(currStop)) {
                    if (visitedBuses.contains(bus)) {
                        continue;
                    }
                    visitedBuses.add(bus);
                    for (int nextStop : routes[bus]) {
                        if (!visitedStops.contains(nextStop)) {
                            visitedStops.add(nextStop);
                            queue.offer(nextStop);
                        }
                    }
                }
            }
        }
        
        return -1;
    }
}

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