LeetCode 2858: Minimum Edge Reversals So Every Node Is Reachable Solution

Master LeetCode problem 2858 (Minimum Edge Reversals So Every Node Is Reachable), 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.

2858. Minimum Edge Reversals So Every Node Is Reachable

Hard

Dynamic Programming Depth-First Search Breadth-First Search Graph

Problem Explanation

Explanation:

To solve this problem, we can use a modified version of BFS traversal to find the minimum number of edge reversals required for each node to reach every other node in the graph. We will start from each node and explore all possible paths by considering edge reversals when necessary. The idea is to keep track of the minimum number of edge reversals needed for each node. We will use a queue to perform the BFS traversal.

Initialize an adjacency list to represent the graph.
For each node, perform a BFS traversal and keep track of the number of edge reversals required to reach every other node.
Update the answer array with the minimum number of edge reversals for each node.

Time Complexity: O(n^2) where n is the number of nodes Space Complexity: O(n) for the adjacency list and answer array

Solution Code

class Solution {
    public int[] minReorder(int n, int[][] edges) {
        List<List<int[]>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        
        for (int[] edge : edges) {
            graph.get(edge[0]).add(new int[]{edge[1], 1}); // Reversal needed
            graph.get(edge[1]).add(new int[]{edge[0], 0}); // No reversal needed
        }
        
        int[] answer = new int[n];
        Queue<int[]> queue = new LinkedList<>();
        boolean[] visited = new boolean[n];
        
        for (int i = 0; i < n; i++) {
            queue.offer(new int[]{i, 0});
            visited[i] = true;
            while (!queue.isEmpty()) {
                int[] curr = queue.poll();
                int node = curr[0];
                int rev = curr[1];
                for (int[] next : graph.get(node)) {
                    int nextNode = next[0];
                    int nextRev = next[1];
                    if (!visited[nextNode]) {
                        visited[nextNode] = true;
                        answer[i] += nextRev == 1 ? 1 : 0;
                        queue.offer(new int[]{nextNode, nextRev});
                    }
                }
            }
            Arrays.fill(visited, false);
        }
        
        return answer;
    }
}

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