LeetCode 1466: Reorder Routes to Make All Paths Lead to the City Zero Solution
Master LeetCode problem 1466 (Reorder Routes to Make All Paths Lead to the City Zero), a medium 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.
1466. Reorder Routes to Make All Paths Lead to the City Zero
Problem Explanation
Explanation:
To solve this problem, we can perform a depth-first search (DFS) starting from the capital city (city 0) and iteratively change the directions of the edges so that all cities connect to city 0. We need to count the number of edges that need to be changed.
- Initialize a graph to represent the connections between cities.
- Perform a DFS starting from city 0 and keep track of the edges that need to be changed.
- For each edge, if it is currently pointing towards city 0, we mark it as an edge that needs to be changed.
- Recursively visit neighboring cities and update the count of edges that need to be changed.
- Finally, return the count of edges that need to be changed.
Time Complexity: O(n) where n is the number of cities. Space Complexity: O(n) for the graph and recursion stack.
:
Solution Code
class Solution {
public int minReorder(int n, int[][] connections) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i < n; i++) {
graph.add(new ArrayList<>());
}
for (int[] connection : connections) {
graph.get(connection[0]).add(connection[1]);
graph.get(connection[1]).add(-connection[0]); // Marking reverse edge
}
return dfs(0, -1, graph);
}
private int dfs(int current, int parent, List<List<Integer>> graph) {
int count = 0;
for (int neighbor : graph.get(current)) {
if (neighbor == parent) continue;
count += neighbor > 0 ? 1 : 0; // Check if edge needs to be changed
count += dfs(Math.abs(neighbor), current, graph);
}
return count;
}
}Try It Yourself
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