LeetCode 3236: CEO Subordinate Hierarchy Solution
Master LeetCode problem 3236 (CEO Subordinate Hierarchy), 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.
3236. CEO Subordinate Hierarchy
Problem Explanation
Explanation:
To solve this problem, we can use a directed graph to represent the relationships between the CEO and subordinates. We first build the graph based on the input data, then perform a depth-first search (DFS) starting from the CEO to calculate the total hierarchy under each subordinate. Finally, we return the hierarchy information in the required format.
- Build a directed graph from the given CEO-subordinate relationships.
- Perform a DFS starting from the CEO to calculate the total hierarchy under each subordinate.
- Return the hierarchy information in the required format.
Time Complexity: O(N) where N is the number of relationships
Space Complexity: O(N)
:
Solution Code
import java.util.*;
class Solution {
public List<String> ceoSubordinateHierarchy(List<List<String>> relationships) {
Map<String, List<String>> graph = new HashMap<>();
for (List<String> relation : relationships) {
String ceo = relation.get(0);
String subordinate = relation.get(1);
graph.putIfAbsent(ceo, new ArrayList<>());
graph.get(ceo).add(subordinate);
}
Map<String, Integer> hierarchy = new HashMap<>();
calculateHierarchy(graph, "CEO", hierarchy);
List<String> result = new ArrayList<>();
for (Map.Entry<String, Integer> entry : hierarchy.entrySet()) {
result.add(entry.getKey() + " " + entry.getValue());
}
return result;
}
private int calculateHierarchy(Map<String, List<String>> graph, String ceo, Map<String, Integer> hierarchy) {
if (!graph.containsKey(ceo)) {
hierarchy.put(ceo, 0);
return 1;
}
int totalHierarchy = 1;
for (String subordinate : graph.get(ceo)) {
totalHierarchy += calculateHierarchy(graph, subordinate, hierarchy);
}
hierarchy.put(ceo, totalHierarchy);
return totalHierarchy;
}
}Try It Yourself
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