LeetCode 1270: All People Report to the Given Manager Solution

Master LeetCode problem 1270 (All People Report to the Given Manager), 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.

1270. All People Report to the Given Manager

Medium

Database

Problem Explanation

Explanation:

To solve this problem, we can start by creating a graph where the key is the manager and the value is a list of employees reporting to that manager. Then, we can perform a depth-first search (DFS) starting from the given manager to find all employees reporting to this manager.

Algorithm:

Create a graph where the key is the manager and the value is a list of employees reporting to that manager.
Perform a depth-first search (DFS) starting from the given manager to find all employees reporting to this manager.

Time Complexity:

The time complexity of this solution is O(V + E), where V is the number of vertices (managers) and E is the number of edges (relationships between managers and employees).

Space Complexity:

The space complexity of this solution is O(V + E) for storing the graph and the visited set.

Solution Code

class Solution {
    public List<Integer> countSubordinates(int n, int[][] manager, int[] informTime) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int i = 0; i < n; i++) {
            graph.putIfAbsent(manager[i][0], new ArrayList<>());
            graph.get(manager[i][0]).add(i);
        }
        return dfs(graph, informTime, 0);
    }
    
    private List<Integer> dfs(Map<Integer, List<Integer>> graph, int[] informTime, int manager) {
        List<Integer> subordinates = new ArrayList<>();
        if (graph.containsKey(manager)) {
            for (int subordinate : graph.get(manager)) {
                List<Integer> subordinatesOfSubordinate = dfs(graph, informTime, subordinate);
                subordinates.add(informTime[manager] + Collections.max(subordinatesOfSubordinate));
            }
        }
        return subordinates;
    }
}

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