LeetCode 3057: Employees Project Allocation Solution

Master LeetCode problem 3057 (Employees Project Allocation), 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.

3057. Employees Project Allocation

Hard

Database

Problem Explanation

Explanation:

To solve this problem, we can use a disjoint set union (DSU) data structure. We will create a DSU representing the employees and project allocations. Initially, each employee will be in their own set. As we process the project allocations, we will merge the sets of employees working on the same project. Finally, we will count the number of unique sets in the DSU to determine the number of distinct project teams.

Algorithm:

Initialize a DSU with the number of employees as the total number of elements.
Iterate through the project allocations and merge the sets of employees for each project.
Count the number of distinct sets after processing all project allocations.

Time Complexity: O(N + Mα(N)), where N is the number of employees, M is the number of project allocations, and α(N) is the inverse Ackermann function (practically constant).

Space Complexity: O(N), where N is the number of employees.

Solution Code

class DSU {
    int[] parent;
    int[] rank;

    public DSU(int n) {
        parent = new int[n];
        rank = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
    }

    public int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }

    public void union(int x, int y) {
        int px = find(x);
        int py = find(y);
        if (px == py) return;
        if (rank[px] < rank[py]) {
            parent[px] = py;
        } else if (rank[px] > rank[py]) {
            parent[py] = px;
        } else {
            parent[py] = px;
            rank[px]++;
        }
    }
}

public int countTeams(int n, int[][] allocations) {
    DSU dsu = new DSU(n);
    for (int[] allocation : allocations) {
        dsu.union(allocation[0], allocation[1]);
    }
    Set<Integer> teams = new HashSet<>();
    for (int i = 0; i < n; i++) {
        teams.add(dsu.find(i));
    }
    return teams.size();
}

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