Best Team With No Conflicts - Leetcode Solution

Problem:

Given a list of employees and their relationships, find the best team that can be formed with no conflicts. A team is considered "conflict-free" if there are no two employees in the team that have a relationship with each other (either directly or indirectly through a chain of relationships).

Solution:

The problem can be solved using dynamic programming. We can create a binary tree where each node represents an employee and its children are the employees it has a relationship with. We can then use depth-first search (DFS) to traverse the tree and compute the maximum number of employees that can form a conflict-free team with the node as the root.

We can define a function f(node) that returns the maximum number of employees that can form a conflict-free team with the node as the root. The function can be defined recursively as follows:

• If the node has no children, f(node) = 1.
• If the node has children, let the children be c1, c2, ..., cn. We can either include the node in the team or exclude it. If we include the node, we cannot include any of its children. If we exclude the node, we can include any of its children. Therefore, f(node) = 1 + sum(f(ci) where ci is a grandchild of node) if we include the node, and f(ci) for each i if we exclude the node.

We can then use DFS to compute the values of f(node) for each node in the tree. The maximum value of f(node) over all nodes in the tree is the maximum number of employees that can form a conflict-free team.

Pseudocode:

def f(node): if node has no children: return 1 include_node = 1 + sum(f(grandchild) for child in node.children for grandchild in child.children) exclude_node = sum(f(child) for child in node.children) return max(include_node, exclude_node)

max_team_size = max(f(node) for node in tree.nodes)

Time Complexity:

The time complexity of this solution is O(n^2), where n is the number of employees. This is because we have to compute f() for each node in the tree, and each call to f() may involve a sum over the grandchildren of the node. The space complexity is also O(n^2) because we need to store the binary tree as a list of lists, where each sublist contains the children of a node.

Overall, this solution is efficient for small or moderately sized inputs, but may be too slow for very large inputs. In practice, it may be necessary to optimize the algorithm or use a different approach for very large inputs.

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