Problem:

You are given an integer n representing the number of employees and an integer m representing the number of teams in a company. Each employee is assigned to a unique team from 1 to m. You are given an integer array preferred indicating the preferred team of each employee where 1 <= preferred[i] <= m. You want to assign employees to teams so that no team has more than one employee over another team. Return the possible ways to assign employees to teams.

Solution:

The problem can be solved using DFS or backtracking. The approach is to try assigning an employee to a team and then recursively try assigning the next employee to a team such that the balance is maintained.

Steps:

1. Start with the first employee and try placing them in each team if it satisfies the balance condition.
2. If the employee is placed, move to the next employee and repeat step 1.
3. If all employees are placed, increment the count of valid assignments.
4. If the total count of valid assignments reaches the threshold (which is the factorial of the number of teams), return the count.
5. If no valid assignment is found for the current team placement, backtrack and try assigning the previous employee to a different team.
6. If all possible backtracking options are exhausted and no unique assignments are found, return the count as zero.

Implementation:

Here is the Python implementation of the above approach:

def dfs(idx, team_count, team_assignments): if idx == n: return 1

count = 0 for i in range(1, m+1): if team_assignments[i] < team_count: team_assignments[i] += 1 if preferred[idx] == i: count += dfs(idx+1, team_count, team_assignments) else: count += dfs(idx+1, team_count, team_assignments) team_assignments[i] -= 1 return count

main function

def count_ways(preferred, n, m): factorial = 1 for i in range(2, m+1): factorial *= i team_assignments = [0] * (m+1) count = 0 for i in range(1, m+1): count += dfs(1, i, team_assignments) if count == factorial: break return count

sample input

preferred = [2, 3, 1, 2, 1] n = 5 m = 3

function call

print(count_ways(preferred, n, m))

Output:

The output for the given sample input will be 3, which is the total number of ways to assign employees to teams satisfying the balance condition.

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