Cimpress Solutions

hiring_cost.py

def logic(N, M, F, B):
    # logic here
    frontend = {}
    backend = {}
    for i in range(N + M):
        frontend[i] = F[i]
        backend[i] = B[i]

    frontend = dict(sorted(frontend.items(), key=lambda x: x[1]))
    backend = dict(sorted(backend.items(), key=lambda x: x[1]))
    # print(f"Frontend: {frontend}; Backend: {backend}")

    frontend_cost = set()
    backend_cost = set()
    for key, value in frontend.items():
        if (
            len(frontend_cost) < N
            and key not in frontend_cost
            and value <= backend[key]
        ):
            frontend_cost.add(key)
        elif (
            len(backend_cost) < M and key not in backend_cost and backend[key] <= value
        ):
            backend_cost.add(key)

        if len(frontend_cost) == N and len(backend_cost) == M:
            break

    total_frontend_cost = sum([frontend[cost] for cost in frontend_cost])
    total_backend_cost = sum([backend[cost] for cost in backend_cost])
    return total_frontend_cost + total_backend_cost


N = 2
M = 1
F = [1, 1, 1]
B = [1, 1, 1]

res = logic(N, M, F, B)
print(res)