Simple recursive solver for set cover problem

SetCover.py

from collections import Counter

def set_cover_recursive(sets, result, element_index=0, prefix=[]):
    if len(prefix)==len(sets[0]):
        result.append(prefix)
        return True
    # Check validity of allocation    
    for i in range(len(sets)):
        if sets[i][element_index] == 1:
            set_cover_recursive(sets, result, element_index+1, prefix+[i])

def get_optimal_sets(sets, solutions):
    optimal = []
    minimal = len(sets)
    for solution in solutions:
        if len(Counter(solution))<minimal:
            optimal.clear()
            minimal = len(Counter(solution))
            optimal.append(solution)
        elif len(Counter(solution))==minimal:
            optimal.append(solution)
            
    return (optimal, minimal)

def set_cover(sets):
    result = []
    set_cover_recursive(sets, result)
    
    return get_optimal_sets(sets, result)

#Example
sets = [
#el0 el1 el2
 [1,  0,  0], #set_0
 [0,  1,  1], #set_1
 [1,  0,  1]  #set_2
]

set_cover(sets) 
#answer ([[0, 1, 1], [2, 1, 1], [2, 1, 2]], 2) means that optimal quantity of sets are 2 and there are three identical solutions
#[0, 1, 1] means that el0 in set_0, el1, el2 in set_1