The problem of getting the product number of pairs that lead to a particular solution has been dealt many times, this articles aims at extending that to 3 numbers and discussing several ways in which this particular problem can be solved. Let’s discuss certain ways in which this can be performed.
Method #1 : Using Nested loops This is the naive method in which this particular problem can be solved and takes outer loop to iterate for each elements and inner loop checks for the remaining difference multiplying the pairs to the result.
Python3
# Python3 code to demonstrate# 3 element product# using nested loops# initializing listtest_list = [4, 1, 3, 2, 6, 12]# initializing productproduct = 24# printing original listprint("The original list : " + str(test_list))# using nested loops# 3 element productres = []for i in range(0, len(test_list)-2): for j in range(i + 1, len(test_list)-1): for k in range(j + 1, len(test_list)): if test_list[i] * test_list[j] * test_list[k] == product: temp = [] temp.append(test_list[i]) temp.append(test_list[j]) temp.append(test_list[k]) res.append(tuple(temp))# print resultprint("The 3 product element list is : " + str(res)) |
The original list : [4, 1, 3, 2, 6, 12] The 3 product element list is : [(4, 1, 6), (4, 3, 2), (1, 2, 12)]
Time Complexity: O(n*n*n) where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n) where n is the number of elements in the list “test_list”.
Method #2 : Using itertools.combination() This particular problem can also be done in a concise manner using the inbuilt function of function. The combination function finds all the combination taking K arguments leading to particular product.
Python3
# Python3 code to demonstrate# Triple Product to K# using itertools.combination()from itertools import combinations# function to get the productdef test(val): prod = 1 for ele in val: prod *= ele return (prod == 24) # initializing listtest_list = [4, 1, 3, 2, 6, 12]# initializing productproduct = 24# printing original listprint("The original list : " + str(test_list))# using itertools.combination()# 3 element product res = list(filter(test, list(combinations(test_list, 3))))# print resultprint("The 3 product element list is : " + str(res)) |
The original list : [4, 1, 3, 2, 6, 12] The 3 product element list is : [(4, 1, 6), (4, 3, 2), (1, 2, 12)]
Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n*n), where n is the number of elements in the list “test_list”.
