Python | Triple Product to K
Last Updated :
09 Apr, 2023
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 list
test_list = [4, 1, 3, 2, 6, 12]
# initializing product
product = 24
# printing original list
print("The original list : " + str(test_list))
# using nested loops
# 3 element product
res = []
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 result
print("The 3 product element list is : " + str(res))
Output : 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 product
def test(val):
prod = 1
for ele in val:
prod *= ele
return (prod == 24)
# initializing list
test_list = [4, 1, 3, 2, 6, 12]
# initializing product
product = 24
# printing original list
print("The original list : " + str(test_list))
# using itertools.combination()
# 3 element product
res = list(filter(test, list(combinations(test_list, 3))))
# print result
print("The 3 product element list is : " + str(res))
Output : 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”.
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