# Python Challenge - Product of Everything Else

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This is my answer, not sure if there is a more efficient way:

def product_of_the_others(array):
new_list =
for i in range(len(array)):
temp_list = array[:i] + array[i+1:]
product = 1
for x in temp_list:
product = product * x
new_list.append(product)
return new_list

print(product_of_the_others([1, 2, 3, 4, 5]))

Space complexity, O(N); time complexity, O(N^2).

``````def product_of_the_others(array):
arr = []
for i, x in enumerate(array):
temp = array[:]
temp[i] = 1
prod = 1
for n in temp:
prod *= n
arr += [prod]
return arr

print(product_of_the_others([1, 2, 3, 4, 5]))
``````

This is how I solved it:

def product_of_the_others(array): # Write your code here position = 0 array_instance = [] new_list = [] for i in array: new_number = 1 array_instance = array.copy() array_instance.pop(position) for i in array_instance: new_number *= i position+=1 new_list.append(new_number) return new_list print(product_of_the_others([1, 2, 3, 4, 5]))
import numpy def remove_from_list(array, index): newList = array.copy() newList.pop(index) return newList def product_of_the_others(array): return [numpy.prod(remove_from_list(array, x)) for x in range(len(array))] print(product_of_the_others([1, 2, 3, 4, 5]))
from collections import deque def product_of_the_others(array): jar = deque(array) tulemus = 1 counter = 0 uus = [] while counter < len(jar): for i in range(1,len(jar)): tulemus *= jar[i] uus.append(tulemus) jar.rotate(-1) counter +=1 tulemus = 1 return uus print(product_of_the_others([1, 2, 3, 4, 5]))

The instructions state we are not permitted to use external libraries.

It’s also possible to use pop() and append() approaches, but I prefer deque.( jar.pop(0) and jar.append().

def product_of_the_others(array): tulemus = 1 counter = 0 uus = [] while counter < len(array): for i in range(1,len(array)): tulemus *= array[i] uus.append(tulemus) es = array.pop(0) array.append(es) counter +=1 tulemus = 1 return uus print(product_of_the_others([1, 2, 3, 4, 5]))

O(n)

``````def product_of_the_others(array):
fullprod = 1;
zeroes = 0
for el in array:
if el != 0:
fullprod *= el
else:
zeroes += 1
return [0 if (zeroes and el != 0 or zeroes > 1 and el == 0) else fullprod/(el if el else 1) for el in array]

print(product_of_the_others([5,5,0,]))
``````
def product_of_the_others(array): tulemus = 1 uus = [] for i in range(len(array)): for j in range (len(array)): if i != j: tulemus *=array[j] uus.append(tulemus) tulemus = 1 return uus print(product_of_the_others([1, 2, 3, 4, 5]))
1 Like
def product_of_the_others(array): product = 1 res = [] for i in range (len(array)): num_taken = array.pop(i) for j in range (len(array)): product *= array[j] res.append(product) array.insert(i, num_taken) product = 1 return res print(product_of_the_others([1, 2, 3, 4, 5]))

I wish this solution will help you all I first duplicates the original array with temp_array = array, but then all chages on the temp_array will effect the array itself as well.

def product_of_the_others(array): products = [] for i in range(len(array)): result = 1 temp_array = [] for x in array: temp_array.append(x) temp_array.remove(array[i]) for y in temp_array: result *= y products.append(result) return products print(product_of_the_others([1, 2, 3, 4, 5]))
def product_of_the_others(array): # Write your code here storage=1 newArray=[] for x in range(len(array)): for y in range(len(array)): if x==y: pass else: storage=storage*array[y] newArray.append(storage) storage=1 return(newArray) print(product_of_the_others([1, 2, 3, 4, 5]))
def product_of_the_others(array): from functools import reduce # functools comes standard; Guido doesn't like reduce(), though! return_list = [] for i in array: to_mult = array.copy() # copy the array, since I'm modifying it. to_mult.remove(i) # remove the current index from the copy. return_list.append(reduce(lambda x,y: x*y, to_mult)) # append the reduced list via multiplication return return_list print(product_of_the_others([1, 2, 3, 4, 5]))

Been toying around with functional programming, so I’m using lambda, map, filter, and reduce when possible. I was able to use a custom product function and list slicing to solve this way.
The space complexity is O(N) and time complexity is O(N**2).
I don’t know if there is a more efficient way

import numpy def product_of_the_others(array): # Write your code here lkjasd = [] yes = [] for i in range(len(array)): yes = [] for j in range(len(array)): if j != i: yes.append(array[j]) lkjasd.append(numpy.prod(yes)) return lkjasd print(product_of_the_others([1,2,3,4,5]))

will post explanation later (codecademy pls don’t kill me)

import numpy as np def product_of_the_others(array): # Write your code here result = [] for idx in range(len(array)): temp_array = array.copy() temp_array.pop(idx) result.append(np.product(temp_array)) return result print(product_of_the_others([1, 2, 3, 4, 5])) print(product_of_the_others([5, 5, 5]))

I tried two versions …

Classic version, which is slower, but works better if the numbers get big. O(n2)
It has a loop in a loop, iterating by index in each.

code for classic version
``````def product_of_the_others(array):
copy = array.copy()
length = len(array)
for i in range(length):
product = 1
for j in range(length):
if j != i:
product = product * array[j]
copy[i] = product
return copy
``````

And a version that uses division (in a loop), which is faster than the other version, but requires logic to deal with edge cases. It’s almost O(n)
(Divides product of all of them by each integer in the list to get the answers.)

code for division version
``````# /* function for getting product of everything in the array */
def multiply(array):
so_far = 1
for x in array:
so_far *= x
return so_far

# /* function to check if everything in the list is an integer */
def all_are_integers(array):
if len(array) < 1:
return False
for x in array:
if not isinstance(x, int):
return False
return True

# /* function where store product, use division */
def product_of_the_others(array):
if (array.count(0) >= 2):
return [0 for x in array]  # /* return all 0s */
product_of_all = multiply(array)
length = len(array)
copy = array.copy()
if all_are_integers(array) and isinstance(product_of_all, int):
div = lambda a, b : a // b
else:
div = lambda a, b : a / b

for i in range(length):
if copy[i] == 0:
product = 1
for j in range(length):
if j != i:
product = product * array[j]
copy[i] = product
else:
copy[i] = div( product_of_all, copy[i] )
return copy
``````

My code is longer than the others. I’m not an efficient coder.

``````from numpy import prod

def product_of_the_others(array):
return [prod(array[:i]+array[i+1:]) for i in range(len(array))]
``````
1 Like