Python Challenge - Product of Everything Else

FAQ
#1

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#2

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]))

#3

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]))
#4

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]))
#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]))
#7
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]))
#8

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

#9

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

#10
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]))
#11

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,]))
#12
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
#13
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 :smiley:

#14

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]))
#15
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]))