Python program to find product of given number of consecutive elements

Given a List, the task is to write a python program that can construct a list with products of consecutive elements for a given number of elements.

Input : test_list = [5, 6, 2, 1, 7, 5, 3], K = 3

Output : [60, 12, 14, 35, 105]

Explanation : 5 * 6 * 2 = 60, 6 * 2 * 1 = 12.. And so on.

Input : test_list = [5, 6, 2, 1, 7, 5, 3], K = 4

Output : [60, 84, 70, 105]

Explanation : 5 * 6 * 2 * 1 = 60, 6 * 2 * 1 * 7 = 84.. And so on.

Method 1 : Using list slicing and loop

In this, we perform task of getting K slice using list slicing and task of getting product is done by an external function call.

Example:

Python3

# getting product
def prod(sub):
    res = 1
    for ele in sub:
        res = ele * res
    return res


# initializing lists
test_list = [5, 6, 2, 1, 7, 5, 3]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 3

res = []
for idx in range(len(test_list) - K + 1):

    # getting product using external function
    res.append(prod(test_list[idx: idx + K]))

# printing result
print("Computed Products : " + str(res))

Output:

The original list is : [5, 6, 2, 1, 7, 5, 3]

Computed Products : [60, 12, 14, 35, 105]

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 2 : Using generator, slicing, reduce() and mul operator

In this, generator is used to compute and return intermediate result. Task of getting sliced multiplication is done using inbuilt function reduce(), and mul operator.

Example:

Python3

from functools import reduce
from operator import mul

# generator function


def sliced_prod(sub, K):
    for idx in range(len(sub) - K + 1):

        # slicing and returning intermediate product
        sliced = sub[idx: idx + K]
        yield reduce(mul, sliced)

# generator function


# initializing lists
test_list = [5, 6, 2, 1, 7, 5, 3]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 3

# calling fnc.
res = list(sliced_prod(test_list, K))

# printing result
print("Computed Products : " + str(res))

Output:

The original list is : [5, 6, 2, 1, 7, 5, 3]

Computed Products : [60, 12, 14, 35, 105]

Time complexity: O(n), where n is the length of the test_list. The generator, slicing, reduce() and mul operator takes O(n) time to create the final list.
Auxiliary Space: O(1), extra space required is not required

Method 3 :  Use the numpy library

Python3

import numpy as np

# initializing lists
test_list = [5, 6, 2, 1, 7, 5, 3]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 3

# using numpy library
result = np.prod(np.array([test_list[i:i+K] for i in range(len(test_list) - K + 1)]), axis=1)

# printing result
print("Computed Products : " + str(result))

OUTPUT : 
The original list is : [5, 6, 2, 1, 7, 5, 3]
Computed Products : [ 60  12  14  35 105]

Time complexity: O(NK), where N is the length of the input list test_list and K is the size of the sliding window.
Auxiliary space: O(NK). We are storing all the intermediate products in an array before computing their product.