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# Python | Consecutive chunks Product

• Last Updated : 21 Mar, 2023

Some of the classical problems in programming domain comes from different categories and one among them is finding the product of subsets. This particular problem is also common when we need to compute the product and store consecutive group product value. Let’s try different approaches to this problem in python language.

Method #1 : Using list comprehension + loop The list comprehension can be used to perform the this particular task to filter out successive groups and product explicit function can be used to get the product of the filtered solution.

## Python3

 `# Python3 code to demonstrate``# Consecutive chunks Product``# using list comprehension + loop` `# getting Product``def` `prod(val) :``    ``res ``=` `1``    ``for` `ele ``in` `val:``        ``res ``*``=` `ele``    ``return` `res` `# initializing list``test_list ``=` `[``4``, ``7``, ``8``, ``10``, ``12``, ``15``, ``13``, ``17``, ``14``]` `# printing original list``print``("The original ``list` `: " ``+` `str``(test_list))` `# using list comprehension + loop``# Consecutive chunks Product``res ``=` `[ prod(test_list[x : x ``+` `3``]) ``for` `x ``in` `range``(``0``, ``len``(test_list), ``3``)]` `# printing result``print``("The chunked product ``list` `is` `: " ``+` `str``(res))`

Output :

```The original list : [4, 7, 8, 10, 12, 15, 13, 17, 14]
The chunked product list is : [224, 1800, 3094]```

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

Method #2 : Using loop + itertools.islice() The task of slicing the list into chunks is done by islice method here and the conventional task of getting the product is done by the explicit product function as the above method.

## Python3

 `# Python3 code to demonstrate``# Consecutive chunks Product``# using itertools.islice() + loop``import` `itertools` `# getting Product``def` `prod(val) :``    ``res ``=` `1``    ``for` `ele ``in` `val:``        ``res ``*``=` `ele``    ``return` `res` `# initializing list``test_list ``=` `[``4``, ``7``, ``8``, ``10``, ``12``, ``15``, ``13``, ``17``, ``14``]` `# printing original list``print``("The original ``list` `: " ``+` `str``(test_list))` `# using itertools.islice() + loop``# Consecutive chunks Product``res ``=` `[prod(``list``(itertools.islice(test_list, i, i ``+` `3``))) ``for` `i ``in` `range``(``0``, ``len``(test_list), ``3``)]` `# printing result``print``("The chunked product ``list` `is` `: " ``+` `str``(res))`

Output :

```The original list : [4, 7, 8, 10, 12, 15, 13, 17, 14]
The chunked product list is : [224, 1800, 3094]```

Time Complexity: O(n) where n is the number of elements in the list “test_list”. loop + itertools.islice() performs n number of operations.
Auxiliary Space: O(n), extra space is required where n is the number of elements in the list

Method #3 : Using reduce() + lambda
The inbuilt reduce function and the lambda function can also be used to perform this particular task. The reduce function is used to perform the product operation over the chunks and lambda is used to filter out the chunks.

## Python3

 `# Python3 code to demonstrate``# Consecutive chunks Product``# using reduce() + lambda``  ` `# getting Product``from` `functools ``import` `reduce``  ` `# initializing list``test_list ``=` `[``4``, ``7``, ``8``, ``10``, ``12``, ``15``, ``13``, ``17``, ``14``]``  ` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))``  ` `# using reduce() + lambda``# Consecutive chunks Product``res ``=` `[``reduce``(``lambda` `x, y : x ``*` `y, test_list[x : x ``+` `3``]) ``for` `x ``in` `range``(``0``, ``len``(test_list), ``3``)]``  ` `# printing result``print``(``"The chunked product list is : "` `+` `str``(res))``#This code is contributed by Edula Vinay Kumar Reddy`

Output

```The original list : [4, 7, 8, 10, 12, 15, 13, 17, 14]
The chunked product list is : [224, 1800, 3094]```

Time complexity of the above code is O(N).
Space complexity of the above code is O(N).

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