# Python | Maximize alternate element List

The problem of getting maximum of a list is quite generic and we might some day face the issue of getting the maximum of alternate elements and get the list of 2 elements containing maximum of alternate elements. Let’s discuss certain ways in which this can be performed.

Method #1 : Using list comprehension + list slicing + max() List slicing clubbed with list comprehension can be used to perform this particular task. We can have list comprehension to get run the logic and list slicing can slice out the alternate character, maximize by the max function.

## Python3

 `# Python3 code to demonstrate` `# Maximize alternate element List` `# using list comprehension + list slicing + max()`   `# initializing list ` `test_list ``=` `[``2``, ``1``, ``5``, ``6``, ``8``, ``10``]`   `# printing original list ` `print``("The original ``list` `: " ``+` `str``(test_list))`   `# using list comprehension + list slicing + max()` `# Maximize alternate element List` `res ``=` `[``max``(test_list[i : : ``2``]) ``for` `i ``in` `range``(``len``(test_list) ``/``/` `(``len``(test_list)``/``/``2``))]`   `# print result` `print``("The alternate elements maximum ``list` `: " ``+` `str``(res))`

Output :

```The original list : [2, 1, 5, 6, 8, 10]
The alternate elements maximum list : [8, 10]```

Time Complexity: O(n), where n is the length of the input list. This is because we’re using the list comprehension + list slicing + max() which has a time complexity of O(n) in the worst case.
Auxiliary Space: O(n), as we’re using additional space res other than the input list itself with the same size of input list.

Method #2 : Using loop This is the brute method to perform this particular task in which we have the maximum of alternate elements in different element indices and then return the output list.

## Python3

 `# Python3 code to demonstrate` `# Maximize alternate element List` `# using loop`   `# initializing list ` `test_list ``=` `[``2``, ``1``, ``5``, ``6``, ``8``, ``10``]`   `# printing original list ` `print``("The original ``list` `: " ``+` `str``(test_list))`   `# using loop` `# Maximize alternate element List` `res ``=` `[``0``, ``0``]` `for` `i ``in` `range``(``0``, ``len``(test_list)):` `    ``if``(i ``%` `2``):` `        ``res[``1``] ``=` `max``(res[``1``], test_list[i])` `    ``else` `:` `        ``res[``0``] ``=` `max``(res[``0``], test_list[i])`   `# print result` `print``("The alternate elements maximum ``list` `: " ``+` `str``(res))`

Output :

```The original list : [2, 1, 5, 6, 8, 10]
The alternate elements maximum list : [8, 10]```

Time Complexity: O(n) where n is the number of elements in the string list. The loop is used to perform the task and it takes O(n) time.
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res test_list.

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