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# Python – Maximum element in consecutive subsets

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

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

## Python3

 `# Python3 code to demonstrate``# Maximum element in consecutive subsets``# using list comprehension + max()` `# 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 + max()``# Maximum element in consecutive subsets``res ``=` `[ ``max``(test_list[x : x ``+` `3``])``        ``for` `x ``in` `range``(``0``, ``len``(test_list), ``3``)]` `# printing result``print``("The grouped maximized ``list` `is` `: " ``+` `str``(res))`

Output :

```The original list : [4, 7, 8, 10, 12, 15, 13, 17, 14]
The grouped maximized list is : [8, 15, 17]```

Time Complexity: O(n*n), where n is the length of the input list. This is because we’re using list comprehension + max() which has a time complexity of O(n*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 max() + itertools.islice() The task of slicing the list into chunks is done by islice method here and the conventional task of getting the maximum is done by the max function as the above method.

## Python3

 `# Python3 code to demonstrate``# Maximum element in consecutive subsets``# using itertools.islice() + max()``import` `itertools` `# 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() + max()``# Maximum element in consecutive subsets``res ``=` `[``max``(``list``(itertools.islice(test_list, i, i ``+` `3``)))``                  ``for` `i ``in` `range``(``0``, ``len``(test_list), ``3``)]` `# printing result``print``("The grouped maximized ``list` `is` `: " ``+` `str``(res))`

Output :

```The original list : [4, 7, 8, 10, 12, 15, 13, 17, 14]
The grouped maximized list is : [8, 15, 17]```

Time Complexity: O(n*n) where n is the number of elements in the string list. The max() + itertools.islice() is used to perform the task and it takes O(n*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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