# Python – Maximum of String Integer list

• Last Updated : 23 Mar, 2023

Sometimes, while working with data, we can have a problem in which we receive a series of lists with data in string format, which we wish to find the max of each string list integer. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using loop + int()
This is the brute force method to perform this task. In this, we run a loop for the entire list, convert each string to integer and perform maximization listwise and store in a separate list.

## Python3

 `# Python3 code to demonstrate working of``# Maximum of String Integer``# using loop + int()` `# initialize list``test_list ``=` `[[``'1'``, ``'4'``], [``'5'``, ``'6'``], [``'7'``, ``'10'``]]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# Maximum of String Integer``# using loop + int()``res ``=` `[]``for` `sub ``in` `test_list:``    ``par_max ``=` `0``    ``for` `ele ``in` `sub:``        ``par_max ``=` `max``(par_max, ``int``(ele))``    ``res.append(par_max)` `# printing result``print``(``"List after maximization of nested string lists : "` `+` `str``(res))`

Output

```The original list : [['1', '4'], ['5', '6'], ['7', '10']]
List after maximization of nested string lists : [4, 6, 10]```

Time Complexity: O(n*n), where n is the length of the input list. This is because we’re using the loop + int() 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() + int() + list comprehension
This is the shorthand with the help of which this task can be performed. In this, we run a loop on lists using list comprehension and extract maximization using max().

## Python3

 `# Python3 code to demonstrate working of``# Maximum of String Integer``# using max() + int() + list comprehension` `# initialize list``test_list ``=` `[[``'1'``, ``'4'``], [``'5'``, ``'6'``], [``'7'``, ``'10'``]]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# Maximum of String Integer``# using max() + int() + list comprehension``res ``=` `[``max``(``int``(ele) ``for` `ele ``in` `sub) ``for` `sub ``in` `test_list]` `# printing result``print``(``"List after maximization of nested string lists : "` `+` `str``(res))`

Output :

```The original list : [['1', '4'], ['5', '6'], ['7', '10']]
List after maximization of nested string lists : [4, 6, 10]```

Method #3 : Using map()

Another approach that can be used to find the maximum of a list of string integers is by using the map() function to convert the strings to integers, and then using the max() function to find the maximum value.

## Python3

 `# Python3 code to demonstrate working of``# Maximum of String Integer``# using map() and max()`` ` `# initialize list``test_list ``=` `[[``'1'``, ``'4'``], [``'5'``, ``'6'``], [``'7'``, ``'10'``]]`` ` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))`` ` `# Maximum of String Integer``# using map() and max()``res ``=` `[``max``(``map``(``int``, sub)) ``for` `sub ``in` `test_list]`` ` `# printing result``print``(``"List after maximization of nested string lists : "` `+` `str``(res))``#This code is contributed by Edula Vinay Kumar Reddy`

Output

```The original list : [['1', '4'], ['5', '6'], ['7', '10']]
List after maximization of nested string lists : [4, 6, 10]```

The time and Auxiliary space of this approach is O(n) where n is the number of elements in the list.

Method #4 : Using map() and lambda function:

## Python3

 `test_list ``=` `[[``'1'``, ``'4'``], [``'5'``, ``'6'``], [``'7'``, ``'10'``]]``res ``=` `list``(``map``(``lambda` `x: ``max``(``map``(``int``, x)), test_list))``print``(``"List after maximization of nested string lists : "` `+` `str``(res))``#This code is contributed by pinjala Jyothi.`

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

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