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# Python | Tuples with maximum key of similar values

Sometimes, while working with Python, we can have a problem in which we need to get all the records. This data can have similar values and we need to find maximum key-value pair. This kind of problem can occur while working with data. Let’s discuss certain ways in which this task can be done.

Method #1: Using max() + groupby() + itemgetter() + list comprehension The combination of above functions can be used to perform this particular task. In this, we first group the like valued elements using groupby() and itemgetter(), and then extract the maximum of those using max() and cumulate the result in list using list comprehension.

## Python3

 `# Python3 code to demonstrate working of``# Tuples with maximum key of similar values``# using max() + groupby() + itemgetter() + list comprehension``from` `operator ``import` `itemgetter``from` `itertools ``import` `groupby` `# initialize list``test_list ``=` `[(``4``, ``3``), (``2``, ``3``), (``3``, ``10``), (``5``, ``10``), (``5``, ``6``)]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# Tuples with maximum key of similar values``# using max() + groupby() + itemgetter() + list comprehension``res ``=` `[``max``(values) ``for` `key, values ``in` `groupby(test_list, key ``=` `itemgetter(``1``))]` `# printing result``print``(``"The records retaining maximum keys of similar values : "` `+` `str``(res))`

Output

```The original list : [(4, 3), (2, 3), (3, 10), (5, 10), (5, 6)]
The records retaining maximum keys of similar values : [(4, 3), (5, 10), (5, 6)]```

Time complexity: O(n log n), where n is the length of the input list. This is because the groupby() function requires the list to be sorted, which takes O(n log n) time, and the max() function needs to iterate through each group, which takes O(n) time.
Auxiliary space: O(n), where n is the length of the input list. This is because the groupby() function creates a new list of groups, which could potentially have n elements if every tuple in the input list has a unique second element. The res list also has n elements in the worst case, if every group contains only one tuple.

Method #2: Using setdefault() + items() + loop + list comprehension The combination of above functions can also achieve this task. In this, we convert the list tuple key-value pairs into a dictionary and assign a default value using setdefault(). The final result is computed using list comprehension.

## Python3

 `# Python3 code to demonstrate working of``# Tuples with maximum key of similar values``# using setdefault() + items() + loop + list comprehension` `# initialize list``test_list ``=` `[(``4``, ``3``), (``2``, ``3``), (``3``, ``10``), (``5``, ``10``), (``5``, ``6``)]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# Tuples with maximum key of similar values``# using setdefault() + items() + loop + list comprehension``temp ``=` `{}``for` `val, key ``in` `test_list:``   ``if` `val > temp.setdefault(key, val):``        ``temp[key] ``=` `val``res ``=` `[(val, key) ``for` `key, val ``in` `temp.items()]` `# printing result``print``(``"The records retaining maximum keys of similar values : "` `+` `str``(res))`

Output :

```The original list : [(4, 3), (2, 3), (3, 10), (5, 10), (5, 6)]
The records retaining maximum keys of similar values : [(4, 3), (5, 10), (5, 6)]```

Time Complexity: O(n*n), where n is the length of the input list. This is because we’re using setdefault() + items() + loop + list comprehension 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 #3: Using dict.fromkeys(): We can use dict.fromkeys() to find the tuples with maximum key of similar values by first creating a dictionary with the unique second elements of the tuples as the keys and the first elements as the values. Then we can iterate through the list of tuples and update the dictionary with the maximum key-value pair.

Here is an example of how to use dict.fromkeys() to find the tuples with maximum key of similar values:

## Python3

 `# initialize list``test_list ``=` `[(``4``, ``3``), (``2``, ``3``), (``3``, ``10``), (``5``, ``10``), (``5``, ``6``)]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# create a dictionary with the unique second elements of the tuples as the keys``temp ``=` `dict``.fromkeys(``set``(x[``1``] ``for` `x ``in` `test_list), ``0``)` `# iterate through the list of tuples and update the dictionary with the maximum key value pair``for` `x ``in` `test_list:``    ``if` `x[``0``] > temp[x[``1``]]:``        ``temp[x[``1``]] ``=` `x[``0``]` `# create the final result``res ``=` `[(val, key) ``for` `key, val ``in` `temp.items()]` `# printing result``print``(``"The records retaining maximum keys of similar values : "` `+` `str``(res))` `# Output:``# The original list : [(4, 3), (2, 3), (3, 10), (5, 10), (5, 6)]``# The records retaining maximum keys of similar values : [(4, 3), (5, 10), (5, 6)]``#This code is contributed by Edula Vinay Kumar Reddy`

Output

```The original list : [(4, 3), (2, 3), (3, 10), (5, 10), (5, 6)]
The records retaining maximum keys of similar values : [(5, 10), (4, 3), (5, 6)]```

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

Method #4: Using defaultdict() and loop

Use the defaultdict class from the collections module to create a dictionary with default value as an empty list. Then, loop through the original list and append each tuple to the list corresponding to its second element (i.e., key) in the dictionary. Finally, loop through the dictionary and for each key, we sort the list of tuples by their first element (i.e., value) in descending order and take the first tuple as the result.

## Python

 `from` `collections ``import` `defaultdict` `# initialize list``test_list ``=` `[(``4``, ``3``), (``2``, ``3``), (``3``, ``10``), (``5``, ``10``), (``5``, ``6``)]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# Tuples with maximum key of similar values``# using defaultdict() and loop``temp ``=` `defaultdict(``list``)``for` `val, key ``in` `test_list:``    ``temp[key].append((val, key))``res ``=` `[(``sorted``(lst, reverse``=``True``)[``0``]) ``for` `lst ``in` `temp.values()]` `# printing result``print``(``"The records retaining maximum keys of similar values : "` `+` `str``(res))`

Output

```The original list : [(4, 3), (2, 3), (3, 10), (5, 10), (5, 6)]
The records retaining maximum keys of similar values : [(5, 10), (4, 3), (5, 6)]```

Time complexity: O(n*logn) where n is the length of the input list because of the sorting step.
Auxiliary space: O(n) for the dictionary.