Python - Cross Pairing in Tuple List

Given 2 tuples, perform cross pairing of corresponding tuples, convert to single tuple if 1st element of both tuple matches.

Input : test_list1 = [(1, 7), (6, 7), (8, 100), (4, 21)], test_list2 = [(1, 3), (2, 1), (9, 7), (2, 17)] 
Output : [(7, 3)] 
Explanation : 1 occurs as tuple element at pos. 1 in both tuple, its 2nd elements are paired and returned.

Input : test_list1 = [(10, 7), (6, 7), (8, 100), (4, 21)], test_list2 = [(1, 3), (2, 1), (9, 7), (2, 17)] 
Output : [] 
Explanation : NO pairing possible.

Method #1 : Using list comprehension

In this, we check for 1st element using conditional statements and, and construct new tuple in list comprehension.

The original list 1 : [(1, 7), (6, 7), (9, 100), (4, 21)]
The original list 2 : [(1, 3), (2, 1), (9, 7), (2, 17)]
The mapped tuples : [(7, 3), (100, 7)]

Time complexity: O(n^2), where n is the length of the longer input list.
Auxiliary Space: O(k), where k is the length of the resulting list, since the resulting list is stored in memory.

Method #2 : Using zip() + list comprehension

In this, the task of pairing is done using zip() and conditional check is done inside list comprehension.

The original list 1 : [(1, 7), (6, 7), (9, 100), (4, 21)]
The original list 2 : [(1, 3), (2, 1), (9, 7), (2, 17)]
The mapped tuples : [(7, 3), (100, 7)]

Method #3 : Using Recursion

1. Define two lists of tuples called test_list1 and test_list2.
2. Create an empty list called result to store the cross pairs.
3. Loop over each tuple tup1 in test_list1.
4. For each tup1, loop over each tuple tup2 in test_list2.
5. Compare the first elements of tup1 and tup2.
6. If the first elements are equal, cross-pair the second elements and append the resulting tuple to the result list.
7. After all pairs have been checked, return the result list.

[(7, 3), (100, 7)]

Time complexity: O(n*m), where n is the length of test_list1 and m is the length of test_list2. This is because the function needs to iterate over each tuple in both lists in the worst case. The comparison of the first elements of each tuple is a constant-time operation. The appending of tuples to the result list has a time complexity of O(1) for each tuple.
Auxiliary space: O(k), where k is the size of the result list. This is because the result list is the only data structure that grows with the size of the input. The space required to store tup1, tup2, and temporary variables used in the loops is constant, regardless of the input size.

METHOD 4:Using dictionary

APPROACH:

The approach uses a dictionary to store the tuples in list 1, with the first element as the key and the second element as the value. Then, it loops through each tuple in list 2 and checks if the first element of the tuple is a key in the dictionary. If it is, a tuple is appended to the mapped tuples list with the value of the key in the dictionary as the first element and the second element of the tuple in list 2 as the second element. Finally, the list of mapped tuples is returned.

ALGORITHM:

1.Initialize an empty dictionary to store the tuples in list 1.
2.Loop through each tuple in list 1.
3.Add the tuple to the dictionary with the first element as the key and the second element as the value.
4.Initialize an empty list to store mapped tuples.
5.Loop through each tuple in list 2.
6.If the first element of the tuple in list 2 is a key in the dictionary, append a tuple of the value of the key in the dictionary and the second element of the tuple in list 2 to the mapped tuples list.
7.Return the mapped tuples list.

[(7, 3), (100, 7)]

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