Sometimes, while working with Python list, one can have a problem in which one needs to find perform the maximization of list in pair form. This is useful as a subproblem solution of bigger problem in web development and day-day programming. Let’s discuss certain ways in which this problem can be solved.
Method #1 : Using loop + max() This is the brute force method to perform this particular task. In this, we just iterate the list till last element in skipped manner to get all the pair maximas in other list in iterative way.
Python3
# Python3 code to demonstrate working of# Consecutive Pair Maximum# Using loop + max()# initializing listtest_list = [5, 8, 3, 5, 9, 10]# printing listprint("The original list : " + str(test_list))# Consecutive Pair Maximum# Using loop + max()res = []for ele in range(0, len(test_list), 2): res.append(max(test_list[ele], test_list[ele + 1]))# Printing resultprint("Pair maximum of list : " + str(res)) |
The original list : [5, 8, 3, 5, 9, 10] Pair maximum of list : [8, 5, 10]
Time Complexity: O(n) where n is the number of elements in the list “test_list”.rloop + max() performs n number of operations.
Auxiliary Space: O(n), extra space is required where n is the number of elements in the list
Method #2 : Using zip() + list comprehension + max() This task can also be performed using the combination of above functionalities. In this, we just iterate the list and the task of combining pairs is performed by zip(). Works only on Python2.
Python3
# Python code to demonstrate working of# Consecutive Pair Maximum# Using zip() + list comprehension + max()# initializing listtest_list = [5, 8, 3, 5, 9, 10]# printing listprint("The original list : " + str(test_list))# Consecutive Pair Maximum# zip() + list comprehension + max()res = [max(i, j) for i, j in zip(test_list, test_list[1:])[::2]]# Printing resultprint("Pair maximum of list : " + str(res)) |
The original list : [5, 8, 3, 5, 9, 10] Pair maximum of list : [8, 5, 10]
Time Complexity: O(n*n), where n is the length of the input list. This is because we’re using zip() + 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 #3 : Using map() + max()
This task can also be performed using the combination of above functionalities. In this, we just iterate the list and the task of combining pairs is performed by map().
Python3
# Python code to demonstrate working of# Consecutive Pair Maximum# Using map() + max() # initializing listtest_list = [5, 8, 3, 5, 9, 10] # printing listprint("The original list : " + str(test_list)) # Consecutive Pair Maximum# map() + max()res = list(map(max, test_list, test_list[1:]))[::2] # Printing resultprint("Pair maximum of list : " + str(res)) |
The original list : [5, 8, 3, 5, 9, 10] Pair maximum of list : [8, 5, 10]
Time complexity: O(n)
Space complexity: O(n)
Method#4: Using Recursive method.
Algorithm:
- The function takes a list lst as input.
- If the length of lst is 1, the function returns a list containing the single element of lst.
- If the length of lst is 2, the function returns a list containing the maximum of the first two elements of lst.
- If the length of lst is greater than 2, the function returns a list containing the maximum of the first two elements of lst, concatenated with the result of recursively calling pair_max_recursive() on the sublist of lst starting from index 2.
- The final list containing the maximum of each consecutive pair of elements is returned by the function.
Python3
def pair_max_recursive(lst): if len(lst) == 1: return [lst[0]] elif len(lst) == 2: return [max(lst[0], lst[1])] else: return [max(lst[0], lst[1])] + pair_max_recursive(lst[2:])# Example usage:# initializing listtest_list = [5, 8, 3, 5, 9, 10] # printing listprint("The original list : " + str(test_list)) res = pair_max_recursive(test_list) # Printing resultprint("Pair maximum of list : " + str(res))#this code contributed by tvsk |
The original list : [5, 8, 3, 5, 9, 10] Pair maximum of list : [8, 5, 10]
Time complexity: O(n), where n is the length of the input list. This is because the function recursively calls itself once per pair of elements in the input list, and each recursive call processes a sublist of length 2 less than the previous call.
Auxiliary space: O(n), where n is the length of the input list. This is because the function uses a new list to store the result of each recursive call, and the maximum depth of recursion is n/2. Therefore, the maximum amount of memory used by the function at any given time is proportional to the length of the input list.
