Python – Extract element from list succeeded by K

Last Updated : 10 May, 2023

Given a list, extract the elements which are having K as the next element.

Input : test_list = [2, 3, 5, 7, 8, 5, 3, 5], K = 3
Output : [2, 5]
Explanation : Elements before 3 are 2, 5.

Input : test_list = [2, 3, 5, 7, 8, 5, 3, 8], K = 8
Output : [7, 3]
Explanation : Elements before 8 are 7, 3.

Method #1: Using loop

In this, we iterate the list and look for each K, and extract the element preceding it.

Python3

# Python3 code to demonstrate working of
# Extract elements succeeded by K
# Using loop
 
# initializing list
test_list = [2, 3, 5, 7, 8, 5, 3, 5]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 5
 
# Using loop to extract elements succeeded by K
res = []
for idx in range(len(test_list) - 1):
 
    # checking for succession
    if test_list[idx + 1] == K:
        res.append(test_list[idx])
 
# printing result
print("Extracted elements list : " + str(res))

Output

The original list is : [2, 3, 5, 7, 8, 5, 3, 5]
Extracted elements list : [3, 8, 3]

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

Method #2: Using list comprehension

Another way to solve this question, in this, we use list comprehension as shorthand to solve the problem of getting elements.

Python3

# Python3 code to demonstrate working of
# Extract elements succeeded by K
# Using list comprehension
 
# initializing list
test_list = [2, 3, 5, 7, 8, 5, 3, 5]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 5
 
# List comprehension used as shorthand
res = [test_list[idx]
       for idx in range(len(test_list) - 1) if test_list[idx + 1] == K]
 
# printing result
print("Extracted elements list : " + str(res))

Output

The original list is : [2, 3, 5, 7, 8, 5, 3, 5]
Extracted elements list : [3, 8, 3]

Time Complexity: O(n) where n is the number of elements in the list “test_list”. list comprehension performs n number of operations.
Auxiliary Space: O(n), extra space is required where n is the number of elements in the list

Method 3: Using the enumerate() function.

Step-by-step approach:

Initialize the original list and K value.
Create an empty list to store the extracted elements.
Loop through the list using the enumerate() function to access both the index and value of each element in the list.
Check if the current element is not the last element of the list and if the next element is equal to K.
If the condition is true, append the current element to the extracted elements list.
After the loop, print the extracted elements list.

Python3

# Python3 code to demonstrate working of
# Extract elements succeeded by K
# Using enumerate()
 
# initializing list
test_list = [2, 3, 5, 7, 8, 5, 3, 5]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 5
 
# create an empty list to store extracted elements
res = []
 
# loop through the list using enumerate()
for idx, val in enumerate(test_list):
    # check if the current element is not the last element of the list and if the next element is equal to K
    if idx < len(test_list) - 1 and test_list[idx + 1] == K:
        # append the current element to the extracted elements list
        res.append(val)
 
# printing result
print("Extracted elements list : " + str(res))

Output

The original list is : [2, 3, 5, 7, 8, 5, 3, 5]
Extracted elements list : [3, 8, 3]

Time complexity: O(n) as it involves looping through the list once.
Auxiliary space: O(k), where k is the number of elements succeeded by K in the given list.

Method #4: Using slicing and zip function

Step-by-step approach:

Start by initializing the test_list and K as given in the problem statement.
Create a new list called “res” to store extracted elements.
Use slicing to create a new list called “next_list” that contains all elements in the original list except the first element.
Use the zip function to create a tuple of each pair of elements in the original list and “next_list”.
Loop through each tuple and check if the second element in the tuple is equal to K.
If it is, append the first element of the tuple to the “res” list.
Finally, print the “res” list to display the extracted elements.

Python3

# initializing list
test_list = [2, 3, 5, 7, 8, 5, 3, 5]
 
# initializing K
K = 5
 
# create an empty list to store extracted elements
res = []
 
# use slicing to create a new list containing all elements except the first element
next_list = test_list[1:]
 
# use the zip function to create a tuple of each pair of elements in the original list and next_list
for x, y in zip(test_list, next_list):
    # check if the second element in the tuple is equal to K
    if y == K:
        # append the first element in the tuple to the extracted elements list
        res.append(x)
 
# print result
print("Extracted elements list : " + str(res))

Output

Extracted elements list : [3, 8, 3]

Time complexity: O(n), where n is the length of the list.
Auxiliary space: O(1), since we only use a few variables to store information about the list.