Given a List, extract Kth occurrence of Even Element.
Input : test_list = [4, 6, 2, 3, 8, 9, 10, 11], K = 3
Output : 8
Explanation : K = 3, i.e 0 based index, 4, 6, 2 and 4th is 8.Input : test_list = [4, 6, 2, 3, 8, 9, 10, 11], K = 2
Output : 2
Explanation : K = 2, i.e 0 based index, 4, 6, and 3rd is 2.
Method #1 : Using list comprehension
In this, we extract list of even elements using % operator and use list index access to get Kth even element.
Python3
test_list = [4, 6, 2, 3, 8, 9, 10, 11]
print["The original list is : " + str[test_list]]
K = 4
res = [ele for ele in
test_list if ele % 2 == 0][K]
print["The Kth Even Number : " + str[res]]
Output
The original list is : [4, 6, 2, 3, 8, 9, 10, 11] The Kth Even Number : 10
Method #2 : Using filter[] + lambda
In this, task of finding even elements is done using filter[] + lambda function.
Python3
test_list =
[4, 6, 2, 3, 8, 9, 10, 11]
print["The original list is : " + str[test_list]]
K = 4
res = list[filter[lambda ele : ele % 2 == 0, test_list]][K]
print["The Kth Even Number : " + str[res]]
Output
The original list is : [4, 6, 2, 3, 8, 9, 10, 11] The Kth Even Number : 10
Sometimes, while working with Python, we can have a problem in which we need to perform the pairing of each character with every other in String. This can have application in many domains including web development and day-day. Lets discuss certain ways in which this task can be performed.
Method #1 : Using loop This task can be performed using loop. This a brute force manner in which this task can be performed. In this, we iterate each character and append the Kth letter to each and construct a list.
Python3
test_str = "geeksforgeeks"
print["The original string is : " + test_str]
K = 4
res = []
for ele in test_str:
res.append[test_str[K] + ele]
print["List after pairing : " + str[res]]
Output :
The original string is : geeksforgeeks List after pairing : ['sg', 'se', 'se', 'sk', 'ss', 'sf', 'so', 'sr', 'sg', 'se', 'se', 'sk', 'ss']
Method #2 : Using join[] + zip[] + cycle[] The combination of above functions can be used to perform this task. In this, we perform the task of joining using join[]. The task of pairing with all is done with zip[] + cycle[].
Python3
from itertools import cycle
test_str = "geeksforgeeks"
print["The original string is : " + test_str]
K = 4
res = list[map[''.join, zip[cycle[test_str[K]], test_str]]]
print["List after pairing : "
+ str[res]]
Output :
The original string is : geeksforgeeks List after pairing : ['sg', 'se', 'se', 'sk', 'ss', 'sf', 'so', 'sr', 'sg', 'se', 'se', 'sk', 'ss']
The Time and Space Complexity for all the methods are the same:
Time Complexity: O[nlogn]
Auxiliary Space: O[n]