# Python – Rear Kth elements

• Last Updated : 10 Jul, 2020

Given a list, the task is to extract all Kth elements from rear end.

Input : test_list = [3, 4, 5, 2, 1], K = 2
Output : [1, 5, 3]
Explanation : Every 2nd elements are extracted from rear starting from 1.

Input : test_list = [3, 4, 5], K = 1
Output : [5, 4, 3]
Explanation : Every elements are extracted from rear starting from 1.

Method #1 : Using loop
The is brute method to solve this problem. In this, we reverse and then perform iteration to get each Kth multiple element.

 `# Python3 code to demonstrate working of ``# Rear Kth elements``# Using loop`` ` `# initializing list``test_list ``=` `[``3``, ``5``, ``7``, ``9``, ``10``, ``2``, ``8``, ``6``] `` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# initializing K ``K ``=` `3`` ` `# Rear Kth elements``res ``=` `[]``test_list.reverse()``for` `idx, ele ``in` `enumerate``(test_list):``     ` `    ``# Extracting elements divisible by K``    ``if` `idx ``%` `K ``=``=` `0``:``        ``res.append(ele)`` ` `# printing result ``print``(``"Rear Kth elements : "` `+` `str``(res))`
Output :
```The original list is : [3, 5, 7, 9, 10, 2, 8, 6]
Rear Kth elements : [6, 10, 5]
```

Method #2 : Using list slicing
This is shorthand version solution to this problem. We use power of list slicing skip to skip Kth element, and also reverse it.

 `# Python3 code to demonstrate working of ``# Rear Kth elements``# Using list slicing`` ` `# initializing list``test_list ``=` `[``3``, ``5``, ``7``, ``9``, ``10``, ``2``, ``8``, ``6``] `` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# initializing K ``K ``=` `3`` ` `# Rear Kth elements``# Starting slicing from Rear (-1) and extracting all Kth elements``res ``=` `test_list[``-``1``::``-``K]`` ` `# printing result ``print``(``"Rear Kth elements : "` `+` `str``(res))`
Output :
```The original list is : [3, 5, 7, 9, 10, 2, 8, 6]
Rear Kth elements : [6, 10, 5]
```

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