# Python – Rear shift K in List

The conventional problem involving the element shifts has been discussed many times earlier, but sometimes we have strict constraints to performing them and knowledge of any possible variation helps. This article talks about one such problem of shifting K’s at end of list, catch here is it checks for just K’s excluding the conventional ‘None’ (False) values. Let’s discuss certain ways in which this can be performed.

Method #1 : Using list comprehension + isinstance() In this method, we perform the operation of shifting in 2 steps. In 1st step we get all the values that we need to get at the front and at the end we just push the K’s to end. The isinstance method is used to filter out the Boolean False entity.

## Python3

 `# Python3 code to demonstrate` `# Rear shift K in List` `# using list comprehension + isinstance()`   `# initializing list` `test_list ``=` `[``1``, ``4``, ``None``, "Manjeet", ``False``, ``4``, ``False``, ``4``, "Nikhil"]`   `# printing original list` `print``("The original ``list` `: " ``+` `str``(test_list))`   `# initializing K ` `K ``=` `4`   `# using list comprehension + isinstance()` `# Rear shift K in List` `temp ``=` `[ele ``for` `ele ``in` `test_list ``if` `ele !``=` `K ``and` `ele ``or` `ele ``is` `None` `or` `isinstance``(ele, ``bool``) ]` `res ``=` `temp ``+` `[K] ``*` `(``len``(test_list) ``-` `len``(temp))`   `# print result` `print``("The ``list` `after shifting K's to end : " ``+` `str``(res))`

Output :

```The original list : [1, 4, None, 'Manjeet', False, 4, False, 4, 'Nikhil']
The list after shifting K's to end : [1, None, 'Manjeet', False, False, 'Nikhil', 4, 4, 4]```

Time Complexity: O(n), where n is the length of the input list. This is because we’re using the list comprehension + isinstance() which has a time complexity of O(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 #2 : Using list comprehension + isinstance() + list slicing This method is similar to the above method, the only modification is that to reduce number of steps, list slicing is used to attach the K’s to perform whole task in just 1 step.

## Python3

 `# Python3 code to demonstrate` `# Shift zeroes at end of list` `# using list comprehension + isinstance() + list slicing`   `# initializing list` `test_list ``=` `[``1``, ``4``, ``None``, "Manjeet", ``False``, ``4``, ``False``, ``4``, "Nikhil"]`   `# printing original list` `print``("The original ``list` `: " ``+` `str``(test_list))`   `# initializing K ` `K ``=` `4`   `# using list comprehension + isinstance() + list slicing` `# Shift zeroes at end of list` `res ``=` `([ele ``for` `ele ``in` `test_list ``if` `ele !``=` `K ``and` `ele ``or` `not` `isinstance``(ele, ``int``)` `       ``or` `isinstance``(ele, ``bool``)]` `        ``+` `[K] ``*` `len``(test_list))[:``len``(test_list)]`   `# print result` `print``("The ``list` `after shifting K's to end : " ``+` `str``(res))`

Output :

```The original list : [1, 4, None, 'Manjeet', False, 4, False, 4, 'Nikhil']
The list after shifting K's to end : [1, None, 'Manjeet', False, False, 'Nikhil', 4, 4, 4]```

Method#3:using list filtering and list concatenation.

## Python3

 `# Python3 code to demonstrate` `# Shift zeroes at end of list` `# using list filtering and list concatenation`   `# initializing list` `test_list ``=` `[``1``, ``4``, ``None``, ``"Manjeet"``, ``False``, ``4``, ``False``, ``4``, ``"Nikhil"``]`   `# printing original list` `print``(``"The original list : "` `+` `str``(test_list))`   `# initializing K` `K ``=` `4`   `# using list filtering and list concatenation` `# Shift zeroes at end of list` `res ``=` `list``(``filter``(``lambda` `x: x !``=` `K, test_list)) ``+` `[K] ``*` `test_list.count(K)`   `# print result` `print``(``"The list after shifting K's to end : "` `+` `str``(res))` `#this code contributed by tvsk`

Output

```The original list : [1, 4, None, 'Manjeet', False, 4, False, 4, 'Nikhil']
The list after shifting K's to end : [1, None, 'Manjeet', False, False, 'Nikhil', 4, 4, 4]```

Time Complexity: O(n)

Auxiliary Space: O(n)

Method #4 using NumPy module:

## Python3

 `import` `numpy as np`   `# initializing list` `test_list ``=` `[``1``, ``4``, ``None``, ``"Manjeet"``, ``False``, ``4``, ``False``, ``4``, ``"Nikhil"``]`   `# printing original list` `print``(``"The original list : "` `+` `str``(test_list))`   `# initializing K` `K ``=` `4`   `temp ``=` `np.array(test_list)` `mask ``=` `np.logical_or(temp ``=``=` `K, np.logical_or(temp ``is` `None``, np.isin(temp, [``False``])))` `temp ``=` `temp[~mask].tolist()` `res ``=` `temp ``+` `[K] ``*` `(``len``(test_list) ``-` `len``(temp))`   `# print result` `print``(``"The list after shifting K's to end : "` `+` `str``(res))`

Output:

```The original list : [1, 4, None, 'Manjeet', False, 4, False, 4, 'Nikhil']
The list after shifting K's to end : [1, None, 'Manjeet', 'Nikhil', 4, 4, 4, 4, 4]```

Time Complexity: O(n)

Auxiliary Space: O(n)

Method #5: Using a loop

## Python3

 `# initializing list` `test_list ``=` `[``1``, ``4``, ``None``, ``"Manjeet"``, ``False``, ``4``, ``False``, ``4``, ``"Nikhil"``]` `print``(``"The original list : "` `+` `str``(test_list))`   `# initializing K` `K ``=` `4`   `# initialize empty lists` `k_list ``=` `[]` `non_k_list ``=` `[]`   `# loop through each element in the list` `for` `elem ``in` `test_list:` `    ``if` `elem ``=``=` `K:` `        ``k_list.append(elem)` `    ``else``:` `        ``non_k_list.append(elem)`   `# concatenate both lists` `res ``=` `non_k_list ``+` `k_list`   `# append K's at the end until the length` `# of the list is equal to the original list` `while` `len``(res) < ``len``(test_list):` `    ``res.append(K)`   `print``(``"The list after shifting K's to end : "` `+` `str``(res))`

Output

```The original list : [1, 4, None, 'Manjeet', False, 4, False, 4, 'Nikhil']
The list after shifting K's to end : [1, None, 'Manjeet', False, False, 'Nikhil', 4, 4, 4]```

Time Complexity: O(n)

Auxiliary Space: O(n)

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