Given a list, write a Python program to swap first and last element of the list.
Examples:
Input : [12, 35, 9, 56, 24]
Output : [24, 35, 9, 56, 12]
Input : [1, 2, 3]
Output : [3, 2, 1]
Approach #1: Find the length of the list and simply swap the first element with (n-1)th element.
Python3
def swapList(newList):
size = len (newList)
temp = newList[ 0 ]
newList[ 0 ] = newList[size - 1 ]
newList[size - 1 ] = temp
return newList
newList = [ 12 , 35 , 9 , 56 , 24 ]
print (swapList(newList))
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Output[24, 35, 9, 56, 12]
Approach #2: The last element of the list can be referred as list[-1]. Therefore, we can simply swap list[0] with list[-1].
Python3
def swapList(newList):
newList[ 0 ], newList[ - 1 ] = newList[ - 1 ], newList[ 0 ]
return newList
newList = [ 12 , 35 , 9 , 56 , 24 ]
print (swapList(newList))
|
Output[24, 35, 9, 56, 12]
Time Complexity: O(1)
Auxiliary Space: O(n), where n is length of list
Approach #3: Swap the first and last element is using tuple variable. Store the first and last element as a pair in a tuple variable, say get, and unpack those elements with first and last element in that list. Now, the First and last values in that list are swapped.
Python3
def swapList( list ):
get = list [ - 1 ], list [ 0 ]
list [ 0 ], list [ - 1 ] = get
return list
newList = [ 12 , 35 , 9 , 56 , 24 ]
print (swapList(newList))
|
Output[24, 35, 9, 56, 12]
Approach #4: Using * operand.
This operand proposes a change to iterable unpacking syntax, allowing to specify a “catch-all” name which will be assigned a list of all items not assigned to a “regular” name.
Python3
list = [ 1 , 2 , 3 , 4 ]
a, * b, c = list
print (a)
print (b)
print (c)
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Now let’s see the implementation of above approach:
Python3
def swapList( list ):
start, * middle, end = list
list = [end, * middle, start]
return list
newList = [ 12 , 35 , 9 , 56 , 24 ]
print (swapList(newList))
|
Output[24, 35, 9, 56, 12]
Approach #5: Swap the first and last elements is to use the inbuilt function list.pop(). Pop the first element and store it in a variable. Similarly, pop the last element and store it in another variable. Now insert the two popped element at each other’s original position.
Python3
def swapList( list ):
first = list .pop( 0 )
last = list .pop( - 1 )
list .insert( 0 , last)
list .append(first)
return list
newList = [ 12 , 35 , 9 , 56 , 24 ]
print (swapList(newList))
|
Output[24, 35, 9, 56, 12]
Approach #6: Using slicing
In this approach, we first check if the list has at least 2 elements.
If the list has at least 2 elements, we swap the first and last elements using slicing by assigning the value of the last element to the first element and the value of the first element to the last element.
We then slice the list from the second element to the second-to-last element and concatenate it with a list containing the first element and the last element in their new positions.
Python3
def swap_first_last_3(lst):
if len (lst) > = 2 :
lst = lst[ - 1 :] + lst[ 1 : - 1 ] + lst[: 1 ]
return lst
inp = [ 12 , 35 , 9 , 56 , 24 ]
print ( "The original input is:" ,inp)
result = swap_first_last_3(inp)
print ( "The output after swap first and last is:" ,result)
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OutputThe original input is: [12, 35, 9, 56, 24]
The output after swap first and last is: [24, 35, 9, 56, 12]
Time Complexity: O(1)
Space Complexity: O(1)