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Python program to swap two elements in a list

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  • Difficulty Level : Easy
  • Last Updated : 29 Dec, 2022
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Given a list in Python and provided the positions of the elements, write a program to swap the two elements in the list.
 

Examples:  

Input : List = [23, 65, 19, 90], pos1 = 1, pos2 = 3
Output : [19, 65, 23, 90]

Input : List = [1, 2, 3, 4, 5], pos1 = 2, pos2 = 5
Output : [1, 5, 3, 4, 2]

Approach #1: Simple swap, using comma assignment 
Since the positions of the elements are known, we can simply swap the positions of the elements.
 

Python3




# Python3 program to swap elements
# at given positions
 
# Swap function
def swapPositions(list, pos1, pos2):
     
    list[pos1], list[pos2] = list[pos2], list[pos1]
    return list
 
# Driver function
List = [23, 65, 19, 90]
pos1, pos2  = 1, 3
 
print(swapPositions(List, pos1-1, pos2-1))

Output: 
 

[19, 65, 23, 90]

Time Complexity: O(1), for using constant operations.
Auxiliary Space: O(1), for using constant extra space.

  
Approach #2 : Using Inbuilt list.pop() function 
Pop the element at pos1 and store it in a variable. Similarly, pop the element at pos2 and store it in another variable. Now insert the two popped element at each other’s original position.
 

Python3




# Python3 program to swap elements
# at given positions
 
# Swap function
def swapPositions(list, pos1, pos2):
     
    # popping both the elements from list
    first_ele = list.pop(pos1)  
    second_ele = list.pop(pos2-1)
    
    # inserting in each others positions
    list.insert(pos1, second_ele) 
    list.insert(pos2, first_ele) 
     
    return list
 
# Driver function
List = [23, 65, 19, 90]
pos1, pos2  = 1, 3
 
print(swapPositions(List, pos1-1, pos2-1))

Output: 
 

[19, 65, 23, 90]

Time Complexity: O(1), for using constant operations.
Auxiliary Space: O(1), for using constant extra space.

  
Approach #3 : Using tuple variable
Store the element at pos1 and pos2 as a pair in a tuple variable, say get. Unpack those elements with pos2 and pos1 positions in that list. Now, both the positions in that list are swapped. 
 

Python3




# Python3 program to swap elements at
# given positions
 
# Swap function
def swapPositions(list, pos1, pos2):
 
    # Storing the two elements
    # as a pair in a tuple variable get
    get = list[pos1], list[pos2]
      
    # unpacking those elements
    list[pos2], list[pos1] = get
      
    return list
 
# Driver Code
List = [23, 65, 19, 90]
 
pos1, pos2  = 1, 3
print(swapPositions(List, pos1-1, pos2-1))

Output: 
 

[19, 65, 23, 90]

Time Complexity: O(1), for using constant operations.
Auxiliary Space: O(1), for using constant extra space.

Approach #4: Using temp variable

Python3




# Python3 program to swap elements
# at given positions
 
# Swap function
def swapPositions(lis, pos1, pos2):
    temp=lis[pos1]
    lis[pos1]=lis[pos2]
    lis[pos2]=temp
    return lis
# Driver function
List = [23, 65, 19, 90]
pos1, pos2 = 1, 3
 
print(swapPositions(List, pos1-1, pos2-1))

Output

[19, 65, 23, 90]

Time Complexity: O(1), for using constant operations.
Auxiliary Space: O(1), for using constant extra space.

Approach #5: Using enumerate

Another approach to swapping elements in a list is to use the enumerate function to get the index and value of each element in the list, and then use a loop to find the elements that need to be swapped and swap them.

For example:

Python3




def swapPositions(lis, pos1, pos2):
    for i, x in enumerate(lis):
        if i == pos1:
            elem1 = x
        if i == pos2:
            elem2 = x
    lis[pos1] = elem2
    lis[pos2] = elem1
    return lis
 
List = [23, 65, 19, 90]
pos1, pos2 = 1, 3
print(swapPositions(List, pos1-1, pos2-1))
#This code is contributed by Edula Vinay Kumar Reddy

Output

[19, 65, 23, 90]

This approach has a time complexity of O(n), since it involves looping through the entire list to find the elements to be swapped. The space complexity is O(1), since it only uses a constant amount of additional space to store the elements to be swapped.

Note that this approach can be made more efficient by using a single loop and breaking out of the loop once the elements have been found and swapped.


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