Sometimes, while working with lists, we can have a problem in which we require to insert a particular element at an index. But, before that it is essential to know that particular index is part of list or not. Let’s discuss certain shorthands that can perform this task error free.
Method #1 : Using len() This task can be performed easily by finding the length of list using len(). We can check if the desired index is smaller than length which would prove it’s existence.
Python3
# Python3 code to demonstrate working of# Check for Nth index existence in list# Using len()# initializing listtest_list = [4, 5, 6, 7, 10]# printing original listprint("The original list is : " + str(test_list))# initializing N N = 6# Check for Nth index existence in list# Using len()res = len(test_list) >= N# printing result print("Is Nth index available? : " + str(res)) |
The original list is : [4, 5, 6, 7, 10] Is Nth index available? : False
Time Complexity : O(n)
Auxiliary Space : O(n), where n is length of list.
Method #2 : Using try-except block + IndexError exception This task can also be solved using the try except block which raises a IndexError exception if we try to access an index not a part of list i.e out of bound.
Python3
# Python3 code to demonstrate working of# Check for Nth index existence in list# Using try-except block + IndexError exception# initializing listtest_list = [4, 5, 6, 7, 10]# printing original listprint("The original list is : " + str(test_list))# initializing N N = 6# Check for Nth index existence in list# Using try-except block + IndexError exceptiontry: val = test_list[N] res = Trueexcept IndexError: res = False # printing result print("Is Nth index available? : " + str(res)) |
The original list is : [4, 5, 6, 7, 10] Is Nth index available? : False
Time Complexity: O(n*n) where n is the length of the list
Auxiliary Space: O(1), constant extra space is required
Method#3: using the in operator
Python3
# initializing listtest_list = [4, 5, 6, 7, 10]# printing original listprint("The original list is : " + str(test_list))# initializing NN = 6# Check for Nth index existence in list# Using the in operatorres = N in range(len(test_list))# printing resultprint("Is Nth index available? : " + str(res))#This code is contributed by Vinay Pinjala. |
The original list is : [4, 5, 6, 7, 10] Is Nth index available? : False
Time Complexity: O(n)
Auxiliary Space: O(1)
