The benefit of actually growing up.
Ggs mate rip pop smoke have a good day I’m just clowning around
Answer:
nums = []
while True:
in = input()
if in:
nums.append(in)
else:
break
if nums:
avg = sum(nums) / len(nums)
for i in range(len(nums)):
if nums[i] == avg:
print(f"index: {i+1}")
print(nums[i])
else:
print(-1) # if there aren't any values in nums
Explanation:
Assuming that you are coding in Python 3x. The last 'else' statement is an edge case that you might want to consider, I don't know what you want to put there, but I'm just going to leave it as -1.
Answer:
Finding kth element is more efficient in a doubly-linked list when compared to a singly-linked list
Explanation:
Assuming that both lists have firs_t and last_ pointers.
For a singly-linked list ; when locating a kth element, you have iterate through a number of k-1 elements which means that locating an element will be done only in one ( 1 ) direction
For a Doubly-linked list : To locate the Kth element can be done from two ( directions ) i.e. if the Kth element can found either by traversing the number of elements before it or after it . This makes finding the Kth element faster because the shortest route can be taken.
<em>Finding kth element is more efficient in a doubly-linked list when compared to a singly-linked list </em>
