Put some more of your time to help improve that skill.try to learn to get better.
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Explanation:
def letterDigitCount(j):
dic={"digits":0,"letters":0}
for a in j:
if a.isdigit():
dic["digits"]+=1
elif a.isalpha():
dic["letters"]+=1
return dic
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>
Try pluggin it into word it usually catches it
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