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My solution:
if __name__ == '__main__': n = int(input()) arr = map(int, input().split()) print (sorted(set(arr))[-2])
'Set' is the key.
Exact same thing I came up with. So clean and simple, not sure why it's not the top ranked. Is there a problem using 'set' to remove the dups?
He used sorted instead of sort for sorting iterable which is not a list ... Sort is faster than sorted because it doesn't have to create copy..
A small variant of this solution:
n = int(input()) a = sorted(map(int, input().split()), reverse=True) print(a[a.count(a[0])])
if __name__ == '__main__': n = int(input()) arr = list(set(sorted(map(int, input().split())))) print(arr[-2])
Why does the above code not pass 3 test cases ?
What does the sorted(set(arr))
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You are viewing a single comment's thread. Return to all comments →
My solution:
'Set' is the key.
Exact same thing I came up with. So clean and simple, not sure why it's not the top ranked. Is there a problem using 'set' to remove the dups?
He used sorted instead of sort for sorting iterable which is not a list ... Sort is faster than sorted because it doesn't have to create copy..
A small variant of this solution:
Why does the above code not pass 3 test cases ?
What does the sorted(set(arr))