reduce(lambda x,y: x if x>y else y, nums)

returns max element in nums list!

Check this link for more info

you do have a point, max() is O(n), it's not very efficient if there are many duplicated max values

I was trying to use max as well... @nnk94087 has really beautiful code man!

I am trying same.

yess u can do this without max funtion using sorting .fristly convert it into list then sort the list and print the second element , that will be the runnerup..

I found the easiest to find the max by this without using any pre defined function. "" def get_max(arr): if len(arr): max = arr[0] for i in arr: if i > max: max = i return max """

n = int(input()) arr = list(map(int, input().split())) negatives = [-1,-2,-3,-4,-5,-6,-7,-8,-9,-10] d = [] for i in arr: if i in negatives: d.append(i) d.sort()
d.reverse() c = [] for i in arr: if i not in c: c.append(i) c.sort() print(c[-2])
*

Hi, I believe this code is correct only in the cases where the max number apears only once in the input.

Considering the following input [1, 2, 2] it would return 2 whereas it should return 1.

nlog(n) though.

I was trying to convert set to list like you did. But I noticd you could just use sorted() for the set ! Weird, but it works just fine ...

input()

myList = sorted(set(map(int, input().strip().split()))) print(myList[len(myList)-2])

hi can you tell me please where the value of n is being used and how ? thanks any ways

Can you please tell what map() is doing here ?

List isn't necessary as sorted takes any object iirc

print raw_input()==0 or sorted(set(map(int, raw_input().split())))[-2]

For those that didn't understand, first input isn't actually necessary because validation of the length of the numbers entered in 2nd line isn't actually being asked

raw_input returns type string, hence it will never be zero(0) and won't print the first input under any circumstance.

Next read the input,split away the spaces and map it to integer type

Make it's set

Sort the set

Print the 2nd last element in the sorted set

I was able to do it in one line. But I will not tell you how! Try to figure it out yourself HA HA

Can anyone plz explain why sorted() works and not the function sort() ?

won't work if 6,6 element are the largest in list

thanks it worked for me.

even without using list

print(sorted(set(arr))[-2])

Can you please explain it. Code worked like magic. I am new in python. So I need to get it clear in my head ASAP!

Good one moglobal!

if enter n=6 and give 10 elements then will get wrong. ur code is not using n

hy can u tell me about what set is exactly works in set(nums)?

Nice, I didn't think of the set function. Here is my aproach:

if __name__ == '__main__':
    n = int(input())
    arr = map(int, input().split())
    
    list = list(arr)
    l = [x for x in list if x != max(list)]
    print(max(l))

dude where are we using n here??

This is a great solution.

This fails if u had more than 2 numbers of maximum value in the given list

It is wonderful broo

it can also be written like this, n = int(raw_input()) nums = map(int, raw_input().split())
print sorted(list(set(nums)),reverse=True)[1]

you dont need to convert it into list! sorted(set(nums))[-2] works fine even with negative numbers

I really think I understand this answer... I was thinking of max but this one makes more sense to me.

I have mostly worked in python 2.7 and am trying to refresh my knowledge of it while updating to 3.0 is there documentation on the sorted() and hot to use it that is easy to understand?

Python 3 version:

print(sorted(set(arr))[-2])

what does the function set()do?

I get an a error list object is not callable , however it works if I remove the list keep sorted(set(nums))[-2]

set defines what ,. sort used to assing in ascending order ,list fn is to print the mapped object. for this set ,how it is assing for this particular output

wow that was smooth... why i didnt think that way... lol i always choose long path

Good answer. It can be done without casting to a list:

    print sorted(set(arr)) [-2]

Although I don't see why you had to apply list to the set.

May be a noob version of your code

n = int(raw_input())
arr = map(int, raw_input().split())

dup =[]
for i in arr:
    if i not in dup:
        dup.append(i)
dup = sorted(dup)
print dup[len(dup) - 2]

that's cool dude

This worked neatly for me. The only thing I'll add are paranthases wrapped around the print statement because I was getting errors without them:

n = int(input()) arr = map(int, input().split()) print (sorted(list(set(arr)))[-2])

Really awesome logic.

Very nice! So smooth!

I didn't understand why you used sorted()... I thought set() already did that.

Sorry... new in python...

Thanks!

I like this solution

This wont work on Python 3.

I don't understand at all.

Can you pls explain how this works??

You're doing a bit more work than you need to. Instead of iterating over a range, just iterate over the number list.

raw_input()
max = max2 = -101
for n in map(int,raw_input().split()):
    if n > max2:
        if n > max:
            max,max2 = n,max
        elif n < max:
            max2 = n
print(max2)

I think this is the best approach too to stay within O(n). My solution was something similar:

n = int(input())
a = [int(x) for x in input().split()]
largest = secondlargest = -100
for x in a:
    if x > largest:
        tmp = largest
        largest = x
        secondlargest = tmp
    elif x > secondlargest and x != largest:
        secondlargest = x
print(secondlargest)

Thanks for the clear and concise comparison!

Is there a reason you wouldn't just store max(arr) in a separate variable first and use that in the list comprehension so that it's not called multiple times? Seems like a simple solution to reduce the number of calls.

No. I made the max value in the list the min value by updating it to -101(see the constraints). So the 2nd max value automatically becomes the max value