that's pretty elegant code!

a = int(input())

b = (raw_input().split(" "))

b.sort() #-----> sort is not working for negative numbers here only any idea ??

print b

b.reverse()

print b

for i in range(len(b)):

if b[i] < b[0]:

    print b[i]

    break

The problem with this implementarion is the complexity. You are doing it in O(n²), while you can do this in O(n). Even sorting the list have a better complexity O(n*log(n)). An example of a patological case is the list [5, 5, 5, 5, 5 ,6] or something like that.

Very nice, but max() is a built-in function just like sorted() :P I've come up with this two liner using sorted(), set() and listcomp instead of map():

n, a = int(raw_input()), list(set([int(x) for x in raw_input().split()]))

print sorted(a)[len(a)-2]

It was a clear solution, you don't need calculate the largest number over and over, just use

lis.remove(z)

Even more Pythonic way to do it would be as follows:

N = int(raw_input())

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

print max([x for x in arr if x != max(arr)])

I just thought to make the occurences of my most maximum the most minimum .

t=int(input())
m=list(map(int,input().split(' ')))
x=max(m)
ind=list()
for i in range(0,len(m)):
    if m[i]==x:
        m[i]=-101        
print(max(m))

So Good !!! Can you please explain the "[:i]" in the end...? Is it to just accept the first i entries?

But your list has been changed. so try something new, you can check my code. n=int(input()) L=(map(int,raw_input().strip().split(" "))) L.sort() a=max(L) b=L[0] for i in L: if i>b and i

there is no use of .strip() , you will get the output without that , if any please specify.Thanks

As for me, no function at all except list initialization.

N = int(input())
a = list(map(int,raw_input().split()))
max1 = max2 = -100
for i in a:
    if i > max1:
        max1 = i;
i in a:
    if i > max2 and i != max1:
        max2 = i;        
print max2

This has very poor time complexity...bcoz you check for max every time and you also use remove which has O(n) time. Hence your code is atleast O(n²).

Max is O(n), where n is the size of the list. https://wiki.python.org/moin/TimeComplexity

This is O(n^2) altogether.

brilliant

Could you pls explain for what reason you used function 'list'? As I know, "map" is already returning list object. Am I wrong?

Can you explain why you did max(lis) == z in the while list?

py3 in one line :)

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

Nice think of code man.

I think it doesn't need to use strip()

ur code will not work if all the no in the list is same

lis = list(map(int,raw_input().strip().split()))[:i]

What actually this statement does? Can anyone please clarify

Can you please elaborate on the [:i] part in 2nd line?

You assumed that the max value does not have a duplicate. What if there are more than one max?

hey,why that extra split()[:i]??

Hi, Would you explain what is [:i] at the end of second line? thank you!

I had pretty much the same code as you but mine didn't work, I think it is because I didn't declare the list properly.

So, my only questions are about line 2. I had only used input().split() and it seemed to work everywhere, except in tests #6, #9 or whatever.

Is it needed to build a list with list(), specify that the values are integers with map()? Also, why the [:i] at the end?

What does [:i] do exactly???

The code seems better than what i tried but there is no check for validating the input taken in lis. That is it must be in between -100 and 100(given in constraints)

why avoid set and sort?

Your function will return second value even if there is none. For example in you are looking for second largest in the list [5,5,5] there is no second largest value.

Mine (in python3)was same idea though a bit longer: n = int(input()) arr = input().split() m1 = max(arr) arr.remove(m1) m2 = max(arr) while m1 == m2: arr.remove(m2) m2 = max(arr) print(max(arr))

Thats my code and I am really not proud of that :/

if __name__ == '__main__':
    n = int(input())
    arr = list(map(int, input().split()))
    rev_arr = sorted(arr, reverse = True)
    for i in range(n):
        if rev_arr[i+1]<rev_arr[i]:
            print (rev_arr[i+1])
            break
        else:
            pass

I add set to remove duplicate maximum element

N=int(input())
s=list(set(map(int , input().split(' '))))
s.remove(max(s))
print(max(s))

arr = map(int, raw_input().split())
arr.sort()
print "elelments are %r"%arr

m=max(arr)
print "maximum number is %r " %m
l = len(arr)
print" length of the list is % r:" %l


for i in range(0,l):
  if arr[i]==m:		
	print"%r is equal to %r"%(arr[i],m)
	del arr[i]
	   
print max(arr)

PLEASE HELP! Why doesnt this get past all the test cases? It runs fine if the input is 1,2,3,4 or 4 3 2 1 or 4 4 3 1..but not the test cases.

n = int(input())
arr = list(map(int, input().split()))

largestNum = max(arr)

while largestNum in arr:
    arr.remove(largestNum)

print(max(arr))

I did kind of same

if __name__ == '__main__':
    n = int(input())
    arr = list(map(int, input().split()))
    mx = max(arr)
    while(mx in arr):
        arr.remove(mx)
    mx = max(arr)
    print(mx)

using filter..

n = int(input())
nums = list(map(int, input().strip().split()))    
z = max(nums)
print(max(list(filter(lambda a: a != z, nums))))

What is the purpose of the line:

while max(lis) == z:

An optimal method without set. Nice!

yes i used sort and set

anyone can explain why he put '[:i]' at the last part of second line? i have tired his codes without that thing, counldnt find any problem

Using list comprehension also works well, but I'm not sure about the efficiency!

n = int(raw_input())
arr = map(int, raw_input().split())
print max([x for x in arr if x != max(arr)])

elegant solution

that's pretty good code

helpful

Isn't it calculating max in the list two times with complexity O(n) + O(n). It is possible to do it in one loop ..

if __name__ == '__main__':
    n = int(input())
    arr = list(map(int, input().split()))
    
    max =  arr[0]
    runner = -9999
    for score in arr[1:]:
        if score > max:
            runner = max
            max = score
        elif score < max and score > runner:
            runner = score
    print(runner)

Very elagant piece of code....

I did first part like yours, but then when creating the logic I was checking against if's. First it was ugly and then it failed with -10,0,10... Looking at your approach it very intelligent... getting the max, then deliting all the max's within the list, then looking for a new max.... awesome :D

Can you please explain
lis = list(map(int,raw_input().strip().split()))[:i]?

Good code! Just, I do not have clear this part of the code:

.split()))[:i]

Could somebody explain it.

Can somebody explain me this line:

lis = list(map(int,raw_input().strip().split()))[:i]

I've never used map(), So I have no idea about it :/ . Let alone map(), I need a clear explanation of the line.

You have an error in code, at least in Python3 version.

print max(lis) is wrong syntax, instead of it you should do print(max(lis))

Otherwise, all great! Cheers!

Bit of a simple question I was also trying to use remove(max(array)). Please could someone explain why
highest = max(array)
while max(array) == highest:
array.remove(max(array))
print(max(array))

works but this won't:
array.remove(max(array))
print(max(array))


Can anyone please tell me if map() returns a list itself then why do we have to typecast it using into a list as done in lis = list(map(......))

Great!

Pretty confused. I did pretty much exactly what you were trying to do

import operator

if name == 'main': n = int(raw_input()) arr = map(int, raw_input().split()) max_value = max(arr) arr = arr.remove(max_value) second_max = max(arr)

print (second_max)

but I get a runtime error with this code.

adding strip was not really necessairy since max work with iterables

if __name__ == '__main__':
        n = int(input())
        arr = list(map(int, input().split()))
        m=max(arr)
        while m in arr :
            arr.remove(m)
        print(max(arr))

Am I wrong that this solution is > O(n)? This problem can be done in linear time, no? Python's max() is O(n) and you call it twice each iteration and only luckily reduce n by 1 each iteration. The set [2,2,2,2,2] would have like 30 comparisons to find max plus 5 for your z initialization and then 0 for the print I believe. 5+5+4+4+3+3+2+2+1+1+0+0 [1,2] would be 2+2+1 [1,7,3] would be 3+3+2

I s this a problem of data structure?

can you please explain me this code?

How about 2 iteration over the array. First to find the maximum ele. Second to find the second max ele.

a pleasure to see your code froma beginner level coder perspective!!

using list comprehension:

python n = int(input()) arr = list(map(int, input().split())) x = max(arr); print(max([i for i in arr if i!=x]))

wouldn't it have a time complexity of O(n^2)?

print(sorted(set(map(int,input().split())))[-2])

fool

Word.

perfect.

what are the bracket for?

This exact one-liner gives me 2 when given the following input:

5
2 3 6 6 5

When the answer should be 5.

That's wrong! The arr can only be used once.

print(sorted(set(map(int,input().split())))[-2])

Jason, I did something similar to yours but took the sorted function and used it to avoid the same pitfall I found tried doing your method (Having a pre defined veeeery low number. Have you thought about using - inf ?)

n = int(input())
arr = map(int, input().split())
arrSorted = sorted(list(arr))
index = 0
index_numero = 0
segundoNumero = arrSorted[0]
while index < len(arrSorted):
	if arrSorted[index] > segundoNumero:
		segundoNumero = arrSorted[index]
		index_numero = index
		index += 1
	else:
		index += 1
print(arrSorted[index_numero-1])

if i > largest: second_largest = largest largest = i

Can you explain this part?

My solution as well.

yep, same. except with python3 you need parents around what you want to print.