How to find second largest number in a list?

Question

So I have to find the second largest number from list. I am doing it through simple loops.

My approach is to divide a list into two parts and then find the largest n

Answer 1

You don't have to sort the input, and this solution runs in O(n). Since your question says you cannot use builtin functions, you can use this

alist=[-45,0,3,10,90,5,-2,4,18,45,100,1,-266,706]
largest, larger = alist[0], alist[0]

for num in alist:
    if num > largest:
        largest, larger = num, largest
    elif num > larger:
        larger = num
print larger

Output

100

Keep track of the largest number and the second largest number (larger variable stores that in the code). If the current number is greater than the largest, current number becomes the largest, largest becomes just larger.

largest, larger = num, largest is a shortcut for

temp = largest
largest = num
larger = temp

Edit: As per OP's request in the comments,

def findLarge(myList):
    largest, larger = myList[0], myList[0]
    for num in myList:
        if num > largest:
            largest, larger = num, largest
        elif num > larger:
            larger = num
    return largest, larger

alist=[-45,0,3,10,90,5,-2,4,18,45,100,1,-266,706]

firstLargest, firstLarger  = findLarge(alist[:len(alist)//2])
secondLargest, secondLarger = findLarge(alist[len(alist)//2:])

print sorted((firstLarger, firstLargest, secondLarger, secondLargest))[-2]