Python - List index out of range

Question

I have been stuck with this question on finding out the second largest number in a list to be read from the console with:

n = size of the list(number of elements) AND arr = The list itself

One of the constraints is -100<=A[i]<=100 which I am unable to integrate within my code as I am getting a LIST INDEX OUT OF RANGE ERROR. For instance, this code fails for the custom input: 3 -10 0 10

but passes for 3 10 0 10

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

sortedlist = sorted(arr)
temp = len(sortedlist)

emptylist = []

i = 0 
for temp in range(2, 11):
    if sortedlist[i]<sortedlist[i+1]:
        i+=1
        emptylist.insert(i, sortedlist[i-1])
b = max(emptylist)
print(emptylist)

Error:

Traceback (most recent call last):

File "solution.py", line 12, in if sortedlist[i]
IndexError: list index out of range

So you sort the list then check if sortedlist[i] < sortedlist[i+1], why don't you trust Python builtins to make their job ? And what is for temp in range(2, 11): supposed to accomplish ? — polku
– polku, Commented May 16, 2017 at 10:23
Because i is getting incremented every time and you have no checking whether it crossed the list length limit or not — kuro
– kuro, Commented May 16, 2017 at 10:23
As for the second case it passes as second element is greater than the third, i is not incremented again. So i always stays 1 — kuro
– kuro, Commented May 16, 2017 at 10:29
Got it! I was unnecessarily putting a constraint on temp between 2 and 11 as the question asked but that wasn't needed. For i in range(0, temp-1) did the job. — rachitmishra25
– rachitmishra25, Commented May 16, 2017 at 10:36

Stuti Rastogi · Accepted Answer · 2017-05-16 10:25:18Z

2

if sortedlist[i] < sortedlist[i+1]:

The problem lies here. In your example n = 3 so i can only be 0, 1 or 2. But in your code, i+1 goes 1, 2 and 3 in the above line, which is out of the array bounds.

You can print i and see that i goes till 2 which is n-1 and so i+1 will be out of bounds. That is one easy way you could debug.

You have to ensure a condition where you do not compare with the next element for the last element.

edited May 16, 2017 at 10:25

answered May 16, 2017 at 10:23

Stuti Rastogi

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3 Comments

rachitmishra25 Over a year ago

Thanks. Got it. Issue was with my constraints for i in the for loop. For i in range(0, temp-1) did the job.

Stuti Rastogi Over a year ago

Great! But just out of curiosity, why are you not using sortedList[-2] to get the second largest element? Would make your job much easier!

rachitmishra25 Over a year ago

I guess I got too obsessed with trying to find the logic behind it.

Balaji Balakrishnan · Accepted Answer · 2017-05-16 10:50:11Z

0

you can just use sortedlist[-2] to get the second largest number.

In case of repeated numbers you have to use set method call over original/sorted list.

sortedlist = sorted(list(set(arr)))

edited May 16, 2017 at 10:50

Balaji Balakrishnan

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answered May 16, 2017 at 10:27

Pavan

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Comments

selvasundarraj · Accepted Answer · 2017-05-16 11:06:11Z

0

try below snippet number_of_inputs = 4 # this is user defined i = 0 collections = list() while True: user_input = input("Enter your numbers \n") if 100 >= user_input >= -100: i += 1 collections.append(user_input) if i == number_of_inputs: print "second largest number is \t ",sorted(collections)[-2] break else: print"this Entry not allowed !!!"

answered May 16, 2017 at 11:06

selvasundarraj

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