CODEWITHSUNDEEP
python
buttHurtBuddy
buttHurtBuddy

Reputation: 35

Find the max/min element and index AFTER a certain index

I have a list like this: 

myList = [10, 7, 11, 5, 8, 9, 6]

I want to find the max element and index AFTER the absolute minimum in the list. I've done it by making a new list with all elements from 5 onwards and then finding the max. But if it's possible, I want to do everything using the original list.

So, how do I find maximum value and index of it that occurs after the absolute minimum?

Upvotes: 2

Views: 1703

Answers (7)

koPytok
koPytok

Reputation: 3733

Like this:

max(myList[myList.index(min(myList)):])

Upvotes: 1

Hai Vu
Hai Vu

Reputation: 40773

Keep in mind that most functions which calls the builtin min and max functions will scan the list more than once, but less than twice: the first scan for min will scan the entire list, the second will scan part of the list. Also, beware of solutions which create slides of the list (hint: look for the colon (:) symbol. Creating slide is alright, unless the size of the list is large then performance will be affected.

My solution will scan the list once, does not create any sub-list:

def minmax(seq):
    min_value = max_value = seq[0]
    for element in seq:
        if element <= min_value:
            min_value = max_value = element
        elif element > max_value:
            max_value = element

    return min_value, max_value

print(minmax([10, 7, 11, 5, 8, 9, 6]))

Output:

(5, 9)

Notes

  • The algorithm is simple: during scan, if we see a new minimum (i.e. element that is smaller than the current min_value), we reset both min_value and max_value. If we see a new maximum, update it.
  • This algorithm assume a list of at least one element. It will crash for empty list.

Upvotes: 0

RoadRunner
RoadRunner

Reputation: 26335

You can also use enumerate() for this:

>>> from operator import itemgetter
>>> myList = [10, 7, 11, 5, 8, 9, 6]
>>> min_index = min(enumerate(myList), key = itemgetter(1))[0] # or key = lambda x: x[1]
>>> max(myList[min_index:])
9

Upvotes: 0

Stefan Pochmann
Stefan Pochmann

Reputation: 28656

O(1) space:

myList = [10, 7, 11, 5, 8, 9, 6]

n = len(myList)
key = myList.__getitem__
imin = min(range(n), key=key)
imax = max(range(imin + 1, n), key=key)

print('maximum', myList[imax], 'is at index', imax)

Prints:

maximum 9 is at index 5

Upvotes: 1

Ajax1234
Ajax1234

Reputation: 71471

You can try this:

myList = [10, 7, 11, 5, 8, 9, 6]
abs_max = max(myList[myList.index(min(myList)):])

Output:

9

Upvotes: 2

Luke McPuke
Luke McPuke

Reputation: 364

If you don't know what the absolute minimum will be in advance, you're going to have to iterate over the entire list, in which case you might as well store your smallest and largest values in separate variables and then use index to find them if you really don't want to create a new list.

*edit: Sagun's answer is more sophisticated.

Upvotes: 0

Sagun Shrestha
Sagun Shrestha

Reputation: 1198

You can do like this

myList = [10, 7, 11, 5, 8, 9, 6]
min_index = myList.index(min(myList)) 
print(max(myList[min_index:]))

Output

9

Upvotes: 2

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