Code To Count How Many Integers Are Strictly Larger Than All The Integers To Their Right Using Python

Given the list of integers, X, https://www.google.com/url?q=https://docs.google.com/document/d/1TjeNYpZ_PbdBISlJPF-_WqULBc1WpuYthLClovjB3Rs/edit?usp%3Dsharing&sa=D&ust=1594

Solution 1:

Here's a solution that does that in O(N), where N is the size of the list.

l = [12,4,4,2,2,3]

defmax_so_far(l):
    m = 0for i in l[::-1]: 
        m = max(m, i)
        yield m

sum([x>y for x,y inzip(l[-2::-1], max_so_far(l))])

Solution 2:

How about this code?

mylist = list(map(int,input().split()))
count = 0for i,item inenumerate(mylist): 
    if i == len(mylist) - 1: #here we excluded the last element for a checkbreak

    check = max(mylist[i+1:])
    if check < item:
        count = count + 1print(count)

Solution 3:

I am understating this problem in two way:

    counter = 0for i in range(len(list_of_number)):
        for j in range(i, len(list_of_number)):
           if list_of_number[j] > list_of_number[i]:
               counter += 1

Here, after it gets the first value, it scans all the list. The code that follow it will check only the number's right neighbour

    counter = 0for i in range(len(list_of_number)-1):
        if list_of_numbers[i+1] > list_of_numbers[i]:
            counter +=1

Solution 4:

How about this?

defcounter(L):
    returnsum([1for i inrange(len(L)-1) if L[i] < L[i+1]])