CODEWITHSUNDEEP
listpython-3.xrecursion
Dhruv Marwha
Dhruv Marwha

Reputation: 1074

Finding whether a list is in descending order or not

I am currently learning recursive functions and i have solved this question partially.

For example,if the input is [4,4,3],the output should be true.The successive element has to be at least as big as the previous one.

Here's my code:

def descending(l):
    x=0
    if len(l)==0 or len(l)==1:
        return True
    for value in range(0,len(l)):
         if l[value]>=l[value]+1:
             x=l[value]+1
             descending(l[x:len(l)])
             return True
         else:
             return False

Do let me know why my logic fails for certain cases.

Upvotes: 0

Views: 5403

Answers (5)

Sai Charan
Sai Charan

Reputation: 11

def decreasing(l):
    if l==[] or len(l) == 1:
        return(True)
    else:
        return( decreasing(l[1:]) if l[0] > l[1] else False)

Upvotes: 1

cdlane
cdlane

Reputation: 41895

For me, the key failure of your code is that descending() returns a value but when you call it recursively, you ignore its return value!

I'm not sure which of the previous answers is more Pythonic than the other but I don't like their style. I'd first clearly lay out your base cases and then end with the recursion:

def descending(array):
    if len(array) <= 1:
        return True

    if array[0] < array[1]:
        return False

    return descending(array[1:])

I also have a personal hangup against return-else-return logic:

if x:
     return a
else:
     return b

vs. simply:

if x:
    return a
return b

Upvotes: 1

Sygmei
Sygmei

Reputation: 465

If you want this function to be recursive, it should look like this :)

def descending(l):
    if len(l) <= 1 or (len(l) == 2 and l[0] >= l[1]):
        return True
    else:
        if l[0] >= l[1]:
            return descending(l[1::])
        else:
            return False

Upvotes: 2

ailin
ailin

Reputation: 501

If u want to use recursion, you can do it like this:

def foo(l):
    if len(l) <= 1:
        return True
    f, *l = l
    return False if f < l[0] else foo(l)

But I strongly not recommend you use recursion for this kind of problems.

Little remark for syntax: f, *l = l will set f to first element of l and set l to rest part of l, it same to: f, l = l[0], l[1:]

Upvotes: 2

joaquinlpereyra
joaquinlpereyra

Reputation: 966

I think your code works, but it could easily be more pythonic.

def descending(l):
    if not l or len(l) == 1:
        return True
    elif l[0]>=l[1]:
         return descending(l[1:])
    else:
         return False

Upvotes: 0

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