Given this sample list:
[5, 3, 9, 10, 8, 2, 7]
How to find the minimum number using recursion? The answer is 2
.
I found this in a question paper while I was doing recursion exercises. I can't figure out a way to solve this. To find it, do I have to sort the list first and then there's nothing to do recursively. Can any one show me a path?
This is a recursive implementation of min
:
l=[5, 3, 9, 10, 8, 2, 7]
def find_min(l,current_minimum = None):
if not l:
return current_minimum
candidate=l.pop()
if current_minimum==None or candidate<current_minimum:
return find_min(l,candidate)
return find_min(l,current_minimum)
print find_min(l)
>>>
2
Take into account that this should not be used in real programs and should be treated as an exercise. The performance will be worse than the built-in min
by several orders of magnitude.
>>> import random
>>> arr=[random.randint(0,8) for r in xrange(10)]
>>> arr
[8, 2, 5, 1, 2, 4, 0, 3, 1, 1]
>>> def func(arr):
if len(arr) == 1:
return arr[0]
else:
return min(arr[0],func(arr[1:]))
>>> f(arr)
0
NB the recursion isn't really needed here.
This answer uses an accumulator to store the min value throughout the recursions.
list = [5, 3, 9, 10, 8, 2, 7]
def min_list(list, min=None):
if len(list) < 1:
return min
return min_list(list[1:], list[0] if min is None or list[0] < min else min)
print(min_list(list))
Thats also working, but only for lists with a length that is a power of two. For other lengths you just have to tweak the split into smaller arrays. The approach is taken from merge sort .
def findCloseToZero(l):
if len(l) == 1:
return l[0]
else:
first = findCloseToZero(l[0:int(len(l)/2)])
sec = findCloseToZero(l[int(len(l)/2):])
return first if abs(first) < abs(sec) else sec
def find_smallest_elem(lst):
k=1
while k != len(lst):
if lst[0] > lst[k]:
return(find_smallest_elem(lst[k:]))
else:
k +=1
return(lst[0])
This seems to works fine
this is my answer using recursion and a line of code
def min_value(n_list):
return n_list[0] if len(n_list) == 1 else min(n_list[0], min_value(n_list[1:]))
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