Reputation: 363
I tried an online challenge which had a question as follows:
You are given an array which increases at first and then starts decreasing. For example:
2 3 4 5 6 7 8 6 4 2 0 -2
. Find the maximum element of these array.
Following is my code using binary search and it gives correct answer in O(log(n)) but I don't know whether there is a better solution or not. Can anyone help me with that?
a= map(int, raw_input().split())
def BS(lo,hi):
mid = lo+ (hi-lo)/2
if a[mid]>=a[mid+1]:
if a[mid]>a[mid-1]:
return mid
else:
return BS(lo,mid)
else:
return BS(mid,hi)
print a[BS(0,len(a)-1)]
Upvotes: 2
Views: 2046
Reputation: 30
Why don't you use the max()
method??
max(lst)
will return the max value in a list
Upvotes: -1
Reputation: 10528
An optimised variant - twice faster in most cases:
# ® Видул Николаев Петров
a = [2, 3, 4, 5, 6, 7, 8, 10, 12, 24, 48, 12, 6, 5, 0, -1]
def calc(a):
if len(a) <= 2:
return a[0] if a[0] > a[1] else a[1]
l2 = len(a) / 2
if a[l2 + 1] <= a[l2] and a[l2] >= a[l2 - 1]:
return a[l2]
if a[l2] > a[l2 + 1]:
return calc(a[:l2+1])
else:
return calc(a[l2:])
print calc(a) # 48
Upvotes: 2
Reputation: 431
i am trying your code with the following input 2 3 4 5 5 8 and the answer should be 8 but the answer is 5 i am posting an image with a few more test cases
i think u cannot run binary search on an unsorted array
the code also gives huge list of exceptions for sorted arrays
Upvotes: 1