32 lines
815 B
Ruby
32 lines
815 B
Ruby
|
# A sorted array is rotated at some unknown point, Search given element in it
|
|||
|
# Time-complexity: O(logn),Auxiliary-space:O(1)
|
|||
|
|
|||
|
# Algorithm: Binary-Search
|
|||
|
|
|||
|
|
|||
|
def search(a,key)
|
|||
|
lo=0
|
|||
|
hi=a.length-1
|
|||
|
while lo<=hi
|
|||
|
mid=lo+(hi-lo)/2
|
|||
|
if a[mid]==key
|
|||
|
return mid
|
|||
|
elsif a[mid]>=a[lo]
|
|||
|
if (a[mid]>key && a[lo]<=key)
|
|||
|
hi=mid-1
|
|||
|
else
|
|||
|
lo=mid+1
|
|||
|
end
|
|||
|
else
|
|||
|
if (a[mid]<key && a[hi]>=key)
|
|||
|
lo=mid+1
|
|||
|
else
|
|||
|
hi=mid-1
|
|||
|
end
|
|||
|
end
|
|||
|
end
|
|||
|
return -1
|
|||
|
end
|
|||
|
|
|||
|
# It doesn’t look possible to search in O(Logn) time in all cases when duplicates are allowed.
|
|||
|
# Consider searching 0 in {2, 2, 2, 2, 2, 2, 2, 2, 0, 2} and {2, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2}.
|