This is a homework question. They say it takes O(logN + logM)
where N
and M
are the arrays lengths.
Let's name the arrays a
and b
. Obviously we can ignore all a[i]
and b[i]
where i > k.
First let's compare a[k/2]
and b[k/2]
. Let b[k/2]
> a[k/2]
. Therefore we can discard also all b[i]
, where i > k/2.
Now we have all a[i]
, where i < k and all b[i]
, where i < k/2 to find the answer.
What is the next step?
I hope I am not answering your homework, as it has been over a year since this question was asked. Here is a tail recursive solution that will take log(len(a)+len(b)) time.
Assumption: The inputs are correct, i.e., k is in the range [0, len(a)+len(b)].
Base cases:
Reduction steps:
Code:
def kthlargest(arr1, arr2, k):
if len(arr1) == 0:
return arr2[k]
elif len(arr2) == 0:
return arr1[k]
mida1 = len(arr1) // 2 # integer division
mida2 = len(arr2) // 2
if mida1 + mida2 < k:
if arr1[mida1] > arr2[mida2]:
return kthlargest(arr1, arr2[mida2+1:], k - mida2 - 1)
else:
return kthlargest(arr1[mida1+1:], arr2, k - mida1 - 1)
else:
if arr1[mida1] > arr2[mida2]:
return kthlargest(arr1[:mida1], arr2, k)
else:
return kthlargest(arr1, arr2[:mida2], k)
Please note that my solution is creating new copies of smaller arrays in every call, this can be easily eliminated by only passing start and end indices on the original arrays.