Working on this problem and post code, my question is whether it is safe to change this line of code
j > 0 and i < m and B[j-1] > A[i]
to
i < m and B[j-1] > A[i]
and also it is safe to change this line of code
i > 0 and j < n and A[i-1] > B[j]
to
i > 0 and A[i-1] > B[j]
I think remove the condition check of j is safe since we already making sure size of A is no bigger than size of B.
Problem statement
There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
Implementation
def median(A, B):
m, n = len(A), len(B)
if m > n:
A, B, m, n = B, A, n, m
if n == 0:
raise ValueError
imin, imax, half_len = 0, m, (m + n + 1) / 2
while imin <= imax:
i = (imin + imax) / 2
j = half_len - i
if j > 0 and i < m and B[j-1] > A[i]:
# i is too small, must increase it
imin = i + 1
elif i > 0 and j < n and A[i-1] > B[j]:
# i is too big, must decrease it
imax = i - 1
else:
# i is perfect
if i == 0: max_of_left = B[j-1]
elif j == 0: max_of_left = A[i-1]
else: max_of_left = max(A[i-1], B[j-1])
if (m + n) % 2 == 1:
return max_of_left
if i == m: min_of_right = B[j]
elif j == n: min_of_right = A[i]
else: min_of_right = min(A[i], B[j])
return (max_of_left + min_of_right) / 2.0
Yes I think, you can remove the condition j > 0, because
j = half_len - i
and you already check that i<m and (m + n + 1) / 2 must be bigger than m since n>=m
same for the second condition j < n. You already make sure that i>0, which ensures that j can at most be (2n+1)/2 - 1 which is smaller than n and thus automatically satisfies your condition