Find a maximal set of non((-))overlapping segments.
expected worst-case time complexity is O(N)
expected worst-case space complexity is O(N)
This can be solved by using greedy search. The beginning of the next segment must come strictly after its predecessor.
def solution(A, B):
if len(A) < 1:
return 0
cnt = 1
prev_end = B[0]
for idx in xrange(1, len(A)):
if A[idx] > prev_end:
cnt += 1
prev_end = B[idx]
return cnt