Short Problem Definition:
Find the earliest time when a frog can jump to the other side of a river.
Link
Complexity:
expected worst-case time complexity is O(N)
expected worst-case space complexity is O(X)
Execution:
Mark seen elements as such in a boolean array. I do not like the idea of returning the first second as 0. But specifications are specifications :)
Solution:
def solution(X, A):
passable = [False] * X
uncovered = X
for idx in xrange(len(A)):
if A[idx] <= 0 or A[idx] > X:
raise Exception("Invalid value", A[idx])
if passable[A[idx]-1] == False:
passable[A[idx]-1] = True
uncovered -= 1
if uncovered == 0:
return idx
return -1