##### 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

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