Codility经典算法题之三十三：FibFrog

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本文探讨了一只青蛙如何利用斐波那契数列跳跃过河的问题。青蛙从河的一边出发，目标到达另一边，它只能跳到有叶子的位置上，并且每次跳跃的距离必须是斐波那契数列中的一个数。文章提供了一个解决方案，通过计算最小跳跃次数来帮助青蛙成功过河。

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Task Description：

The Fibonacci sequence is defined using the following recursive formula:

F(0) = 0 F(1) = 1 F(M) = F(M - 1) + F(M - 2) if M >= 2

A small frog wants to get to the other side of a river. The frog is initially located at one bank of the river (position −1) and wants to get to the other bank (position N). The frog can jump over any distance F(K), where F(K) is the K-th Fibonacci number. Luckily, there are many leaves on the river, and the frog can jump between the leaves, but only in the direction of the bank at position N.

The leaves on the river are represented in an array A consisting of N integers. Consecutive elements of array A represent consecutive positions from 0 to N − 1 on the river. Array A contains only 0s and/or 1s:

0 represents a position without a leaf;
1 represents a position containing a leaf.

The goal is to count the minimum number of jumps in which the frog can get to the other side of the river (from position −1 to position N). The frog can jump between positions −1 and N (the banks of the river) and every position containing a leaf.

For example, consider array A such that:

A[0] = 0 A[1] = 0 A[2] = 0 A[3] = 1 A[4] = 1 A[5] = 0 A[6] = 1 A[7] = 0 A[8] = 0 A[9] = 0 A[10] = 0

The frog can make three jumps of length F(5) = 5, F(3) = 2 and F(5) = 5.

Write a function:

class Solution { public int solution(int[] A); }

that, given an array A consisting of N integers, returns the minimum number of jumps by which the frog can get to the other side of the river. If the frog cannot reach the other side of the river, the function should return −1.

For example, given:

A[0] = 0 A[1] = 0 A[2] = 0 A[3] = 1 A[4] = 1 A[5] = 0 A[6] = 1 A[7] = 0 A[8] = 0 A[9] = 0 A[10] = 0

the function should return 3, as explained above.

Assume that:

N is an integer within the range [0..100,000];
each element of array A is an integer that can have one of the following values: 0, 1.

Complexity:

expected worst-case time complexity is O(N*log(N));
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Solution：

def fibonacci(n):
fib = [1,1]
while fib[-1] < n:
fib.append(fib[-1] + fib[-2])
return fib[1:-1]

def solution(A):
jumps = fibonacci(len(A) + 2)

A.insert(0,1)#insert 1 into the first place
A.append(1) #insert 1 into the last place
lenA = len(A)

reach = [0] * lenA

for pos in range(1,lenA):
jumpmin = lenA
for jump in jumps: #[1, 2, 3, 5, 8]
leftsteps = pos - jump
if leftsteps >= 0:
if A[leftsteps] == 1 and reach[leftsteps] + 1 < jumpmin:
jumpmin = reach[leftsteps] + 1
else:
break

reach[pos] = jumpmin

return reach[-1] if reach[-1] != lenA else -1