Fastest algorithm to hop through an array

Question

Start with an array A of positive numbers. Start at index 0. From index i, you can move to index i+x for any x <= A[i]. The goal is to find the minimum number of moves needed

Answer 1

Since you can chose any x in [1,A[i]] I guess there is a pretty simple solution :

start at 0:

select the next reachable element from which you can reach the farther element. i.e chose i that maximize i+A[i+x] for x in [1,A[i]]

until you arrive at the end of the list.

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Example:

{2 , 4 , 1 , 2 , 3 , 2 , 4 , 2}

start at 0

from 0 you can get to 1 or to 2:

• from 1 you can get to 4
• from 2 you can get to 3

therefore max(0+A[0+x]) is for i = 1

chose 1 from 1 you can get to 2 3 4:

• from 4 you can get to 7
• from 3 you can get to 5
• from 2 you can get to 3

therefore max(1+A[1+x]) is for i = 4

chose 4

you can reach 7

stop

the resulting list is : 

0,1,4,7

As explained in my comments I think it's O(N), because from i you reach i+x+1 in at least 2*x operations.

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'Pseudo' proof

You start at 0 (it's optimal)

then you select i that maximize(0+A[0+x]) (i.e that maximize the reachability for the next element)

from that i you can reach any other element that is reachable from all other elements reachable from 0 (it's a long sentence, but it means : who can do more, can do less, therefore if i is not optimal,it's as good as optimal)

So i is optimal

then following step by step this reasoning, it proves the optimality of the method.

If someone knows how to formulate that more mathematically, feel free to update it.