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Suppose we have a list of sorted numbers called stones and this is representing the positions of stones on a river that we are trying to cross. To cross the river, we must finish at the last stone. Now in each step, we can jump (k - 1, k, or k + 1) steps ahead where k is the distance of the last jump. We have to check whether we can cross the river or not.

So, if the input is like stones = [0, 1, 3, 4, 5, 6, 8, 9, 13], then the output will be True, as we can start from 0, then jump 1 unit to go stone 1, then 2 units to go 3, after that 2 units to 5, then 3 units to go 8, and finally 5 units to go 13, and this is the final place.

To solve this, we will follow these steps −

- start := A[0], end := last element of A
- A := a set of all unique elements of A
- Define a function check() . This will take pos:= start, prev:= 0
- if pos is same as end, then
- return True

- for each jump in [prev - 1, prev, prev + 1], do
- if jump >= 1, then
- next_pos := jump + pos
- if next_pos is in A and check(next_pos, jump) is true, then
- return True

- if jump >= 1, then
- return False
- From the main method call check() and return result

Let us see the following implementation to get better understanding −

class Solution: def solve(self, A): start, end = A[0], A[-1] A = set(A) def check(pos=start, prev=0): if pos == end: return True for jump in [prev - 1, prev, prev + 1]: if jump >= 1: next_pos = jump + pos if next_pos in A and check(next_pos, jump): return True return False return check() ob = Solution() stones = [0, 1, 3, 4, 5, 6, 8, 9, 13] print(ob.solve(stones))

[0, 1, 3, 4, 5, 6, 8, 9, 13]

True

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