human
Solve in Python:
Given an array of integers arr, you are initially positioned at the first index of the array.
In one step you can jump from index i to index:

i + 1 where: i + 1 < arr.length.
i - 1 where: i - 1 >= 0.
j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.
Notice that you can not jump outside of the array at any time.
 
Example 1:
Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2:
Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You don't need to jump.

Example 3:
Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.

Example 4:
Input: arr = [6,1,9]
Output: 2

Example 5:
Input: arr = [11,22,7,7,7,7,7,7,7,22,13]
Output: 3

 
Constraints:

1 <= arr.length <= 5 * 10^4
-10^8 <= arr[i] <= 10^8
assistant
#Here is the Python code you requested.

import collections
class Solution:
    def minJumps(self, arr: List[int]) -> int:
        if len(arr) == 1:
            return 0
        
        d = collections.defaultdict(list)
        for index, element in enumerate(arr):
            d[element].append(index)

        queue = collections.deque([(0, 0)])
        s = set()
        s.add(0)

        while len(queue):
            currIndex, jumps = queue.popleft()

            for nextIndex in [currIndex + 1, currIndex - 1] + d[arr[currIndex]][::-1]:
                if nextIndex < len(arr) and nextIndex > -1 and nextIndex != currIndex and nextIndex not in s:
                    if nextIndex == len(arr) - 1:
                        return jumps + 1
                    s.add(nextIndex)
                    queue.append((nextIndex, jumps + 1))

        return -1