Jump game iv

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        n = len(arr)
        if n <= 1:
            return 0
        g = collections.defaultdict(list)
        for i in range(len(arr)):
            g[arr[i]].append(i)

        q = collections.deque([0])
        visited = set()
        steps = 0

        while q:
            size = len(q)
            for i in range(size):
                curr = q.popleft()
                if curr == n - 1:
                    return steps
                nextNodes = g[arr[curr]]
                for j in range(len(nextNodes)):
                    if nextNodes[j] not in visited:
                        visited.add(nextNodes[j])
                        q.append(nextNodes[j])
                nextNodes.clear()
                for j in [curr - 1, curr + 1]:
                    if 0 <= j < len(arr) and j not in visited:
                        visited.add(j)
                        q.append(j)
            steps += 1
        return -1

Jump Game IV

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 do not 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.

Constraints:

1 <= arr.length <= 5 * 10⁴
-10⁸ <= arr[i] <= 10⁸