Problem discription:

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.

Constraints:

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

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.

The key problems for me here are how to identify the terminating condition(s):

class Solution:
    def minJumps(self, arr) -> int:
        n = len(arr)
        if n <= 1:
            return 0

        graph = {}
        for i in range(n):
            if arr[i] in graph:
                graph[arr[i]].append(i)
            else:
                graph[arr[i]] = [i]

        curs = set([0])  # store layers from start
        visited = {0, n-1}
        step = 0

        other = set([n-1]) # store layers from end

        # when current layer exists
        while curs:
            # search from the side with fewer nodes
"""
A good techinque
"""
            if len(curs) > len(other):
                curs, other = other, curs
            nex = set()

            # iterate the layer
            for node in curs:

                # check same value
                for child in graph[arr[node]]:
                    if child in other:
                        return step + 1
"""
Important here, if the neighbour node exists on the other path, that means the two paths 
have colided with each other.
"""
                    if child not in visited:
                        visited.add(child)
                        nex.add(child)

                # clear the list to prevent redundant search
                graph[arr[node]].clear()

                # check neighbors
                for child in [node-1, node+1]:
                    if child in other:
                        return step + 1
                    if 0 <= child < len(arr) and child not in visited:
                        visited.add(child)
                        nex.add(child)

            curs = nex
            step += 1

        return -1