<p>Given an array of&nbsp;integers <code>arr</code>, you are initially positioned at the first index of the array.</p>

<p>In one step you can jump from index <code>i</code> to index:</p>

<ul>
	<li><code>i + 1</code> where:&nbsp;<code>i + 1 &lt; arr.length</code>.</li>
	<li><code>i - 1</code> where:&nbsp;<code>i - 1 &gt;= 0</code>.</li>
	<li><code>j</code> where: <code>arr[i] == arr[j]</code> and <code>i != j</code>.</li>
</ul>

<p>Return <em>the minimum number of steps</em> to reach the <strong>last index</strong> of the array.</p>

<p>Notice that you can not jump outside of the array at any time.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [100,-23,-23,404,100,23,23,23,3,404]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You need three jumps from index 0 --&gt; 4 --&gt; 3 --&gt; 9. Note that index 9 is the last index of the array.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [7]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Start index is the last index. You do not need to jump.
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>8</sup> &lt;= arr[i] &lt;= 10<sup>8</sup></code></li>
</ul>

import heapq

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        val2idx_unvisited = dict()
        for i, val in enumerate(arr):
            if val not in val2idx_unvisited:
                val2idx_unvisited[val] = set()
            val2idx_unvisited[val].add(i)
        
        # Finding Optimal Path
        # min_dist = [math.inf] * len(arr)
        # min_dist[0] = 0
        # heap = [(0, 0)]  # (dist, idx)
        # visited = [False] * len(arr)
        # while len(heap) > 0:
        #     dist_u, u = heapq.heappop(heap)
        #     visited[u] = True
        #     # print('visit', u, 'with dist', dist_u)
        #     # print(min_dist)
        #     # print(heap)
        #     if u == len(arr) - 1:
        #         return dist_u
        #     if u < len(arr) - 1 \
        #         and dist_u + 1 < min_dist[u + 1] \
        #         and not visited[u + 1]:
        #             min_dist[u + 1] = dist_u + 1
        #             heapq.heappush(heap, (dist_u + 1, u + 1))
        #     if u > 0 \
        #         and dist_u + 1 < min_dist[u - 1] \
        #         and not visited[u - 1]:
        #             min_dist[u - 1] = dist_u + 1
        #             heapq.heappush(heap, (dist_u + 1, u - 1))
        #     if arr[u] in val2idx_unvisited:
        #         for v in val2idx_unvisited[arr[u]]:
        #             if v == u:
        #                 continue

        #             # assert not visited[v], f'{u}, {v} violates'
        #             if dist_u + 1 < min_dist[v]:
        #                 min_dist[v] = dist_u + 1
        #                 heapq.heappush(heap, (dist_u + 1, v))
        #         del val2idx_unvisited[arr[u]]

        # ==========================

        # BFS suits better since each edge has the same length
        # later updates cannot give a shorter distance

        queue = [0]
        visited = [False] * len(arr)
        visited[0] = True
        dist = [math.inf] * len(arr)
        dist[0] = 0
        while len(queue) > 0:
            u = queue.pop(0)
            dist_u = dist[u]
            # print(dist)

            if u == len(arr) - 1:
                return dist_u
            if u < len(arr) - 1 \
                and not visited[u + 1]:
                    dist[u + 1] = dist_u + 1
                    visited[u + 1] = True
                    queue.append(u + 1)
            if u > 0 \
                and not visited[u - 1]:
                    dist[u - 1] = dist_u + 1
                    visited[u - 1] = True
                    queue.append(u - 1)
            if arr[u] in val2idx_unvisited:
                for v in val2idx_unvisited[arr[u]]:
                    if v == u or visited[v]:
                        continue
                    dist[v] = dist_u + 1
                    visited[v] = True
                    queue.append(v)
                del val2idx_unvisited[arr[u]]

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