<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>

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        idxes: Dict[int, List[int]] = dict()
        for idx, a in enumerate(arr):
            if a not in idxes:
                idxes[a] = list()
            idxes[a].append(idx)
            
        shortest = [float('inf') for _ in range(len(arr))]
        visited = set()
        pq = [(0, 0)] # (cost, index)
        while pq:
            cost, i = heapq.heappop(pq)
            if i == len(arr) - 1:
                return cost
            if i - 1 >= 0 and cost+1 < shortest[i-1]:
                heapq.heappush(pq, (cost+1, i-1))
                shortest[i-1] = min(shortest[i-1], cost+1)
            if i + 1 < len(arr) and cost+1 < shortest[i+1]:
                heapq.heappush(pq, (cost+1, i+1))
                shortest[i+1] = min(shortest[i+1], cost+1)
            if arr[i] not in visited:
                for neigh in idxes[arr[i]]:
                    if neigh == i:
                        continue
                    if cost+1 < shortest[neigh]:
                        heapq.heappush(pq, (cost+1, neigh))
                        shortest[neigh] = min(shortest[neigh], cost+1)
            visited.add(arr[i])
        return -1
758