<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>

from queue import Queue

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        buckets = {}
        
        for i in range(len(arr)):
            if arr[i] not in buckets:
                buckets[arr[i]] = [i]
            else:
                buckets[arr[i]].append(i)
        # print(buckets)
        q = Queue()
        q.put((0,0))
        visited = set()
        while q:
            i, depth = q.get()
            # print(i,depth)
            visited.add(i)
            if i==len(arr)-1:
                return depth

            for j in buckets[arr[i]]:
                if j != i:
                    if j not in visited:
                        q.put((j,depth+1))
                    else:
                        break
            
            if i< len(arr)-1 and i+1 not in visited:
                q.put((i+1,depth+1))
            if i > 0 and i-1 not in visited:
                q.put((i-1, depth+1))
            

1356