<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:
        n = len(arr)
        adj = defaultdict(set)
        for i, num in enumerate(arr):
            if i > 0 and i < n - 1 and arr[i] == arr[i - 1] and arr[i] == arr[i + 1]:
                continue
            adj[num].add(i)
        memo = [math.inf] * n
        memo[0] = 0
        q = deque([(0, 0)])
        while q:
            for _ in range(len(q)):
                front, cost = q.popleft()
                memo[front] = min(memo[front], cost)
                if front < n - 1 and memo[front + 1] > cost + 1:
                    q.append((front + 1, cost + 1))
                if front > 0 and memo[front - 1] > cost + 1:
                    q.append((front - 1, cost + 1))
                for neighbor in adj[arr[front]]:
                    if memo[neighbor] > cost + 1:
                        q.append((neighbor, cost + 1))
                adj[arr[front]].clear()
        return memo[-1]
                
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