<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:
        # use bfs
        # tc: O(n) sc:O(n)

        # create graph
        d = {}
        for i in range(len(arr)):
            if arr[i] not in d:
                d[arr[i]] = [i]
            else:
                d[arr[i]].append(i)
        
        visited = set([0])
        q = deque([0])
        step = 0

        while q:
            for _ in range(len(q)):
                # expand
                cur = q.popleft()
                if cur == (len(arr) - 1):
                    return step
                
                # generate
                for i in d[arr[cur]]:
                    if i not in visited:
                        q.append(i)
                        visited.add(i)
                d[arr[cur]] = []

                if (cur + 1) < len(arr) and (cur + 1) not in visited:
                    q.append(cur + 1)
                    visited.add(cur + 1)
                if (cur - 1) >= 0 and (cur - 1) not in visited:
                    q.append(cur - 1)
                    visited.add(cur - 1)
            step += 1

        return -1
        
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