<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:

        # 100 => [0, 4]
        valueMap = collections.defaultdict(set)
        que = collections.deque()

        for i in range(len(arr)):
            n = arr[i]
            valueMap[n].add(i)

        que.append((0, 0))
        visited = set()
        while len(que) > 0:
            i, steps = que.popleft()
            visited.add(i)

            if i == len(arr) - 1:
                return steps

            childrenIndexes = valueMap[arr[i]]
            childrenIndexes.add(i-1)
            childrenIndexes.add(i+1)

            for chi in childrenIndexes:
                if -1 < chi < len(arr) and chi not in visited:
                    que.append((chi, steps+1))
                    visited.add(chi)

            valueMap[arr[i]].clear()
        return 0
"""

  0 => [1, 4]
  1 => [0, 2]
  2 => [1, 3]
  3 => [2, 4, 9]
  4 => [3, 5, 0]
  5 => [3, 4, 7]
  6 => [5, 7]
  7 => [6, 8, 5]
  8 => [7, 9]
  9 => [8, 3]

"""
644