<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:
        queue = deque([0])
        mapp = defaultdict(list)
        n = len(arr)
        visited = set()
        jump = 0

        for i in range(n):
            mapp[arr[i]].append(i)
        
        while(queue):
            # Process all nodes (indices) at the current level
            for _ in range(len(queue)):
                curr_index = queue.popleft()

                # If we reached the last index, return the number of jumps
                if curr_index == n - 1:
                    return jump
                
                # Try to move to index curr_index - 1 and curr_index + 1
                for next_index in [curr_index - 1, curr_index + 1]:
                    if 0 <= next_index < n and next_index not in visited:
                        visited.add(next_index)
                        queue.append(next_index)

                # Check all indices with the same value as arr[curr_index]
                val = arr[curr_index]
                if val in mapp:  # If the value has corresponding indices
                    for next_index in mapp[val]:
                        if next_index not in visited:
                            visited.add(next_index)
                            queue.append(next_index)
                    # Clear the list to prevent redundant future checks
                    del mapp[val]

            # Increment jump counter after processing all nodes at the current level
            jump += 1
        return jump
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