<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 collections import deque, defaultdict

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        n = len(arr)
        if n == 1:
            return 0

        # Build a mapping from value to all indices with that value
        val_indices = defaultdict(list)
        for i, val in enumerate(arr):
            val_indices[val].append(i)

        visited = set()
        queue = deque()
        queue.append((0, 0))  # (current index, steps taken)
        visited.add(0)

        while queue:
            index, steps = queue.popleft()

            # If we've reached the last index, return the steps taken
            if index == n - 1:
                return steps

            # Explore neighboring indices
            neighbors = []

            # Jump to index + 1
            if index + 1 < n and index + 1 not in visited:
                neighbors.append(index + 1)

            # Jump to index - 1
            if index - 1 >= 0 and index - 1 not in visited:
                neighbors.append(index - 1)

            # Jump to all indices with the same value
            same_value_indices = val_indices[arr[index]]
            for nei in same_value_indices:
                if nei not in visited:
                    neighbors.append(nei)

            # Mark neighbors as visited and add them to the queue
            for nei in neighbors:
                visited.add(nei)
                queue.append((nei, steps + 1))

            # Clear the list to prevent redundant processing
            val_indices[arr[index]].clear()

        # If we haven't returned by now, it's not possible to reach the end
        return -1

        
587