<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 functools import cache
from collections import defaultdict

from collections import deque, defaultdict
from typing import List

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        if len(arr) == 1:
            return 0
        
        # Create a map from value to indices where that value occurs
        graph = defaultdict(list)
        for i, val in enumerate(arr):
            graph[val].append(i)
        
        # BFS setup
        q = [(0, 0)]  # (index, steps)
        visited = set([0])
        
        while q:
            index, steps = q.pop(0)
            
            # Explore the three possible moves
            
            # 1. Move to index + 1
            if index + 1 < len(arr) and index + 1 not in visited:
                if index + 1 == len(arr) - 1:
                    return steps + 1
                visited.add(index + 1)
                q.append((index + 1, steps + 1))
            
            # 2. Move to index - 1
            if index - 1 >= 0 and index - 1 not in visited:
                visited.add(index - 1)
                q.append((index - 1, steps + 1))
            
            # 3. Move to any index j where arr[i] == arr[j]
            for j in reversed(graph[arr[index]]):
                if j != index and j not in visited:
                    if j == len(arr) - 1:
                        return steps + 1
                    visited.add(j)
                    q.append((j, steps + 1))
            
            # Clear the list to prevent future unnecessary checks
            graph[arr[index]].clear()
        
        return -1  # just in case, though we should always reach the last index
843