<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:
        return minJumps(arr)


def minJumps( arr: List[int]) -> int:
    """
    - Group nodes by node value
    - Consider each node and each group at most once

    Consider group only once won't lose the ability to find optimal solution.
    Also, limit each group once reduce TC from N^2 to N
    """

    size = len(arr)

    touched: set[int] = set()  # nodes
    used: set[int] = set()  # groups

    group_of: dict[int, set[int]] = defaultdict(set)
    for idx, val in enumerate(arr):
        group_of[val].add(idx)

    def neighbor_of(idx: int) -> int:
        yield from (i for i in [idx - 1, idx + 1] if i in range(size))

    def by_group(val: int) -> Iterator[int]:
        if val in used:
            return
        used.add(val)
        for next_ in group_of[val]:
            yield next_
            
    # level and node ID (array index)
    pending: deque[tuple[int, int]] = deque([(0, 0)])
    while pending:
        level, node = pending.popleft()
        if node == size - 1:
            return level

        candidates = itertools.chain(by_group(arr[node]), neighbor_of(node))
        for next_ in candidates:
            if next_ in touched:
                continue
            pending.append((level + 1, next_))
            touched.add(next_)


    assert False

801