<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 greedy_solution(self, nums):
        already_checked = {}
        next_batch = set({len(nums)-1})
        dist = 0
        warp_dict = {i: set() for i in nums}
        for i in range(len(nums)):
            warp_dict[nums[i]].add(i)
        while 0 not in already_checked:
            current_batch = next_batch
            next_batch = set()
            for i in current_batch:
                if i in already_checked:
                    continue
                already_checked[i] = dist
                if i == 0:
                    break
                if i-1 >= 0:
                    next_batch.add(i-1)
                if i+1 < len(nums)-1:
                    next_batch.add(i+1)
                if nums[i] in warp_dict:
                    next_batch.update(warp_dict[nums[i]])
                    del warp_dict[nums[i]]
            dist += 1
        return already_checked[0]
    def minJumps(self, arr: List[int]) -> int:
        return self.greedy_solution(arr)
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