<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:
        # Time limit exceeded
        if not arr or len(arr) == 1:
            return 0
        idx = {}
        # {100:[index, index]}
        n = len(arr)
        for i in range(n):
            if arr[i] not in idx:
                idx[arr[i]] = []
            idx[arr[i]].append(i)
        queue = [(arr[0], 0, 0)]
        # num, index, times
        seen = set()
        while queue:
            num, loc, depth = queue.pop(0)
            seen.add(loc)
            for i in [-1, 1]:
                new_loc = loc + i
                if new_loc not in seen:
                    if new_loc == n-1:
                        return depth+1
                    elif 0 <= new_loc < n:
                        seen.add(new_loc)
                        queue.append((arr[new_loc], new_loc, depth +1))
            for j in idx[num]:
                if j not in seen:
                    loc_new = j
                    seen.add(j)
                    if loc_new == n-1:
                        return depth+1 
                    else:
                        queue.append((arr[j], loc_new, depth+1))  
            idx[num] = []

        # h={}
        # for i,e in enumerate(arr):
        #     if e not in h:
        #         h[e] = []
        #     h[e].append(i)
        # q = [(0,0)]
        # while q:
        #     n,d = q.pop(0)
        #     if n == len(arr)-1:
        #         return d
        #     if n+1 == len(arr)-1:
        #         return d+1
        #     if n+1 < len(arr) and h.get(arr[n+1]):
        #         q.append((n+1,d+1))
        #     if n-1 >= 0 and h.get(arr[n-1]):
        #         q.append((n-1,d+1))
        #     for i in h[arr[n]]:
        #         if i != n:
        #             q.append((i,d+1))
        #         if i == len(arr)-1:
        #             return d+1
        #     h[arr[n]] = []

                



                

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