<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>

#snitch,hangover,

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        num_to_idx = collections.defaultdict(list)
        for i, num in enumerate(arr):
            num_to_idx[num].append(i)

        visited = set()

        q = collections.deque()
        q.append((0, 0)) #index,steps

        while q:
            index, distance = q.popleft()
            if index == len(arr) - 1:
                return distance 
            visited.add(index)
            value = arr[index]
            for nei in num_to_idx[value]:
                if nei not in visited:
                    visited.add(nei)
                    q.append((nei, distance + 1))

            if index + 1 < len(arr) and index + 1 not in visited:
                q.append((index + 1, distance + 1))


            if index - 1 >= 0 and index - 1 not in visited:
                q.append((index - 1, distance + 1))
            
            num_to_idx[value].clear() # why we need t clear this 
            # for example we have {100,100,....,100},Now we had already process the 100 we will removing from the map. So, that we do not process this again, there will be cases where we can get a tle if we keep the processes index in map for example if all these are 100. We'll keep on processing the i - 1 i + 1 i - 1 i + 1 and every other index multiple no of times and we'll get a tle. So, for that reason we are just removing from our map.

        return -1 

# class Solution:
#     def minJumps(self, arr: List[int]) -> int:
#         if len(arr) == 1:
#             return 0
#         stack = deque([0])
#         graph = defaultdict(list)
#         for idx, num in enumerate(arr):
#             graph[num].append(idx)

#         steps = 0
#         while stack:
#             steps += 1
#             for _ in range(len(stack)):
#                 idx = stack.popleft()
#                 if idx - 1 >= 0 and arr[idx - 1] in graph:
#                     stack.append(idx - 1)

#                 if idx + 1 == len(arr) - 1:
#                     return steps

#                 if idx + 1 < len(arr) and arr[idx + 1] in graph:
#                     stack.append(idx + 1)

#                 for jump_idx in graph.pop(arr[idx], []):
#                     if jump_idx == len(arr) - 1:
#                         return steps
#                     if jump_idx != idx:
#                         stack.append(jump_idx)

#         return steps
28963