<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 collections import defaultdict
from collections import deque
class Solution:
    '''
    def minJumps(self, arr: List[int]) -> int:
        dic = defaultdict(list)
        for i in range(len(arr)):
            dic[arr[i]].append(i)
        visited = [0] * len(arr)
        q = deque()
        q.append((0, 0))
        visited[0] = 1
        counter = 0
        while q:
            v, numSteps = q.popleft()
            visited[v] == 1
            if v == len(arr) - 1:
                return numSteps
            for node in dic[arr[v]]:
                if visited[node] == 0:
                    visited[node] = 1
                    q.append((node, numSteps + 1))
            if v + 1 < len(arr) and visited[v + 1] == 0:
                visited[v + 1] = 1
                q.append((v + 1, numSteps + 1))
            if v - 1 >= 0 and visited[v - 1] == 0:
                visited[v - 1] = 1
                q.append((v - 1, numSteps + 1))
                '''
    def minJumps(self, arr: List[int]) -> int:

        # set pos[x][0] = True as unvisited indicator
        pos = collections.defaultdict(lambda: [True])

        # Store positions i based on value x=arr[i]
        # reverse is essential as we are finding shortest path backward
        for i, x in reversed(list(enumerate(arr))):
            pos[x].append(i)

        # res = result table, q = BFS queue
        n = len(arr)
        res = list(range(n-1, -1, -1))
        q = collections.deque()

        # start the search backward
        for k in range(n-1, -1, -1):
            q.append(k)
            while q:
                i = q.popleft()
                if pos[arr[i]][0]: # if unvisited
                    pos[arr[i]][0] = False

                    # create jump portal for all positions with same value
                    for j in pos[arr[i]][1:]:
                        res[j] = min(res[j], res[i]+1)
                        q.append(j)

                    # add adjcent posisitons to queue
                    for j in pos[arr[i]][1:]:
                        if j > 0:
                            res[j-1] = min(res[j-1], res[j]+1)
                            q.append(j-1)
                        if j < n-1:
                            res[j+1] = min(res[j+1], res[j]+1)
                            q.append(j+1)
        return res[0]





            
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