<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 dijkstra(self, G, startingNode):
        visited = set()
        parentsMap = {}
        pq = []
        nodeCosts = defaultdict(lambda: float('inf'))
        nodeCosts[startingNode] = 0
        heapq.heappush(pq, (0, startingNode))
    
        while pq:
            # go greedily by always extending the shorter cost nodes first
            _, node = heapq.heappop(pq)
            visited.add(node)
    
            for adjNode in G[node]:
                if adjNode in visited:	continue
                    
                newCost = nodeCosts[node] + 1
                if nodeCosts[adjNode] > newCost:
                    parentsMap[adjNode] = node
                    nodeCosts[adjNode] = newCost
                    heapq.heappush(pq, (newCost, adjNode))
            
        return parentsMap, nodeCosts
        
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        intervals = sorted(intervals, key=lambda x: (x[0], x[1]))
        if len(intervals) == 1:
            return intervals
        result = []

        p1 = 0
        current = intervals[0]
        while p1 < len(intervals):
            if (intervals[p1][0] <= current[1] <= intervals[p1][1]):
                current[1] = intervals[p1][1]
            elif intervals[p1][0] > current[0] and intervals[p1][1] < current[1]:
                pass
            else:
                result.append(current)
                current = intervals[p1]

            p1 += 1

        result.append(current)

        return result

    def minJumps(self, arr: List[int]) -> int:
        i_to_j = defaultdict(list)
        for i, n in enumerate(arr):
            i_to_j[n].append([i, i+1])
        
        i_to_j = {k: self.merge(v) for (k, v) in i_to_j.items()}

        new_i_to_j = defaultdict(set)
        for k, v in i_to_j.items():
            for (a, b) in v:
                new_i_to_j[k].add(a)
                new_i_to_j[k].add(b - 1)
        i_to_j = {k:sorted(list(v)) for (k, v) in new_i_to_j.items()}

        how_many = len(set(arr))


        graph = defaultdict(list)
        for (i, n) in enumerate(arr):
            if i - 1 >= 0:
                graph[i].append(i-1)
            if i + 1 < len(arr):
                graph[i].append(i+1)
            for j in i_to_j[n][:how_many]:
                graph[i].append(j)
            for j in i_to_j[n][-how_many:]:
                graph[i].append(j)
        
        result = self.dijkstra(graph, 0)
        return result[1][len(arr) - 1]
        # return 0
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