<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 Solution1:
    def getNeighbors(self, nums, i, indexMapping):
        neighbors = set()
        neighbors.add(i+1)
        neighbors.update(indexMapping[nums[i]])
        neighbors.remove(i)
        if i > 0:
            neighbors.add(i-1)

        return neighbors
    
    def minJumps1(self, arr: List[int]) -> int:
        """
        """
        if len(arr) == 1:
            return 0

        indexMapping = collections.defaultdict(list)
        for i,n in enumerate(arr):
            indexMapping[n].append(i)

        level = 0
        seen = set([0])
        bfs = deque([0])
        while bfs:
            levelSize = len(bfs)
            for i in range(levelSize):
                currentIndex = bfs.popleft()
                neighbors = self.getNeighbors(arr, currentIndex, indexMapping)
                for neighborIndex in neighbors:
                    if neighborIndex not in seen:
                        if neighborIndex == len(arr)-1:
                            return level + 1

                        seen.add(neighborIndex)
                        bfs.append(neighborIndex)

            level += 1

        # Added for completeness, there is always a solution since worse case, just advance one
        # step at at time.
        return -1

    def minJumps2(self, arr: List[int]) -> int:
        """
        I need to figure out the correct heuristic, it's a combination of level and
        reach, but not sure how.
        """
        if len(arr) == 1:
            return 0

        indexMapping = collections.defaultdict(list)
        for i,n in enumerate(arr):
            indexMapping[n].append(i)

        nodeScore = dict()
        for i in range(len(arr)):
            nodeScore[i] = max(self.getNeighbors(arr, i, indexMapping))

        seen = set([0])
        bfsHeap = [(0, -nodeScore[0], 0)]
        while bfsHeap:
            currentLevel, _, currentIndex = heapq.heappop(bfsHeap)
            neighbors = self.getNeighbors(arr, currentIndex, indexMapping)
            for neighborIndex in neighbors:
                if neighborIndex not in seen:
                    if neighborIndex == len(arr)-1:
                        return currentLevel + 1

                    seen.add(neighborIndex)
                    heapq.heappush(bfsHeap, (currentLevel+1, -nodeScore[neighborIndex], neighborIndex))

        # Added for completeness, there is always a solution since worse case, just advance one
        # step at at time.
        return -1
        
class Solution:
    def minJumps(self, arr: List[int]) -> int:
        n = len(arr)
        if n <= 1:
            return 0

        graph = {}
        for i in range(n):
            if arr[i] in graph:
                graph[arr[i]].append(i)
            else:
                graph[arr[i]] = [i]

        curs = [0]  # store current layers
        visited = {0}
        step = 0

        # when current layer exists
        while curs:
            nex = []

            # iterate the layer
            for node in curs:
                # check if reached end
                if node == n-1:
                    return step

                # check same value
                for child in graph[arr[node]]:
                    if child not in visited:
                        visited.add(child)
                        nex.append(child)

                # clear the list to prevent redundant search
                graph[arr[node]].clear()

                # check neighbors
                for child in [node-1, node+1]:
                    if 0 <= child < len(arr) and child not in visited:
                        visited.add(child)
                        nex.append(child)

            curs = nex
            step += 1

        return -1
530