<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:
        # memo = dict()
        # def dp(i):
            
        #     #Base cases
        #     if i == len(arr)-1:
        #         return 0
        #     elif i in memo:
        #         return memo[i]
            
        #     #Generate memo entries
        #     memo[i] = float("inf") #If we reach this state again in subs, don't reconsider
        #     left = right = special = float("inf")

        #     #Left
        #     if (i-1) >= 0:
        #         left = 1+dp(i-1) #We are counting steps here

        #     #Right
        #     if (i+1) < len(arr):
        #         right = 1 + dp(i+1)

        #     #Special
        #     for j in range(len(arr)): #Propogate forward from the whole list
        #         if arr[i] == arr[j] and i != j:
        #             special = min(special, 1+dp(j)) #Update if shorter distance found

        #     memo[i] = min(left, right, special)
        #     return memo[i]

        # return dp(0)

        # Precompute the indices of elements with the same value
        if len(arr) == 1:
            return 0

        # Map value -> list of indices where that value appears
        value_indices = {}
        for i in range(len(arr)):
            if arr[i] in value_indices:
                value_indices[arr[i]].append(i)
            else:
                value_indices[arr[i]] = [i]

        # BFS setup
        queue = [(0, 0)]  # (index, number of steps)
        visited = [False] * len(arr)
        visited[0] = True

        while queue:
            index, steps = queue.pop(0)

            # Check if we've reached the last index
            if index == len(arr) - 1:
                return steps

            # Explore adjacent nodes: left and right
            if index - 1 >= 0 and not visited[index - 1]:
                visited[index - 1] = True
                queue.append((index - 1, steps + 1))

            if index + 1 < len(arr) and not visited[index + 1]:
                visited[index + 1] = True
                queue.append((index + 1, steps + 1))

            # Explore all indices with the same value
            if arr[index] in value_indices:
                for next_index in value_indices[arr[index]]:
                    if not visited[next_index]:
                        visited[next_index] = True
                        queue.append((next_index, steps + 1))

                # After processing, we clear the list to avoid redundant checks
                value_indices[arr[index]] = []

        return -1  # In case no solution is found (shouldn't happen in valid inputs)
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