# 1345. Jump Game IV

#### Hard

***

Given an array of integers `arr`, you are initially positioned at the first index of the array.

In one step you can jump from index `i` to index:

* `i + 1` where: `i + 1 < arr.length`.
* `i - 1` where: `i - 1 >= 0`.
* `j` where: `arr[i] == arr[j]` and `i != j`.

Return *the minimum number of steps* to reach the **last index** of the array.

Notice that you can not jump outside of the array at any time.

&#x20;

**Example 1:**

```
Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.
```

**Example 2:**

```
Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You do not need to jump.
```

**Example 3:**

```
Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.
```

&#x20;

**Constraints:**

* `1 <= arr.length <= 5 * 104`
* `-108 <= arr[i] <= 108`

```python
from collections import defaultdict
class Solution:
    def minJumps(self, arr: List[int]) -> int:
        d = defaultdict(list)
        _ = [d[x].append(i) for i, x in enumerate(arr)]
        q = [(0,0)]
        visited = {
            
        }
        while q:
            pos, step = q.pop(0)
            if pos == (len(arr)-1):
                return step
            if pos-1 >= 0 and (pos-1) not in visited:
                visited[pos-1] = 1
                q.append((pos-1, step+1))
            if pos+1 <= len(arr) and (pos+1) not in visited:
                visited[pos+1] = 1
                q.append((pos+1, step+1))
            for index in d[arr[pos]]:
                if index not in visited:
                    visited[index] = 1
                    q.append((index, step+1))
            d[arr[pos]] = list()
        return -1
```