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.

**Example 1:**

Input:arr = [100,-23,-23,404,100,23,23,23,3,404]Output:3Explanation: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:0Explanation:Start index is the last index. You don't need to jump.

**Example 3:**

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

**Example 4:**

Input:arr = [6,1,9]Output:2

**Example 5:**

Input:arr = [11,22,7,7,7,7,7,7,7,22,13]Output:3

**Constraints:**

`1 <= arr.length <= 5 * 10`

^{4}`-10`

^{8}<= arr[i] <= 10^{8}

class Solution {
public int minJumps(int[] arr) {
}
}