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# Jump Game V

Given an array of integers `arr` and an integer `d`. In one step you can jump from index `i` to index:

* `i + x` where: `i + x < arr.length` and `0 < x <= d`.
* `i - x` where: `i - x >= 0` and `0 < x <= d`.

In addition, you can only jump from index `i` to index `j` if `arr[i] > arr[j]` and `arr[i] > arr[k]` for all indices `k` between `i` and `j` (More formally `min(i, j) < k < max(i, j)`).

You can choose any index of the array and start jumping. Return *the maximum number of indices* you can visit.

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

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/01/23/meta-chart.jpeg)

```
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
```

**Example 2:**

```
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
```

**Example 3:**

```
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies. 
```

**Example 4:**

```
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
```

**Example 5:**

```
Input: arr = [66], d = 1
Output: 1
```

**Constraints:**

* `1 <= arr.length <= 1000`
* `1 <= arr[i] <= 10^5`
* `1 <= d <= arr.length`

```java
class Solution {
    static class Pair {
        int h, i;

        // h -> height, i-> index
        Pair(int h, int i) {
            this.h = h;
            this.i = i;
        }
    }

    public int maxJumps(int[] nums, int d) {
        Pair[] arr = new Pair[nums.length];
        for (int i = 0; i < nums.length; i++)
            arr[i] = new Pair(nums[i], i);
        // Sorting on basis of height
        // This helps in running our recursion from less dependent variables to more
        // dependent variables
        Arrays.sort(arr, (a, b) -> a.h - b.h);
        int[] dp = new int[nums.length];
        Arrays.fill(dp, -1);
        int max = 0;
        for (int i = 0; i < arr.length; i++) {
            int index = arr[i].i;
            // Base case -> If we start at a index, we can count it as 1
            dp[index] = 1;
            int lower = Math.max(0, index - d);
            int upper = Math.min(nums.length - 1, index + d);
            // Making sure all the buildings are lower, if we were to make a jump
            for (int j = index - 1; j >= lower && nums[j] < nums[index]; j--)
                if (dp[j] != -1)
                    dp[index] = Math.max(dp[index], dp[j] + 1);
            for (int j = index + 1; j <= upper && nums[j] < nums[index]; j++)
                if (dp[j] != -1)
                    dp[index] = Math.max(dp[index], dp[j] + 1);
            max = Math.max(max, dp[index]);
        }
        return max;
    }
}
```