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Kotlin-based LeetCode algorithm problem solutions, regularly updated
1340\. Jump Game V
Hard
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.
**Constraints:**
* `1 <= arr.length <= 1000`
* 1 <= arr[i] <= 105
* `1 <= d <= arr.length`