# 1344. 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`

对于数组arr和整数d，能从任何一个点开始，每步能最远跳到j 属于\[i - d, i + d]，且(i, j]中任何一个点arr\[k]不能小于等于arr\[i]，问这么走最多能覆盖多少元素。每步只能从值高走到低，因此对arr做排序并从小到大遍历，dp\[i]表示从i出发最多能覆盖多少点，对i的action为从\[i - d, i + d]找dp最多的点j，让i蹦到j并以此更新dp\[i]。

时间复杂度O(NlgN + ND)。也可以用top-bottom 的DFS + memo，把dfs的中间结果存在dp里，时间复杂度O(ND)。

参考自[这](https://link.zhihu.com/?target=https%3A//leetcode.com/problems/jump-game-v/discuss/496620/Python-DP)。

```cpp
class Solution {
public:
    int maxJumps(vector<int>& arr, int d) {
        const int N = arr.size();
        vector<int> dp(N, 1);
        vector<vector<int>> nums(N, vector<int>(2));
        for (int i = 0; i < N; ++i) {
            nums[i][0] = arr[i];
            nums[i][1] = i;
        }
        sort(nums.begin(), nums.end(), [](const auto &a, const auto &b) {
            return a[0] < b[0];
        });
        int res = 0;
        for (const auto &num : nums) {
            int i = num[1];
            for (int j = i + 1; j <= min(i + d, N - 1); ++j) {
                if (arr[j] >= arr[i]) break;
                dp[i] = max(dp[i], dp[j] + 1);
            } 
            for (int j = i - 1; j >= max(i - d, 0); --j) {
                if (arr[j] >= arr[i]) break;
                dp[i] = max(dp[i], dp[j] + 1);
            } 
            res = max(res, dp[i]);
        }
        return res;
    }
};
```


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