Problem Statement:

You are given a 0-indexed array nums of n integers and an integer target.

You are initially positioned at index 0. In one step, you can jump from index i to any index j such that:

0 <= i < j < n
-target <= nums[j] - nums[i] <= target

Return the maximum number of jumps you can make to reach index n - 1.

If there is no way to reach index n - 1, return -1.

Example 1:

Input: nums = [1,3,6,4,1,2], target = 2
Output: 3
Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:
- Jump from index 0 to index 1. 
- Jump from index 1 to index 3.
- Jump from index 3 to index 5.
It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 3 jumps. Hence, the answer is 3.

Example 2:

Input: nums = [1,3,6,4,1,2], target = 3
Output: 5
Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:
- Jump from index 0 to index 1.
- Jump from index 1 to index 2.
- Jump from index 2 to index 3.
- Jump from index 3 to index 4.
- Jump from index 4 to index 5.
It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 5 jumps. Hence, the answer is 5.

Example 3:

Input: nums = [1,3,6,4,1,2], target = 0
Output: -1
Explanation: It can be proven that there is no jumping sequence that goes from 0 to n - 1. Hence, the answer is -1.

Constraints:

2 <= nums.length == n <= 1000
-10⁹ <= nums[i] <= 10⁹
0 <= target <= 2 * 10⁹

Solution:

First step in such problem is to write a recursive function and check validity for small inputs.

 
class Solution {
public:
    int maximumJumps(vector<int>& nums, int target, int cur=0) 
    {
        if (cur==nums.size()-1) return 0;
        int res = -1;
        for (int i=cur+1; i<nums.size(); i++) 
            if (abs(nums[i]-nums[cur])<=target && maximumJumps(nums, target, i)>=0)
                res = max(res, 1+maximumJumps(nums, target, i));
        return res;
    }
};

Next step is to add memoization:

 
class Solution {
   vector<int> memo = vector<int>(10001,-2);
public:
    int maximumJumps(vector<int>& nums, int target, int cur=0) 
    {
        if (cur==nums.size()-1) return 0;
        if (memo[cur]!=-2) return memo[cur];
        int res = -1;
        for (int i=cur+1; i<nums.size(); i++) if (abs(nums[i]-nums[cur])<=target)
        {
            int res_i = maximumJumps(nums, target, i);
            if (res_i!=-1) res = max(res, 1+res_i);
        }            
        return memo[cur] = res;
    }
};

We get AC at this point.

Bottom-up DP is another way to get AC. But sometimes this might be difficult.

 
class Solution {
public:
    int maximumJumps(vector<int>& nums, int target) 
    {
        int n = nums.size();
        vector<int>dp(n,-1);
        dp[n-1] = 0;
        for (int i=n-2; i>=0; i--)
        {
            for (int j=i+1; j<n; j++) 
                if (abs(nums[j]-nums[i])<=target && dp[j]>=0)
                    dp[i] = max(dp[i], 1+dp[j]);                
        }
        return dp[0];
    }
};