2770

2770. Maximum Number of Jumps to Reach the Last Index (Medium)

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

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Solution 1. DP

// OJ: https://leetcode.com/problems/maximum-number-of-jumps-to-reach-the-last-index
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int maximumJumps(vector<int>& A, int target) {
        int N = A.size();
        vector<int> dp(N, -1);
        dp[0] = 0;
        for (int i = 1; i < N; ++i) {
            for (int j = i - 1; j >= 0; --j) {
                if (dp[j] == -1 || abs(A[i] - A[j]) > target) continue;
                dp[i] = max(dp[i], dp[j] + 1);
            }
        }
        return dp[N - 1];
    }
};