Classic Dynamic Programming VI

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Another DP, this time doing a simple N^2 traversal storing the maximum jumps per index. N=10^3 so it is very tractable. Code is down below, ACC.

Maximum Number of Jumps to Reach the Last Index - LeetCode

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
Accepted
11,003
Submissions
45,387

public int MaximumJumps(int[] nums, int target)
{
    int[] maxPerIndex = new int[nums.Length];

    maxPerIndex[0] = 0;
    for (int i = 1; i < nums.Length; i++)
    {
        maxPerIndex[i] = -1;
        for (int j = 0; j < i; j++)
        {
            if (maxPerIndex[j] >= 0 &&
                Math.Abs(nums[i] - nums[j]) <= target && 
                maxPerIndex[j] + 1 > maxPerIndex[i])
            {
                maxPerIndex[i] = maxPerIndex[j] + 1;
            }
        }
    }

    return maxPerIndex[maxPerIndex.Length - 1];
}

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