Classic Dynamic Programming VI
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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