LeetCode in Kotlin
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] <= 1090 <= target <= 2 * 109
Solution
class Solution {
private class Pair(var prev: Int, var len: Int)
fun maximumJumps(nums: IntArray, target: Int): Int {
val arr = arrayOfNulls<Pair>(nums.size)
arr[0] = Pair(0, 0)
for (i in 1 until nums.size) {
arr[i] = Pair(-1, 0)
for (j in i - 1 downTo 0) {
if (Math.abs(nums[i] - nums[j]) <= target &&
arr[j]!!.prev != -1 && arr[j]!!.len + 1 > arr[i]!!.len
) {
arr[i]!!.prev = j
arr[i]!!.len = arr[j]!!.len + 1
}
}
}
return if (arr[nums.size - 1]!!.len > 0) arr[nums.size - 1]!!.len else -1
}
}