LeetCode in Kotlin
1696. Jump Game VI
Medium
You are given a 0-indexed integer array nums and an integer k.
You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive.
You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array.
Return the maximum score you can get.
Example 1:
Input: nums = [1,-1,-2,4,-7,3], k = 2
Output: 7
Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
Example 2:
Input: nums = [10,-5,-2,4,0,3], k = 3
Output: 17
Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
Example 3:
Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
Output: 0
Constraints:
1 <= nums.length, k <= 105-104 <= nums[i] <= 104
Solution
import java.util.ArrayDeque
import java.util.Deque
class Solution {
fun maxResult(nums: IntArray, k: Int): Int {
val deque: Deque<IntArray> = ArrayDeque()
deque.offer(intArrayOf(0, nums[0]))
for (i in 1 until nums.size) {
val max = deque.peek()[1]
val next = intArrayOf(i, max + nums[i])
while (deque.isNotEmpty() && deque.peekLast()[1] <= next[1]) {
// PURGE FROM THE END
deque.pollLast()
}
deque.offer(next)
if (deque.peekFirst()[0] <= i - k) {
// PURGE FROM THE HEAD
deque.pollFirst()
}
}
return deque.peekLast()[1]
}
}