LeetCode 1696: Jump Game VI Solution

Master LeetCode problem 1696 (Jump Game VI), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1696. Jump Game VI

Medium

Array Dynamic Programming Queue Heap (Priority Queue)Monotonic Queue

Problem Explanation

Explanation

To solve this problem, we will use a deque (double-ended queue) to keep track of the maximum score we can achieve at each index. We will iterate through the array and update the deque with the maximum score reachable from the current index. At each step, we will pop elements from the front of the deque that are out of the current window size k.

At the end of the iteration, the last element of the deque will hold the maximum score we can achieve to reach the last index.

Algorithm:

Initialize a deque to store the maximum scores.
Iterate through the array:
- Calculate the maximum score reachable from the current index by adding the current element with the maximum score from the previous k elements.
- Remove elements from the front of the deque that are out of the current window size k.
- Update the deque with the maximum score achievable at the current index.
Return the last element of the deque which represents the maximum score achievable to reach the last index.

Time Complexity:

The time complexity of this solution is O(n) where n is the length of the input array nums.

Space Complexity:

The space complexity is O(n) where n is the length of the input array nums.

Solution Code

import java.util.ArrayDeque;
import java.util.Deque;

class Solution {
    public int maxResult(int[] nums, int k) {
        int n = nums.length;
        Deque<Integer> deque = new ArrayDeque<>();
        deque.offerLast(0);
        for (int i = 1; i < n; i++) {
            while (deque.peekFirst() < i - k) {
                deque.pollFirst();
            }
            nums[i] += nums[deque.peekFirst()];
            while (!deque.isEmpty() && nums[i] >= nums[deque.peekLast()]) {
                deque.pollLast();
            }
            deque.offerLast(i);
        }
        return nums[n - 1];
    }
}

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