<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>k</code>.</p>

<p>You are initially standing at index <code>0</code>. In one move, you can jump at most <code>k</code> steps forward without going outside the boundaries of the array. That is, you can jump from index <code>i</code> to any index in the range <code>[i + 1, min(n - 1, i + k)]</code> <strong>inclusive</strong>.</p>

<p>You want to reach the last index of the array (index <code>n - 1</code>). Your <strong>score</strong> is the <strong>sum</strong> of all <code>nums[j]</code> for each index <code>j</code> you visited in the array.</p>

<p>Return <em>the <strong>maximum score</strong> you can get</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [<u>1</u>,<u>-1</u>,-2,<u>4</u>,-7,<u>3</u>], k = 2
<strong>Output:</strong> 7
<strong>Explanation:</strong> You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [<u>10</u>,-5,-2,<u>4</u>,0,<u>3</u>], k = 3
<strong>Output:</strong> 17
<strong>Explanation:</strong> You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length, k &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

from collections import deque

class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        n = len(nums)
        if n == 0:
            return 0
        
        # Initialize the dp array to store the maximum score at each index
        dp = [0] * n
        dp[0] = nums[0]
        
        # Initialize a deque to store indices
        deq = deque([0])
        
        for i in range(1, n):
            # Remove indices that are out of the current window
            if deq and deq[0] < i - k:
                deq.popleft()
            
            # Update the dp array with the maximum score from the previous k steps
            dp[i] = dp[deq[0]] + nums[i]
            
            # Maintain the deque in decreasing order of dp values
            while deq and dp[deq[-1]] <= dp[i]:
                deq.pop()
            deq.append(i)
        
        return dp[-1]

# Example usage:
solution = Solution()
nums = [1, -1, -2, 4, -7, 3]
k = 2
print(solution.maxResult(nums, k))  # Output: 7
29400