1696. Jump Game VI

Description

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

 

Solutions

Solution: Dynamic Programming + Queue

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
const maxResult = function (nums, k) {
  const dp = [...nums];
  const queue = [];
  const size = nums.length;

  for (let index = 1; index < size; index++) {
    const previous = dp[index - 1];

    while (queue.length && queue.at(-1) < previous) queue.pop();
    queue.push(previous);
    if (index - k - 1 >= 0 && queue[0] === dp[index - k - 1]) queue.shift();
    dp[index] += queue[0];
  }
  return dp[size - 1];
};