1696. Jump Game VI

Problem

Tags: Array, Dynamic Programming, Queue, Sliding Window, Heap (Priority Queue), Monotonic Queue

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 <= 10^5
• -10^4 <= nums[i] <= 10^4

Code

JS

// 1696. Jump Game VI (1/22/53408)
// Runtime: 144 ms (68.42%) Memory: 52.20 MB (86.84%) 

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var maxResult = function(nums, k) {
    const n = nums.length;
    
    // dp 陣列用來存到 i 處最大分數
    let dp = new Array(n);
    // queue 用來存候選位置
    let queue = [];
    
    // 初始化，分別放入 i = 0 的值與位置
    dp[0] = nums[0];
    queue.push(0);
    
    // 遍歷 nums
    for (let i = 1; i < n; i++) {
        // 把 queue 中超過可到達範圍的位置移除
        while (queue.length && queue[0] < i - k) queue.shift();
        
        // i 位置的最大分數會是 queue 中離 i 最遠的最大分數加上 i 位置的分數
        dp[i] = dp[queue[0]] + nums[i];
        
        // 把 queue 中位置的最大分數比 i 位置的最大分數小的位置移除
        while (queue.length && dp[i] >= dp[queue[queue.length - 1]]) queue.pop();
        
        // 把 i 位置放入 queue
        queue.push(i);
    }
    
    // 回傳到最後一個位置的最大分數
    return dp[n - 1];
};