@vaporeon_ here is another hack with a "beautiful" one liner:

Problem Statement
You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:

• 0 <= j <= nums[i] and
• i + j < n

Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].

Example

• Input: nums = [2,3,1,1,4]
• Output: 2
• Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.

Code

const jump = (nums, i = 0, gorp = {}) => {
    if (i === nums.length - 1) return 0;
    if (i > nums.length - 1 || nums[i] === 0) return Infinity;
    if (gorp[i] !== undefined) return gorp[i];

    let min = Infinity;
    for (let j = 1, res; j <= nums[i]; ++j) res = 1 + jump(nums, i + j, gorp), min = res < min ? res : min;

    return gorp[i] = min;
};