Bumby the Loaf @wallhackio@cat.family

@pikuma@mastodon.gamedev.place i love that main is recursive lmao

@cam very difficult choice

@cam the paradox of choice

caesar's greatest miscalculation was inviting a courier to his camp who had recently looted a rocket launcher

@cam oh no the art is cute oh fuck

@murz google's random number generator

@murz 5, because I ran a random number generator from 3-5 and it picked 5

@aschmitz vaporeon showed me a snippet of code from unix source that used the variable "gorp" and i thought it was funny so now i use it

@aschmitz I had no idea it was a word lol

@vaporeon_ never in production code lol

?????

@vaporeon_ it's even worse than you think! javascript has what is effectively a preprocessing phase where it looks for every single value initialized as a var and initializes them, and then the interpreter runs on the code like you would expect

this means you can use a variable before it is declared if you declare it with var! real javascript code out in the wild actually does this!!!!

@vaporeon_ @aescling no clue i just think the idea of storing project versions on laserdisc funny

@aescling @vaporeon_ with let the algo ran the test cases ~700ms and with var it shaved ~100ms

@aescling @vaporeon_ ....it runs faster with this change

@soft PREACH

@vaporeon_ oooooh I could have done for (var min = Infinity, j = 1, res; j <= nums[i]; ++j) res = 1 + jump(nums, i + j, gorp), min = res < min ? res : min;

javascript variables declared with var and not block scoped, so min is available outside of the for loop teehee

@aescling @vaporeon_ I just ran it and what is here is correct

@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;
};

@coriander @wohali I was gonna write this comment too, lol