Why not simplify TwoCrystalBalls to this?

[demakoff]

I think with only one counter (i) it will be more easy to get the logic, and less memory used.
Looks like works in the same way...

export default function two_crystal_balls(breaks: boolean[]): number {
    const jump = Math.floor(Math.sqrt(breaks.length));

    let i = 0;

    for (; i < breaks.length; i += jump) {
        if (breaks[i]) break;
    }

    for (i -= jump + 1; i < breaks.length; i++ ){
        if (breaks[i]) return i;
    }

    return -1;
}

[katayamahide]

Hi @demakoff,
I really like your simple solution, it looks nice and clear.
I think there are two points to be fixed, though.

1. unexpected result of i -= jump + 1;

let i = 4,
const jump = 2

i -= jump + 1;  
console.log(i) // might expect 3, but actually got 1. 

2. In case breaks = [true], the program will return -1;
   If above issue is fixed, the program will return -1 in case the breaks array has only one value, which is true.

So I think.

export default function two_crystal_balls(breaks: boolean[]): number {
    const jump = Math.floor(Math.sqrt(breaks.length));

-   let i = 0;
+   let i = jump;

    for (; i < breaks.length; i += jump) {
        if (breaks[i]) break;
    }

+    i -= jump;
+    for (; i < breaks.length; i++ ){
-    for (i -= jump + 1; i < breaks.length; i++ ){
        if (breaks[i]) return i;
    }

    return -1;
}

Or stick to initializing with zero,

export default function two_crystal_balls(breaks: boolean[]): number {
    const jump = Math.floor(Math.sqrt(breaks.length));

    let i = 0;
    for (; i < breaks.length; i += jump) {
        if (breaks[i]) break;
    }

+     for (i &&= i - jump + 1; i < breaks.length; i++ ){
-     for (i -= jump + 1; i < breaks.length; i++ ){

        if (breaks[i]) return i;
    }

    return -1;
}

Looking forward to seeing your response!