I was so ready to give up on this Even with @dusanki solution, I didn't understand why the maths works.

It was only when I found on the internet an explanation with pictures that it all clicked.

I will probably implement it now. Just for closure. But I feel like Hackerrank has stolen a part of my life 😛

so what kind of education do you need to be able solve this? i've been in IT 18 years and i don't think i could've ever solved this without internet assistance

I don't know who needs to hears this, but as a 8 YoE software engineer, this took me an entire weekend to solve ^^'.

A little spoiler-free tip I wish I had:

• you will need to use the modulo whenever you can, because the amount of combination explodes pretty quickly. modulo is distributive over +, * and - operations, aka applying it to all those operators, is equivalent to only doing it once at the end.
• even with modulo, in Typescript/Javascript you will need to watch out for some exploding results that exceed the maximum javascript integer (2^53-1)

Having some difficult with a valid JavaScript version. The values that are failing are when combinations get over MOD (10^9 +7) Has anybody got a JavaScript version working or can spot the error in this code? Thanks

*(legoBlocks(8,10) on my code is giving 634608018.

Hacker rank is giving 634597424*

function legoBlocks(h, w) {
    // Write your code here
    let mod = (Math.pow(10, 9) + 7)

    //different combinations for a width when height is 1
    let t = [0];
    if (w >= 1) t.push(1)
    if (w >= 2) t.push(2)
    if (w >= 3) t.push(4)
    if (w >= 4) t.push(8)
    while (w >= 5 && t.length <= w){
        let start = t.length -4
        let arr = t.slice(start, t.length)
        let sum = arr.reduce((partialSum, a) => partialSum + a, 0)
        t.push(sum)
    }

    //different combinations for width given height h
    //e.g. total[height=2] is t^2, total[height=3] is t^3
    let total = t.slice()
    for (let outer=2; outer<=h; outer++) {
        for (let i = 0; i <= w; i++) {
            total[i] = (t[i] * total[i])
        }
    }


    let solid = []
    solid.length = w+1;
    solid.fill(0)
    solid[1] = 1

    //s[x] = total[x] - (SUM (where i starts at 1 and ends at x-1)  s[i] * total[w-i]))
    if (w>1){
        for (let ww=2; ww<=w; ww++){
            let unsolid_sum = 0;
            for (let i=1; i<ww; i++){

                unsolid_sum += (((solid[i]) * (total[ww-i])))
            }
            solid[ww] = ((total[ww] - unsolid_sum))
        }
    }

return solid[w] % mod

}

This question wrinkled my brain until I looked at Willis42's code. Here's some extra annotation in case it helps.

def legoBlocks(n, m): # the ways to build a line of m from 0 up to 4 using legobricks from 1-4 # this takes the prev sums and adds one more way to it each time permutations = [0, 1, 2, 4, 8] mod = 10**9+7 # just keeps our numbers from getting too large - wraps if so

# for the rest, use dp to keep compiling the already set base cases. We dont need to add one more way here (like we did above) because we already compiled all the ways for bricks from 1-4 now if m is >=5 we just need to get the last 4 subproblems we already compiled and sum them. Remember, don't add "one more" bc we're not adding a "new way" bc we only consider lego bricks from 1-4. Permutations array already holds all the ways for 1-4, we just want to grab the already computed prev 4 ways each time. 

for i in range(5, m+1): 
    permutations.append((permutations[i-1]  + permutations[i-2] + permutations[i-3] + permutations[i-4])% mod)

# at this point we have all possible permutations of bricks from 1-4 up to m
# now we need to calculate n (height) 
# we need to loop through all our permutations for each m, and raise them to the power of n to get all possible combinations up to n for each mth row.

for i in range(len(permutations)): 
    permutations[i] = (permutations[i] ** n) % mod

# now we have all possible but we need to introduce our condition to reduce by "bad" walls 
# start with setting "good" permutations as 0 and 1 ( we dont have to worry about reducing till start stacking) 
good_permutations = [0,1]

for i in range(2, m+1): # starting at 2 we are stacking rows
    current = permutations[i] # grab the current one assuming all are "good"
    for j in range(1, i): # now loop from each last known "good" section and subtract the right assuming all bad. as we move, we are only keeping known "good" sections and building on them            
        current -= (good_permutations[j] * permutations[i-j])
    good_permutations.append(current % mod)
return good_permutations[m]