I did not find this easy. Very tedious or maybe I didn't discover the "trick." This leads to frustration and diminished enjoyment. That's why I hope this helps someone and they can improve on this solution (and share their discovery with all). HackerRank Project Eulrer 90

from itertools import combinations
can = lambda d, x: x in d or (x in [6,9] and (6 in d or 9 in d))
N,M = map(int,input().split())
sq = [[int(x) for x in f"{i*i:0{M}d}"] for i in range(1, N+1)]
def check_pair(d1, d2, cache={}):
    key = (d1, d2)
    if key not in cache:
        cache[key] = all(can(d1, s[0]) and can(d2, s[1]) or 
                         can(d1, s[1]) and can(d2, s[0]) for s in sq)
    return cache[key]

def PE90(n, m):
    dice = list(combinations(range(10), 6))
    
    if m == 1:
        req = {s[0] for s in sq}
        return sum(all(can(d, x) for x in req) for d in dice)
        
    if m == 2:
        return sum(check_pair(dice[i], dice[j]) 
                  for i in range(len(dice)) for j in range(i, len(dice)))
    
    return sum(all(any(can(dice[p[0]], s[0]) and can(dice[p[1]], s[1]) 
              and can(dice[p[2]], s[2]) for p in ((i,j,k), (i,k,j), (j,i,k),
              (j,k,i), (k,i,j), (k,j,i))) for s in sq)
              for i in range(len(dice)) 
              for j in range(i, len(dice))
              for k in range(j, len(dice)))

print(PE90(N,M))

I would make something more explicit.

one way this can be achieved is by placing 056789 on one cube and 123489 on the other cube

One could think that it counts as 2 solutions:

• 056789 on the first one, 123489 on the second
• 123489 on the first one, 056789 on the second

This is not the case. It counts as 1 solution.

So for 9 2 the answer is 1217.

For 1 2 the answer is 13461. It can be also computed manually:

0	1	3136
0	01	3920
01	01	2485
01	1	3920

If the question had used the term dice instead of cube it would have been fun to add the input S for the number of sides on the dices.

A single cube only has 210 arrangements.

Hint: Generating all combinations of 3 dices works for this problem in Python. At first I thought there must be some other approach to pass all test cases, but it turns out if you find the right way to implement your combinations loop which avoids iterating through duplicate choices of dices, it would be fine.

Dice with different orders are considered as the same variant. (die1, die2) = (die2, die1).

n=10 m=3 ans=294197