Given the sequence of numbers on the top face of a die rolled 90^\circ at a time, find one possible configuration of numbers on the faces of the die or determine that it doesn’t exist.

QUICK EXPLANATION

Bruteforce all possible answers, check if no step corresponds to rotation to the same face or the opposite face.

EXPLANATION

This problem can be misleading, since it may seem at first that the orientation of the die matters. Actually, it doesn’t, since if the number on the top of the die is x and the number on the bottom face is y, then it’s possible to get any of the remaining 4 numbers can be on top after exactly one step; of course, it’s impossible to get y on top after one step and we can’t keep x on top either.

We don’t actually know the pairs of numbers on opposite faces, but there aren’t many choices, so we can bruteforce them - pick three unordered pairs and ensure they contain each number 1 through 6 exactly once. There are only \frac{1}{3!}\frac{6\cdot5}{2}\frac{4\cdot3}{2}\frac{2\cdot1}{2}=15 such triples of disjoint pairs.

Therefore, all we need to do is try all possible configurations of numbers on the die and for each of them, check if the conditions A_i \neq A_{i+1}, A_i \neq o(A_{i+1}) hold for all 1 \le i < N. We actually found all possible configurations this way, not just one.

There are O(1) configurations, so the time complexity is O(N), same with memory complexity. We can improve this to O(1) memory either by checking which configurations satisfy the constraints as we read the input, or by remembering only how many times each pair (A_i,A_{i+1}) appeared.

I did this question trying all possible position reccursively. It passed the first sub task case but showed wrong ans for the 2nd subtask.Please tell me where I went wrong.

We know that the next number coming on the top will never be the opposite of the previous number , for eg, if the input is,
1 3 4 ,then opp(1) will not be eq to 3 . opp(3) will not be equal to 4 and 1 . opp(4) will not be equal to 3. So this is what is done in the 2D array g, Here we are denoting 1 to show that the numbers will not be opposite to each other .[ for the above example g[1][3]=1 ,g[3][1]=1… etc ]. next_permutation gives the next lexographic combination. if there is an array a=[1,2,3,4,5] ,next_permutation(a,a+5) will give me [2,1,3,4,5] and in next iteration will give [2,3,1,4,5 ] and so on. In this array p we will have the final result. So we keep on shuffling untill all elements index pos is not equal to that number itself . the next thing is ,
if opp[x]=y then opp[y] will be x. So that what we are checking using the condition (p[p[i]]==i) , if this condition is satisfied then i will be greater than 6, now atlast we will check if the array p is legit by cross-checking it with the 2D array ‘g’ ,

Here is the explanation for the setter’s code : #include <bits/stdc++.h>
using namespace std;

int N, A[2333], g[7][7], p[7];
int doit() {
int i;
memset(g, 0, sizeof g);
scanf("%d", &N);
for (i = 1; i <= N; i++) {
scanf("%d", &A[i]);
if (i > 1) g[A[i]][A[i - 1]] = g[A[i - 1]][A[i]] = 1; //checking when solution isnt possible
}
for (i = 1; i <= 6; i++) p[i] = i; // Create for next_permutation
for (i = 1; i <= 6; i++) if (g[i][i]) return printf("-1\n");
do {
for (i = 1; i <= 6; i++) if (p[i] == i) break; // index and number must not be the same
if (i > 6) { // When this has happened
for (i = 1; i <= 6; i++) if (p[p[i]] != i) break; // When solution will be arrived , p[p[2]] = p[1] == 2 ;must be equal to 2
if (i > 6) {
for (i = 1; i <= 6; i++) if (g[i][p[i]]) break; // If in matrix we have 1 for any , then try new
if (i > 6) {
for (i = 1; i <= 6; i++) printf("%d ", p[i]); // If all the obstacles are overcome, print the final array
return printf("\n");
}
}
}
} while (next_permutation(p + 1, p + 7));
printf("-1\n");
}
int main() {int t; scanf("%d", &t); while (t--) doit(); return 0;}