Problem Statement   Fox Ciel is creating a new binary operation.
The operation will be denoted * and it will be defined on the finite set S = {0, 1, 2, ..., n1}.
I.e., for each ordered pair of elements of S the operation will return some element of S.
The operation can be described by its "multiplication table": a square table where in row i and column j we have the result returned by the operation i*j.
Ciel has already filled in row 0 of this table: for each j we know that 0 * j = firstRow[j].
Ciel's operation doesn't have to be symmetric: it is possible that for some i and j the operations i*j and j*i return two different values.
However, Ciel does want her operation to be associative: for all a,b,c in S, (a*b)*c must be equal to a*(b*c).
(Note that this includes cases when some or all of a,b,c are equal to each other.)
You are given the int[] firstRow with n elements.
If there is a way to fill in the rest of the multiplication table so that the resulting operation is associative, find one such way and return a int[] with n*n elements: all elements of the entire multiplication table, listed in row major order.
If there are multiple solutions you may return any of them.
If there are no solutions, return {1}.
(I.e., if there are no solutions, return a int[] of length 1 containing the element 1.)   Definition   Class:  MultiplicationTable  Method:  getMultiplicationTable  Parameters:  int[]  Returns:  int[]  Method signature:  int[] getMultiplicationTable(int[] firstRow)  (be sure your method is public) 
    Constraints    firstRow will contain between 2 and 50 elements, inclusive.    Each element in firstRow will be between 0 and firstRow1, inclusive.   Examples  0)     Returns: {1, 2, 0, 2, 0, 1, 0, 1, 2 }  The example return value corresponds to the following multiplication table:
*  0 1 2
+
0  1 2 0
1  2 0 1
2  0 1 2
In other words, this operation * can be defined using the following formula: i*j = (i+j+1) mod 3.
We can easily verify that for any a,b,c both (a*b)*c and a*(b*c) evaluates to the same value: namely, to the value (a+b+c+2) mod 3. 

 1)     Returns: {1, 1, 1, 1, 1, 1, 1, 1, 1 }  This time we can use: i*j = 1. 

 2)     Returns: {1 }  Row 0 already tells us that 0*0 = 0, 0*1 = 0, and 0*2 = 1.
Let's take a=0, b=0, and c=2.
On one hand we already know that (a*b)*c = (0*0)*2 = 0*2 = 1.
On the other hand we also know that a*(b*c) = 0*(0*2) = 0*1 = 0.
Hence, regardless of how we fill in the rest of the table, the resulting operation will never be associative. 

 3)     4)     Returns: {2, 3, 0, 1, 3, 0, 1, 2, 0, 1, 2, 3, 1, 2, 3, 0 }  
 5)     6)     Returns: {0, 1, 2, 0, 1, 2, 0, 1, 2 }  

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