import java.util.*;
public class Solution {
    public static List<List<Integer>> matrixMultiplication(List<List<List<Integer>>> matrices) 
    
    {
        List<List<Integer>> finalMatrix = new ArrayList<List<Integer>>();

        List<Integer> temp = new ArrayList<Integer>();
        int rowsNextMatrix=0;
        int size=0;
        int matricesSize=matrices.size();
        int numElementsInList=0;
        int counter=0;
        int i;
        List<List <Integer>> matrix;
        boolean hasStartMatrix=true;
        boolean hasPerformedValidation=false;
        int numElementsInRow=0;

        int matrixNumber=0;
        int numMatrixes=0;

        //I am using this size based on principle that
        //number columns Matrix A needs to be equal 
        //number rows Matrix B
        //This is easiest way to ascertain a suitable dimension
        //of Matrix B
        //really this would need to be in a try and catch statement should there only be one matrix!
        int []matrixValue = new int[matrices.get(1).size()];

        
        //The following structure will ensure that
        //each each matrix can be obtained from the matrices
        //We have to remember that inner Lists can still be accessed via their index locations. This can be obtained via .get
        //We know that similar to other transposing matrix challenge, we can obtain the size (matrices) which will output the number of elements..
        //We know in this case this deals with List<List<Integer>>
        //We can now formulate a do while loop which would process the List<List<Integer>> within matrices until it reaches the size of matrices.

        //This will present all matrix in true form in order to visualise and assist with coding 
        System.out.println("***ALL MATRIX*******");
        do
        {
            matrix=matrices.get(numMatrixes);
            System.out.println("\n\nMatrix" +"("+numMatrixes+"):");

            for (List<Integer> row: matrix)
            {
                System.out.println(row);
            }
            numMatrixes++;
            System.out.println("\n");
            }while(numMatrixes<matrices.size());

       do
        {
            matrix=matrices.get(counter);
            System.out.println("\n\nMatrix" +"("+counter+"):");

        for (List<Integer> ee: matrix)
        {
            for (int k: ee)
            {
                //I am now choosing to look ahead at the next matrix as soon as a number appears on the screen.
                //The only exception of course is the final matrix since it would have been processed by penultimate matrix.

                if(counter!=matricesSize-1)
                {
                
                //In this matrix we know that numbers are retrieved left to right.
                //I am going to attempt completing the multiplication in a similar manner to a human.
                /*
                [1, 2, 3] 
                [4, 5, 6]

                [7,    8] 
                [9,   10]
                [11, 12]
                */

            //This processes next matrix
            //we know that whilst it processes row 1 of Matrix A
            //it will ALWAYS be on column 1 of Matrix B
            for (List<Integer> rows: matrices.get((counter+1)))
            {
        
    }
                }


                if (matrixNumber<counter && hasStartMatrix)
                {
                    System.out.println("This is first integer in next matrix: " + k);
                    hasStartMatrix=false;   
                }

                numElementsInList++;
                numElementsInRow++;
                System.out.println(k);
            }
            

            //Using this technique will also ensure that I can check if the matrix is valid dimension and terminate the application
            //WE KNOW THAT NUMBER columns in Matrix A should match rows Matrix B
	    //NOTE for simplicity, I have assumed columms in Matrix A are uniform across its rows.
	    //We know if it had a jagged array, it would be a mismatch against the number rows in Matrix B	
           //Once I have finished the entire challenge, I can potentially extend my validation at this level
           //Since in this challenge there is every possibility of human error populating the matrix!

            if(counter!=(matricesSize-1) && !hasPerformedValidation)
            {
                for (List<Integer> rows: matrices.get((counter+1)))
                {
                    rowsNextMatrix++;
                }

                    if (numElementsInRow!=rowsNextMatrix)
                    {
            System.out.println("columns in Matrix A" + "("+numElementsInRow+")"+ "should match rows Matrix B " +  "("+rowsNextMatrix+")");
                        System.exit(0);
                    }
                    else
                    {
                        System.out.println("columns in Matrix A " + "("+numElementsInRow+")" + " rows Matrix B: " + "("+rowsNextMatrix+")");
                    }
                    hasPerformedValidation=true;
            }

            numElementsInRow=0;

            //break;
        }
        matrixNumber=counter;
        counter++;
        hasStartMatrix=true;
        hasPerformedValidation=false;
        rowsNextMatrix=0;

        }while (counter<matricesSize);

        System.out.println("num elements: " + numElementsInList);

        return finalMatrix;
    }
}