Java：二维数组的总和，其中M [i] [j] =（int）i / j

Question

T - number of test cases | 1<=T<=10 and n - number of elements | 1<=n<=1000000

Eg

if (T >= 1 && T <= 10) {
    for (int i = 0; i < T; i++) {
                int n = sc.nextInt();
                if (n > 0 && n <= 1000000) {
                    array = new int[n][n];
                    System.out.print("\n" + sumOfArray(array, n));
                }
             }
          }

Need to find the sum of M[i][j], where M[i][j] = (int) i/j;

I have written the code, but for n>10000, I start getting OOM, (for obvious reason).

If someone can help me with it, it'll be great. Need a whole new approach on solving the problem.

Eg.

Input   Output
2       
2       4
4       17

Answer 1

Here It is obvious that you don't need to store the values in the matrices because It is not possible to have that much space ( Array[10000][10000] ) available to allocate. So you need to think somehow in a mathematical way.

Consider a 4x4 Matrix and represent each element in the term of i,j .

1,1 1,2 1,3 1,4
2,1 2,2 2,3 2,4
3,1 3,2 3,3 3,4
4,1 4,2 4,3 4,4

Now we can represent here that what is stored in each of these elements.

1/1 1/2 1/3 1/4   (In Integers)     1 0 0 0
2/1 2/2 2/3 2/4   ============>     2 1 0 0
3/1 3/2 3/3 3/4                     3 1 1 0
4/1 4/2 4/3 4/4                     4 2 1 1

Tackle this matrix by dividing it into columns and solve each of the columns . For the first column series would be 1+2+3+4 .Then for the column number two(2) the series would be 0+1+1+2 .

Notice here that for ith column first i-1 values are zeros and then i values are same in the column. Then value is increased. Again it will be same for i values. Again increases by 1 and so on.

So in ith column value get increased on the jth element where j%i==0 .

So you can implement this logic in 1-D array and Complexity of this approach will be O(n logn) for each testcase.

Code:

import java.util.Scanner;

public class Main
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);

        int testcases=sc.nextInt();

        while(testcases-- >0)
        {
            int n=sc.nextInt();

            long array[]=new long[n+1]; //Take long array to avoid overflow

            for(int i=1;i<=n;i++)
            {
                for(int j=i;j<=n;j+=i)
                {
                    array[j]++;          //This will store that which elements get increased
                                         //from zero how many times
                }
            }

            //Now we can do summation of all elements of array but we need to do prefix sum here

            long sum=0;
            for(int i=1;i<=n;i++)
            {  
                array[i]+=array[i-1];
                sum+=array[i];
            }

            System.out.println(sum);
        }
    }
}