class Solution 
{
    //Function to find minimum number of operations that are required 
    //to make the matrix beautiful.
    static int findMinOperation(int matrix[][], int n)
    {
        int sumRow[] = new int[n];
        int sumCol[] = new int[n];
        Arrays.fill(sumRow, 0);
        Arrays.fill(sumCol, 0);
        
        //calculating sumRow[] and sumCol[] array.
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                sumRow[i] += matrix[i][j];
                sumCol[j] += matrix[i][j];
                  
            }
        }
        
        //finding maximum sum value in either row or in column.
        int maxSum = 0;
        for (int i = 0; i < n; ++i)
        {
            maxSum = Math.max(maxSum, sumRow[i]);
            maxSum = Math.max(maxSum, sumCol[i]);
        } 
        
        int count = 0;
        for (int i = 0, j = 0; i < n && j < n;) 
        {
            //finding minimum increment required in either row or column.
            int diff = Math.min(maxSum - sumRow[i], maxSum - sumCol[j]);
            
            //adding difference in corresponding cell, 
            //sumRow[] and sumCol[] array.
            matrix[i][j] += diff;
            sumRow[i] += diff;
            sumCol[j] += diff;
            
            //updating the result.
            count += diff;
            
            //if ith row is satisfied, incrementing i for next iteration.
            if (sumRow[i] == maxSum)
                ++i;
            
            //if jth column is satisfied, incrementing j for next iteration.
            if (sumCol[j] == maxSum)
                ++j;
        }
        //returning the result.
        return count;
    }
}