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```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;
}
}```