Here is problem solution in python, java, c++, c and javascript programming - https://programmingoneonone.com/hackerrank-organizing-containers-of-balls-problem-solution.html

Python code(Easy)

def organizingContainers(container): # Write your code here n = len(container) row_sums = [sum(row) for row in container] col_sums = [sum(container[i][j] for i in range(n)) for j in range(n)] if sorted(row_sums) == sorted(col_sums): return "Possible" else: return "Impossible"

Solution in C#, first thing I did was realize that there is very little sorting necessary in this problem. What we have to do is find the sums of the x rows and the sums of the y columns. Once we've done that we sort both arrays and compare the arrays to ensure they match 1-to-1. If they do not, they are impossible to sort otherwise they are possible. I left my debugging code for brevity.

Essentially this swap sort method only works if each bucket has enough storage in it to hold the maximum count of a given item. (i.e. if there are two items of type A, there must be a corresponding bucket with two slots) this must be true for each item.

 public static string organizingContainers(List<List<int>> container)
{
    List<int> xSum = new List<int>();
    List<int> ySum = new List<int>();
    int localXSum = 0;
    int localYSum = 0;

    for(int i = 0; i < container.Count; i++){
        localXSum = 0;
        localYSum = 0;           
        for(int j = 0; j < container[i].Count; j++){
            localXSum +=  container[i][j];
            localYSum += container [j][i];
        }
        xSum.Add(localXSum);
        ySum.Add(localYSum);
    }

    xSum.Sort();
    ySum.Sort();

    foreach(int pos in xSum){
        Console.Write(pos + " ");
    }

    Console.WriteLine();

    foreach(int pos in ySum){
        Console.Write(pos + " ");
    }

    Console.WriteLine();

    for(int i = 0; i < xSum.Count; i++){
        if(xSum[i] != ySum[i]){
            return "Impossible";
        }
    }

    return "Possible";
}

c++ solution:

string organizingContainers(vector<vector<int>> container) {
    int sizeRow = container.size();
    
    multiset<int> columnSums;
    multiset<int> rowSums;

    vector<int> partialColumnSums(sizeRow, 0);
    for (int i = 0; i < sizeRow; i++) {
        
        int currentRowSum = 0;
        for (int j = 0; j < sizeRow; j++) {
            currentRowSum += container.at(i).at(j);
            partialColumnSums[j] += container.at(i).at(j);
        }
        
        rowSums.insert(currentRowSum);
    }
    
    for (int partialColumnSum : partialColumnSums) {
        columnSums.insert(partialColumnSum);
    }
    
    if (columnSums == rowSums) {
        return "Possible";
    }
    
    return "Impossible";
}

here is my solution

$total = count($container);
	
	$capacityArray = array();
	$indexArray = array();

	for ($i=0; $i < $total; $i++) { 
		$sum = array_sum($container[$i]);
		$capacityArray[] = $sum;
		$indexSum = 0;
		for ($j=0; $j < $total ; $j++) { 
			$indexSum+= $container[$j][$i];
		}
		$indexArray[] = $indexSum;
	}

sort($capacityArray);
sort($indexArray);

$result = '';
if ($capacityArray == $indexArray) {
	$result = 'Possible';
}
else
{
	$result = 'Impossible';
}
echo $result;