Organizing Containers of Balls | HackerRank (JavaScript) 

Are you referring to something like this? function organizingContainers(containers) { const n = containers.length; // keep # of balls per container const containerCounts = new Array(n).fill(0); // keep # of types of balls const ballTypes = new Array(n).fill(0); for(let i = 0; i

I totally get where you're coming from. It wasn't clear when I was first doing this either. Let's look at an example that I hope is clarifying: 5 1 3 1 5 2 2 3 5 Each row is a container. - The first row/container has 9 balls in it, which means it can only ever hold 9 balls (given the swap mechanic). Let's call this container A. - The second row/container has 8 balls in it (container B). - The third row/container has 10 balls in it (container C). Each column is a unique color - There are 8 balls belonging to the first column/color. Let's call this color Red. - There are 9 balls belonging to the second column/color (color Blue). - There are 10 balls belonging to the third column/color (color Yellow). We have one container that can fit each color: - Container A can hold color Blue - Container B can hold color Red - Container C can hold color Yellow However, if we just compare the row and column sum by their original index we'd have a problem, since there is no order enforced on which containers or colors are listed first. - row 0 (container A) holds 9 balls, but column 0 (color Red) only has 8 balls in total. - row 1 (container B) holds 8 balls, but column 1 (color Blue) only has 9 balls in total. - row 2 (container C) holds 10 balls and column 2 (color Yellow) has 10 balls in total, so this happens to work out. If we sort the counts for rows and columns then we can line them up by their new sorted index. - sorted row list 0 holds 8 balls (this corresponds to container B), which can hold sorted column list 0 (color Red). - sorted row list 1 holds 9 balls (this corresponds to container A), which can hold sorted column list 1 (color Blue). - sorted row list 2 holds 10 balls (this corresponds to container C), which can hold sorted column list 2 (color Yellow). I hope this helps. Thanks for watching!

@@GlitchedFailure Thank you very much!

I have used Python3 rather than JS. Please Help me out buddy my head's gonna burst i dont understand why my code isn't working

@@pragadeeshsb3249 Taking the sum won't work, since the containers should be considered separately. Adding 1 + 3 = 4 and 2 + 2 = 4 suggests the same sum, but the containers won't match up.

@@GlitchedFailure Let me consider 3x3 matrix 1 1 1 1 2 1 3 2 1 Rows are container First row has 0-1 1-1 2-1 and so on Now for container 0 possible swap is 2 That is sum of sum of first row - first element Now, total 0 other than considered container would be sum of first coloum - first element Since my code is adding first element on both the cases it does not change the result If no of swaps == total 0 in other container For all containers if the above case is satisfied isn't that possible. Im not understanding where have I gone wrong

Hmm, let's verify we're referring to that 3x3 matrix in the same way. 1 1 1 1 2 1 3 2 1 Each row is a container. - The first row/container has 3 balls in it, which means it can only ever hold 3 balls (given the swap mechanic). - The second row/contianer has 4 balls in it - The third row/contianer has 6 balls in it. Each column is a unique color - There are 5 balls belonging to the first column/color - There are 5 balls belonging to the second column/color - There are 3 balls belonging to the third column/color In order for each container to hold each unique ball color, the containers must have the exact amount of space needed for them. In this case, the first color ball needs a container that fits exaclty 5 balls, but there are none. The same is true for the other colors. This is an "Impossible" situation. ================== Let's consider the following 3x3 matrix: 1 2 2 1 1 1 1 2 1 Each row is a container. - The first row/container has 5 balls in it, which means it can only ever hold 5 balls (given the swap mechanic). - The second row/contianer has 3 balls in it - The third row/contianer has 4 balls in it. Each column is a unique color - There are 3 balls belonging to the first column/color - There are 5 balls belonging to the second column/color - There are 4 balls belonging to the third column/color In order for each container to hold each unique ball color, the containers must have the exact amount of space needed for them. In this case, the first color ball needs a container that fits exaclty 3 balls, which we have. The same is true for the other colors and the containers for 4 and 5 balls respectively. This is a "Possible" situation.

@@GlitchedFailure thanks a lot buddy. Your explanation was amazing subscribed your channel ;)