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Before we start, did you try solving the workbook yourself?

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If not, please click the link in the resources folder and follow the instructions for this workbook.

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In this solution, we're going to revisit Workbook 6.13.

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Instead of printing our rows one by one, we're going to use a nested loop.

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So here it tells us the outer loop needs to define this code.

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Simple enough, we'll say four and I is equal to zero.

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I is smaller than Price's dot length, so price is not.

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Length is going to represent the number of rows three plus plus.

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Where I is going to end up equaling zero one and two.

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So let's copy this over.

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When I equals zero, we're going to print baking.

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When I equals one, we print a beverage, one, I equals two, we print cereals.

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And now every time the outer loop runs through a particular row, we need an inner loop to run through

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every single element in that row.

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So here we'll say int j is equal to zero.

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J is smaller than prices.

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I length.

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J plus plus.

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So when the outer loop runs for one row, the inner loop is going to run through every single element

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in that row.

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And for every element we run through, we can print it.

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And after the inner loop runs to completion, we print a new line.

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Now, this might seem a little complex.

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It might seem a bit scary, but don't worry, once we visualize the runtime, it's all going to make

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a lot of sense.

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So here we've got a two RD array.

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The two RD array has three rows and each row has five elements.

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The outer loop is going to start with equaling zero, and it's going to keep running so long as I is

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smaller than the number of rows.

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So we start at equaling zero case zero.

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That means we're printing prices for baking.

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All right.

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And now each run from the outer loop is going to run the inner loop to completion.

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So this one is going to start off at J equaling zero, and it will keep running so long as J is smaller

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than the length of a row zero row I.

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So J is going to start at zero.

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Here I is zero and J is zero.

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So we're printing 1299.

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Here.

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I still zero, of course, but J is one.

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So we print 899.

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I still zero Jace to 999.

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I still zero J is three 1049.

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i0j equals four 1199.

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So I hope you can now see that each run from the outer loop runs the inner loop to completion.

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And now that we've printed every single element in row zero, we can print a new line.

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All right.

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And now this is the second run of the outer loop I now equals one, which means now we're going to print

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all of the prices for beverage.

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And one run of the outer loop is going to run the inner loop to completion.

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Once again, the inner loop starts off with J equaling zero.

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The loop is going to keep running so long as J is smaller than the length of row one because here I

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is one.

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Here we index row 1j0 $0.99.

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Row one element one 199.

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Row one element 2 to 49.

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Role.

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One element three 149.

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Row one element 4 to 99.

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All right.

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So now, after the inner loop runs to completion, we print a line of space.

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And this is the third time that our outer loop runs.

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This time I equals two, which means we're going to be printing all of the elements for row two.

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Once again, each run of the outer loop is going to run the inner loop to completion.

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J starts off at zero.

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The loop is going to keep running so long as J is smaller than the length of RO two.

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So long as J is smaller than five.

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J starts off at zero.

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Here we are indexing element zero 0 to 899.

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Zero two element to one 790 902.

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Element two 949.

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Row two element three 999 and RO two element four 1099.

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And after the outer loop runs the inner loop to completion, then we print a new line and now I a plus

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plus increases I to three.

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Three is not smaller than the number of rows three so our outer loop is going to break.

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Press continue terminating the runtime and we are done.

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I hope you enjoyed this breakpoint session.

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No matter how complex something may seem once you put your breakpoints and you start visualizing the

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runtime, it gives you a very clear picture of how your code is running.