Weights and values may be any non-negative integer. Yes, it's weird to think about cakes that weigh nothing or duffel bags that can't hold anything. But we're not just super mastermind criminals—we're also meticulous about keeping our algorithms flexible and comprehensive.

cake_tuples = [(7, 160), (3, 90), (2, 15)] capacity = 20 # Returns 555 (6 of the middle type of cake and 1 of the last type of cake) max_duffel_bag_value(cake_tuples, capacity)

Write a function max_duffel_bag_value that takes a list of cake type tuples and a weight capacity , and returns the maximum monetary value the duffel bag can hold.

You brought a duffel bag that can hold limited weight, and you want to make off with the most valuable haul possible.

# Weighs 7 kilograms and has a value of 160 shillings (7, 160) # Weighs 3 kilograms and has a value of 90 shillings (3, 90)

Each type of cake has a weight and a value, stored in a tuple with two indices:

While Queen Elizabeth has a limited number of types of cake, she has an unlimited supply of each type.

You are a renowned thief who has recently switched from stealing precious metals to stealing cakes because of the insane profit margins. You end up hitting the jackpot, breaking into the world's largest privately owned stock of cakes—the vault of the Queen of England.

We can do this in time and space, where n is the number of types of cakes and k is the duffel bag's capacity!

Does your function work if any of the cakes weigh 0 kg? Think about a cake whose weight and value are both 0.

Does your function work if the duffel bag's weight capacity is 0 kg?

The brute force approach is to try every combination of cakes, but that would take a really long time—you'd surely be captured. What if we just look at the cake with the highest value?

We could keep putting the cake with the highest value into our duffel bag until adding one more would go over our weight capacity. Then we could look at the cake with the second highest value, and so on until we find a cake that’s not too heavy to add. Will this work?

Nope. Let's say our capacity is 100 kg and these are our two cakes: [(1, 30), (50, 200)] With our approach, we’ll put in two of the second type of cake for a total value of 400 shillings. But we could have put in a hundred of the first type of cake, for a total value of 3000 shillings! Just looking at the cake's values won’t work. Can we improve our approach?

Well, why didn’t it work?

We didn’t think about the weight! How can we factor that in?

What if instead of looking at the value of the cakes, we looked at their value/weight ratio? Here are our example cakes again: [(1, 30), (50, 200)] The second cake has a higher value, but look at the value per kilogram. The second type of cake is worth 4 shillings/kg (200/50), but the first type of cake is worth 30 shillings/kg (30/1)! Ok, can we just change our algorithm to use the highest value/weight ratio instead of the highest value? We know it would work in our example above, but try some more tests to be safe.

We might run into problems if the weight of the cake with the highest value/weight ratio doesn’t fit evenly into the capacity. Say we have these two cakes: [(3, 40), (5, 70)] If our capacity is 8 kg, no problem. Our algorithm chooses one of each cake, giving us a haul worth 110 shillings, which is optimal. But if the capacity is 9 kg, we're in trouble. Our algorithm will again choose one of each cake, for a total value of 110 shillings. But the actual optimal value is 120 shillings—three of the first type of cake! So even looking at the value/weight ratios doesn’t always give us the optimal answer! Let’s step back. How can we ensure we get the optimal value we can carry?

Try thinking small. How can we calculate the maximum value for a duffel bag with a weight capacity of 1 kg? (Remember, all our weights and values are integers.)

If the capacity is 1 kg, we’ll only care about cakes that weigh 1 kg (for simplicity, let's ignore zeroes for now). And we'd just want the one with the highest value. We could go through every cake, using a greedy approach to keep track of the max value we’ve seen so far.

Here’s an example solution: def max_duffel_bag_value_with_capacity_1(cake_tuples): max_value_at_capacity_1 = 0 for cake_weight, cake_value in cake_tuples: if cake_weight == 1: max_value_at_capacity_1 = max(max_value_at_capacity_1, cake_value) return max_value_at_capacity_1 Ok, now what if the capacity is 2 kg? We’ll need to be a bit more clever.

It’s pretty similar. Again we’ll track a max value, let’s say with a variable max_value_at_capacity_2. But now we care about cakes that weigh 1 or 2 kg. What do we do with each cake? And keep in mind, we can lean on the code we used to get the max value at weight capacity 1 kg.

If the cake weighs 2 kg, it would fill up our whole capacity if we just took one. So we just need to see if the cake's value is higher than our current max_value_at_capacity_2 . If the cake weighs 1 kg, we could take one, and we'd still have 1 kg of capacity left. How do we know the best way to fill that extra capacity? We can use the max value at capacity 1. We’ll see if adding the cake's value to the max value at capacity 1 is better than our current max_value_at_capacity_2 . Does this apply more generally? If we can use the max value at capacity 1 to get the max value at capacity 2, can we use the max values at capacity 1 and 2 to get the max value at capacity 3? Looks like this problem might have overlapping subproblems!

Let's see if we can build up to the given weight capacity, one capacity at a time, using the max values from previous capacities. How can we do this?

Well, let’s try one more weight capacity by hand—3 kg. So we already know the max values at capacities 1 and 2. And just like we did with max_value_at_capacity_1 and max_value_at_capacity_2, now we’ll track max_value_at_capacity_3 and loop through every cake: max_value_at_capacity_3 = 0 for cake_weight, cake_value in cake_tuples: # Only care about cakes that weigh 3 kg or less ... What do we do for each cake?

If the current cake weighs 3 kg, easy—we see if it’s more valuable than our current max_value_at_capacity_3. What if the current cake weighs 2 kg?

Well, let's see what our max value would be if we used the cake. How can we calculate that?

If we include the current cake, we can only carry 1 more kilogram. What would be the max value we can carry?

We already know the max_value_at_capacity_1! We can just add that to the current cake’s value!

Now we can see which is higher—our current max_value_at_capacity_3, or the new max value if we use the cake: max_value_using_cake = max_value_at_capacity_1 + cake_value max_value_at_capacity_3 = max(max_value_at_capacity_3, max_value_using_cake) Finally, what if the current cake weighs 1 kg?

Basically the same as if it weighs 2 kg: max_value_using_cake = max_value_at_capacity_2 + cake_value max_value_at_capacity_3 = max(max_value_at_capacity_3, max_value_using_cake) There’s gotta be a pattern here. We can keep building up to higher and higher capacities until we reach our input capacity. Because the max value we can carry at each capacity is calculated using the max values at previous capacities, we'll need to solve the max value for every capacity from 0 up to our duffel bag's actual weight capacity. Can we write a function to handle all the capacities?

To start, we'll need a way to store and update all the max monetary values for each capacity. We could use a dictionary, where the keys represent capacities and the values represent the max possible monetary values at those capacities. Dictionaries are built on arrays, so we can save some overhead by just using a list. def max_duffel_bag_value(cake_tuples, weight_capacity): # List to hold the maximum possible value at every # integer capacity from 0 to weight_capacity # starting each index with value 0 max_values_at_capacities = [0] * (weight_capacity + 1) What do we do next?

We’ll need to work with every capacity up to the input weight capacity. That’s an easy loop: # Every integer from 0 to the input weight_capacity for current_capacity in xrange(weight_capacity + 1): ... What will we do inside this loop? This is where it gets a little tricky.

We care about any cakes that weigh the current capacity or less. Let's try putting each cake in the bag and seeing how valuable of a haul we could fit from there.

So we'll write a loop through all the cakes (ignoring cakes that are too heavy to fit): for cake_weight, cake_value in cake_tuples: # If the cake weighs as much or less than the current capacity # see what our max value could be if we took it! if cake_weight <= current_capacity: # Find max_value_using_cake ... And put it in our function body so far: def max_duffel_bag_value(cake_tuples, weight_capacity): # We make a list to hold the maximum possible value at every # duffel bag weight capacity from 0 to weight_capacity # starting each index with value 0 max_values_at_capacities = [0] * (weight_capacity + 1) for current_capacity in xrange(weight_capacity + 1): for cake_weight, cake_value in cake_tuples: # If the cake weighs as much or less than the current capacity # see what our max value could be if we took it! if cake_weight <= current_capacity: # Find max_value_using_cake ... How do we compute max_value_using_cake?

Remember when we were calculating the max value at capacity 3kg and we "hard-coded" the max_value_using_cake for cakes that weigh 3 kg, 2kg, and 1kg? # cake weighs 3 kg max_value_using_cake = cake_value # cake weighs 2 kg max_value_using_cake = max_value_at_capacity_1 + cake_value # cake weighs 1 kg max_value_using_cake = max_value_at_capacity_2 + cake_value How can we generalize this? With our new function body, look at the variables we have in scope: max_values_at_capacities current_capacity cake_weight cake_value Can we use these to get max_value_using_cake for any cake?

Well, let's figure out how much space would be left in the duffel bag after putting the cake in: remaining_capacity_after_taking_cake = current_capacity - cake_weight So max_value_using_cake is: the current cake's value, plus the best value we can fill the remaining_capacity_after_taking_cake with remaining_capacity_after_taking_cake = current_capacity - cake_weight max_value_using_cake = cake_value + max_values_at_capacities[remaining_capacity_after_taking_cake] We can squish this into one line: max_value_using_cake = cake_value + max_values_at_capacities[current_capacity - cake_weight] Since remaining_capacity_after_taking_cake is a lower capacity, we'll have always already computed its max value and stored it in our max_values_at_capacities! Now that we know the max value if we include the cake, should we include it? How do we know?

Let's allocate a variable current_max_value that holds the highest value we can carry at the current capacity. We can start it at zero, and as we go through all the cakes, any time the value using a cake is higher than current_max_value, we'll update current_max_value! current_max_value = max(max_value_using_cake, current_max_value) What do we do with each value for current_max_value? What do we need to do for each capacity when we finish looping through all the cakes?

We save each current_max_value in the max_values_at_capacities list. We'll also need to make sure we set current_max_value to zero in the right place in our loops—we want it to reset every time we start a new capacity. So here's our function so far: def max_duffel_bag_value(cake_tuples, weight_capacity): # We make a list to hold the maximum possible value at every # duffel bag weight capacity from 0 to weight_capacity # starting each index with value 0 max_values_at_capacities = [0] * (weight_capacity + 1) for current_capacity in xrange(weight_capacity + 1): # Set a variable to hold the max monetary value so far for # the current weight capacity current_max_value = 0 for cake_weight, cake_value in cake_tuples: # If the current cake weighs as much or less than the # current weight capacity it's possible taking the cake # would get a better value if cake_weight <= current_capacity: # Should we use the cake or not? # If we use the cake, the most kilograms we can include in # addition to the cake we're adding is the current capacity # minus the cake's weight. We find the max value at that # integer capacity in our list max_values_at_capacities max_value_using_cake = ( cake_value + max_values_at_capacities[current_capacity - cake_weight] ) # Now we see if it's worth taking the cake. How does the # value with the cake compare to the current_max_value? current_max_value = max(max_value_using_cake, current_max_value) # Add each capacity's max value to our list so we can use them # when calculating all the remaining capacities max_values_at_capacities[current_capacity] = current_max_value Looking good! But what's our final answer?

Our final answer is max_values_at_capacities[weight_capacity]! Okay, this seems complete. What about edge cases?

Remember, weights and values can be any non-negative integer. What about zeroes? How can we handle duffel bags that can’t hold anything and cakes that weigh nothing?

Well, if our duffel bag can’t hold anything, we can just return 0. And if a cake weighs 0 kg, we return infinity. Right?

Not that simple! What if our duffel bag holds 0 kg, and we have a cake that weighs 0 kg. What do we return?

And what if we have a cake that weighs 0 kg, but its value is also 0. If we have other cakes with positive weights and values, what do we return?

If a cake’s weight and value are both 0, it’s reasonable to not have that cake affect what we return at all. If we have a cake that weighs 0 kg and has a positive value, it’s reasonable to return infinity, even if the capacity is 0. For returning infinity, we have several choices. We could return: Python 2.7 's float('inf') . Return a custom response, like the string 'infinity' . Raise an exception indicating the answer is infinity. What are the advantages and disadvantages of each option?