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You created a game that is more popular than Angry Birds.

Each round, players receive a score between 0 and 100, which you use to rank them from highest to lowest. So far you're using an algorithm that sorts in time, but players are complaining that their rankings aren't updated fast enough. You need a faster sorting algorithm.

Write a function that takes:

  1. a list of unsorted_scores
  2. the highest_possible_score in the game

and returns a sorted list of scores in less than time.

For example:

unsorted_scores = [37, 89, 41, 65, 91, 53] HIGHEST_POSSIBLE_SCORE = 100 # Returns [91, 89, 65, 53, 41, 37] sort_scores(unsorted_scores, HIGHEST_POSSIBLE_SCORE)

We’re defining n as the number of unsorted_scores because we’re expecting the number of players to keep climbing.

And, we'll treat highest_possible_score as a constant instead of factoring it into our big O time and space costs because the highest possible score isn’t going to change. Even if we do redesign the game a little, the scores will stay around the same order of magnitude.

Multiple players can have the same score! If 10 people got a score of 90, the number 90 should appear 10 times in our output list.

We can do this in time and space.

is the time to beat. Even if our list of scores were already sorted we'd have to do a full walk through the list to confirm that it was in fact fully sorted. So we have to spend at least time on our sorting function. If we're going to do better than , we're probably going to do exactly .

What are some common ways to get runtime?

One common way to get runtime is to use a greedy algorithm. But in this case we're not looking to just grab a specific value from our input set (e.g. the "largest" or the "greatest difference")—we're looking to reorder the whole set. That doesn't lend itself as well to a greedy approach.

Another common way to get runtime is to use counting. We can build a list score_counts where the indices represent scores and the values represent how many times the score appears. Once we have that, can we generate a sorted list of scores?

What if we did an in-order walk through score_counts. Each index represents a score and its value represents the count of appearances. So we can simply add the score to a new list sorted_scores as many times as count of appearances.

We use counting sort.

def sort_scores(unsorted_scores, highest_possible_score): # List of 0s at indices 0..highest_possible_score score_counts = [0] * (highest_possible_score+1) # Populate score_counts for score in unsorted_scores: score_counts[score] += 1 # Populate the final sorted list sorted_scores = [] # For each item in score_counts for score in xrange(len(score_counts) - 1, -1, -1): count = score_counts[score] # For the number of times the item occurs for time in xrange(count): # Add it to the sorted list sorted_scores.append(score) return sorted_scores

time and space, where n is the number of scores.

Wait, aren't we nesting two loops towards the bottom? So shouldn't it be time? Notice what those loops iterate over. The outer loop runs once for each unique number in the list. The inner loop runs once for each time that number occurred.

So in essence we're just looping through the n numbers from our input list, except we're splitting it into two steps: (1) each unique number, and (2) each time that number appeared.

Here's another way to think about it: in each iteration of our two nested loops, we append one item to sorted_scores. How many numbers end up in sorted_scores in the end? Exactly how many were in our input list! n.

If we didn't treat highest_possible_score as a constant, we could call it k and say we have time and space.

Note that by optimizing for time we ended up incurring some space cost! What if we were optimizing for space?

We chose to generate and return a separate, sorted list. Could we instead sort the list in place? Does this change the time complexity? The space complexity?

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