Hackerrank solution python find the runner up score 8

certainly! the "find the runner-up score!" problem on hackerrank is a common exercise that helps you practice working with lists and understanding basic python operations. here’s a step-by-step tutorial on how to solve it.

### problem statement
you are given a list of scores from a competition. your task is to find the second highest score in the list.

### steps to solve the problem

1. **read the input**: you will first need to read the number of scores and then the scores themselves.
2. **find unique scores**: since there might be duplicate scores, you should find the unique scores.
3. **sort the scores**: once you have the unique scores, sort them in descending order.
4. **get the runner-up score**: the runner-up score will be the second element in this sorted list.

### example input and output
**input:**

**output:**

### code example
here’s how you can implement this in python:

### explanation of the code
1. **input handling**: the first line reads the number of scores, but it's not used directly since we can read the scores themselves.
2. **using `map` and `list`**: the `map(int, input().split())` takes the input string, splits it into individual elements, converts them to integers, and then creates a list.
3. **set for uniqueness**: `set(scores)` removes any duplicate scores.
4. **sorting**: the `sorted()` function sorts the unique scores in descending order using the `reverse=true` argument.
5. **accessing the runner-up**: the second element in the sorted list (index 1) is the runner-up score.
6. **output**: finally, the runner-up score is printed.

### edge cases
- if there is only one unique score, the problem doesn’t specify what to do. you might want to handle this case according to your needs.
- ensure that the input is valid (i.e., contains at least 2 unique scores).

### test the code
to test the code, you can run it in a local python environment or any online python interpreter. just input the number of scores followed by the scores themselve ...

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