Weekly Challenge: Peak Visitors

Weekly Challenge 345

Each week Mohammad S. Anwar sends out The Weekly Challenge, a chance for all of us to come up with solutions to two weekly tasks. My solutions are written in Python first, and then converted to Perl. It's a great way for us all to practice some coding.

Challenge, My solutions

Task 1: Peak Positions

Task

You are given an array of integers, @ints.

Find all the peaks in the array, a peak is an element that is strictly greater than its left and right neighbours. Return the indices of all such peak positions.

My solution

This is relatively straight forward. I start by comparing the first two elements and add 0 to the peaks list (array in Perl) if the first value is higher.

I then compare the middle values, adding the position to the peaks list if it's before and after values are lower. Finally I compare the last two values.

def peak_positions(ints: list[int]) -> list[int]:

    peaks = []



    if ints[0] > ints[1]:

        peaks.append(0)



    for pos in range(1, len(ints) - 1):

        if ints[pos] > ints[pos - 1] and ints[pos] > ints[pos + 1]:

            peaks.append(pos)



    if ints[-1] > ints[-2]:

        peaks.append(len(ints) - 1)



    return peaks

Examples

$ ./ch-1.py 1 3 2

(1)



$ ./ch-1.py 2 4 6 5 3

(2)



$ ./ch-1.py 1 2 3 2 4 1

(2, 4)



$ ./ch-1.py 5 3 1

(0)



$ ./ch-1.py 1 5 1 5 1 5 1

(1, 3, 5)

Task 2: Last Visitor

Task

You are given an integer array @ints where each element is either a positive integer or -1.

We process the array from left to right while maintaining two lists:

1. 
   @seen stores previously seen positive integers (newest at the front)


2. 
   @ans stores the answers for each -1

Rules:

1. If $ints[i] is a positive number -> insert it at the front of @seen
   


2. If $ints[i] is -1:
   
   
   
   
   1. Let $x be how many -1s in a row we’ve seen before this one.
   
   
   2. If $x < len(@seen) -> append seen[x] to @ans
      
   
   
   3. Else -> append -1 to @ans
      
   
   
   
   
   
   

At the end, return @ans.

My solution

I actually don't understand this task. However, I've followed the instructions in the task, and get the expected results. I renamed x as neg_count to make it more meaningful.

def last_visitor(ints: list[int]) -> list[int]:

    seen = []

    ans = []

    neg_count = 0



    for i in ints:

        if i > 0:

            seen.insert(0, i)

            neg_count = 0

        elif i == -1:

            if neg_count < len(seen):

                ans.append(seen[neg_count])

            else:

                ans.append(-1)

            neg_count += 1

        else:

            raise ValueError("Input integers must be positive or -1.")



    return ans

Examples

$ ./ch-2.py 5 -1 -1

(5, -1)



$ ./ch-2.py 3 7 -1 -1 -1

(7, 3, -1)



$ ./ch-2.py 2 -1 4 -1 -1 

(2, 4, 2)



$ ./ch-2.py 10 20 -1 30 -1 -1

(20, 30, 20)



$ ./ch-2.py -1 -1 5 -1

(-1, -1, 5)