The last odd

Weekly Challenge 306

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: Odd Sum

Task

You are given an array of positive integers, @ints.

Write a script to return the sum of all possible odd-length sub-arrays of the given array. A sub-array is a contiguous sub-sequence of the array.

My solution

So I was starting to see if there was a pattern of how many times each number was counted. If we had 9 digits, they are counted 5, 8, 11, 12, 13, 12, 11, 8 and 5 times. As this didn't follow any easy pattern, I decided to do the task as described. Nothing wrong with that, computers have lots of CPU cycles available :)

For this task, I have two loops. The outer loop has the variable length. It starts with 1 and increments by 2 until the length of the list (array in Perl). The inner loop has the variable start to record the starting position of the sub-array. It starts with 0 (the first item in the ints list) and ends with the length of the list minus the value of length.

For each loop, I add the sum of the integers from start to start + length - 1 to the solutions value. I finish by returning the solutions value.

def odd_sum(ints: list) -> int:
    solution = 0

    l = len(ints)

    for length in range(1, l+1, 2):
        for start in range(0, l-length+1):
            solution += sum(ints[start:start+length])

    return solution

Examples

$ ./ch-1.py 2 5 3 6 4
77

$ ./ch-1.py 1 3
4

Task 2: Last Element

Task

You are given a array of integers, @ints.

Write a script to play a game where you pick two biggest integers in the given array, say x and y. Then do the following:

1. if x == y then remove both from the given array
2. if x != y then remove x and replace y with (y - x)

At the end of the game, there is at most one element left.

Return the last element if found otherwise return 0.

My solution

This task is a little more straight forward, with Mohammad practically writing the code for us. While the number of items in the ints list is greater than 1, I perform the following.

1. Sort the list numerically, lowest first
2. Remove the last item off the list, and assign it to x.
3. Remove the last item off the list, and assign it to y.
4. If x and y are different, append the difference to the ints list.

Lastly I return the remaining value, or 0 if the list is empty.

def last_element(ints: list) -> int:
    while len(ints) > 1:
        ints.sort()
        x = ints.pop()
        y = ints.pop()

        if x != y:
            ints.append(x-y)

    return ints[0] if ints else 0

Examples

$ ./ch-2.py 3 8 5 2 9 2
1

$ ./ch-2.py 3 2 5
0