Day 1 - Advent of Code 2017

Description

Part 1

Given a sequence of digits, determine which digits recur immediately in the sequence. Return the sum of all such digits.

Consider the sequence circular: the final digit in the sequence can recur as the first digit in the sequence.

Part 2

Given a circular sequence of an even number of digits, determine which digits recur at an interval of half the total length of the sequence. (For a sequence of ten digits, the recurrence interval would be five digits.)

My solution

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def solution_part1(digit):
    sum = 0
    try:
        for idx, elt in enumerate(digit):
            if elt == digit[idx + 1]:
                sum += int(elt)
    except IndexError:
        if elt == digit[0]:
            sum += int(elt)
    return sum


def solution_part2(digit):
    sum = 0
    interval = int(len(digit) / 2)
    for idx, elt in enumerate(digit):
        try:
            if elt == digit[idx + interval]:
                sum += int(elt)
        except IndexError:
            if elt == digit[idx % interval]:
                sum += int(elt)
    return sum

Tests

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def test_solution1():
    assert solution_part1('1122') == 3
    assert solution_part1('1111') == 4
    assert solution_part1('1234') == 0
    assert solution_part1('91212129') == 9


def test_solution2():
    assert solution_part2('1212') == 6
    assert solution_part2('1221') == 0
    assert solution_part2('123424') == 4
    assert solution_part2('123123') == 12
    assert solution_part2('12131415') == 4