First clear solution for How Much Gold by kurosawa4434 - python coding challenges

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First solution in Clear category for How Much Gold by kurosawa4434

from re import match
METALS = ('gold', 'tin', 'iron', 'copper')


def checkio(alloys):

    wk_alloys = {}
    for k, v in alloys.items() :
        match_obj = match(r'(.+)-(.+)', k)
        metal1 = match_obj.group(1)
        metal2 = match_obj.group(2)
        other_metals = list(set(METALS) - set((metal1, metal2)))
        wk_alloys[k] = v
        wk_alloys[other_metals[0] + '-' + other_metals[1]] = 1 - v

    a = wk_alloys['gold-tin'] if 'gold-tin' in wk_alloys else wk_alloys['tin-gold']
    b = wk_alloys['gold-iron'] if 'gold-iron' in wk_alloys else wk_alloys['iron-gold']
    c = wk_alloys['tin-iron'] if 'tin-iron' in wk_alloys else wk_alloys['iron-tin']

    return (a + b - c) / 2

July 21, 2016

Comments:

brubru777 on Jan. 8, 2017, 2:22 p.m.
Clever solution. To get the metals, a simpler solution would be <pre class='brush: python'>metal1, metal2 = k.split('-') </pre> You could also declare <pre class='brush: python'>METALS</pre> as a set at the beginning for efficiency.

kurosawa4434 on Jan. 8, 2017, 3:44 p.m.
Thank you for reviewing. My improved code is: <pre class='brush: python'>from fractions import Fraction METALS = {'gold', 'tin', 'iron', 'copper'} def checkio(alloys): wk_alloys = {} for k, v in alloys.items(): metals = k.split('-') other_metals = METALS - set(metals) wk_alloys[tuple(sorted(metals))] = v wk_alloys[tuple(sorted(other_metals))] = 1 - v a = wk_alloys[('gold', 'iron')] b = wk_alloys[('gold', 'tin')] c = wk_alloys[('iron', 'tin')] return (a + b - c) / 2 </pre>

Mysta on Aug. 11, 2026, 9:04 p.m.
As I had some problems to solve that mission I'm impressed by each published solution even if I don't understand it in detail. Some explaining comments would be appreciated.

kurosawa4434 on Aug. 12, 2026, 2:25 a.m.
The total amount of tin and copper can be derived from the total amount of gold and iron. (The same applies to other cases.) This idea is the core of the solution. We simply need to apply the appropriate data structure to this.

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