Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for [old] Worth of Words by mozurin
var VALUES = {'e': 1, 'a': 1, 'i': 1, 'o': 1, 'n': 1, 'r': 1,
't': 1, 'l': 1, 's': 1, 'u': 1, 'd': 2, 'g': 2,
'b': 3, 'c': 3, 'm': 3, 'p': 3, 'f': 4, 'h': 4,
'v': 4, 'w': 4, 'y': 4, 'k': 5, 'j': 8, 'x': 8,
'q': 10, 'z': 10};
function worthOfWords(words)
{
return words.map(
e => [e, [...e].map(c => VALUES[c]).reduce((s, v) => s + v)]
).sort(
(a, b) => b[1] - a[1]
)[0][0];
}
June 17, 2018
Comments: