Probably the shortest const twoNumbersEqual = (a, k) => !!a.filter(x =&...

re: Daily Coding Problem #1 VIEW POST

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Probably the shortest

const twoNumbersEqual = (a, k) => !!a.filter(x => a.includes(k-x)).length

twoNumbersEqual([10, 15, 3, 7], 17) // == true
twoNumbersEqual([10, 13, 4, 7], 17) // == true
twoNumbersEqual([10, 15, 3, 9], 17) // == false

Casey Webb

Nov 27 '18

slight improvement with a short circuit (find vs filter):

const twoNumbersEqual = (a, k) => !!a.find(x => a.includes(k-x))

Casey Webb

Nov 27 '18

I just realized both of these solutions fail in the case of twoNumbersEqual([5], 10). This can be fixed by passing the index to includes, as such...

(a, k) => !!a.find((x, i) => a.includes(k-x, i+1))

This should also theoretically be more performant.

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