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Discussion on: Daily Challenge #25 - Double Cola

coyote profile image
LBcoyote • Edited on

Pure mathematical solution in JS. For larger n's it can be faster than the loop version by roughly a factor of log₂ n.

function who_is_next(names,n){
    let p=2**Math.floor(Math.log2(1+--n/names.length))
    return names[Math.floor((n-(p-1)*names.length)/p)]
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