Isn&#x27;t radix sort o(n) ?<p>&gt; The algorithm will only work if w ≥ log n.<p>What does that mean ? It doesn&#x27;t make sense to me. The algorithm will <i>work</i> regardless of the magnitude of <i>w</i> relative to log(n).<p>Radix sort will be faster than n log(n) if <i>w</i> &lt; log(n). I would add that it is only worth to use if <i>w</i> &lt;&lt; log(n). In fact radix sort process the sorting by groups of bits, so <i>w</i> is usually smaller than the number of bits.