An O(n² ) algorithm is never more efficient than an O(n log n) algorithm.

Question

Quizplus · Accepted Answer

The efficiency of an algorithm depends on the context and the size of the input. For small inputs or specific cases, an O(n²) algorithm might perform better or similarly to an O(n log n) algorithm. The notation O(n²) seems to be a typographical error or non-standard notation, making direct comparison difficult without clarification. However, assuming it represents a polynomial or other higher-order growth rate, there could still be scenarios where it is more efficient, especially if the constants involved are small or the practical implementation details favor the former.

An O(n2 ) Algorithm Is Never More Efficient Than an O(n

An O(n² ) Algorithm Is Never More Efficient Than an O(n