Reorder the following efficiencies from the
smallest to the largest:

(a) n·log2(n)
(b) n + n2 + n3
(c) 24
(d) n0.5

Question

Reorder the following efficiencies from the
    smallest to the largest:
    
      (a) n·log2(n)
      (b) n + n2 + n3
      (c) 24
      (d) n0.5

Accepted Answer

Ans.
    
      24 < n0.5
      < n·log2(n)
      < n + n2 + n3
    
  

  
    💡 Explanation:

    
      To arrange the expressions correctly, compare their growth rates as
      n becomes very large.
    

    
      
        24 = 16 is a constant. It does not
        increase with n, so it has complexity
        O(1) and is the smallest.
      
      
        n0.5 = √n grows with
        n, but more slowly than a linear or log-linear
        expression.
      
      
        n·log2(n) grows faster than
        √n, but slower than a cubic polynomial.
      
      
        In n + n2 + n3, the
        highest-power term n3 dominates for
        large values of n. Therefore, the expression has
        complexity O(n3) and is the largest.
      
    

    
      Therefore, the correct growth order is:
      
        constant → square root → log-linear → cubic
      .
    

    
      Memory tip:
      
        O(1) < O(√n) < O(n log n) < O(n3)

Reorder the following efficiencies from the smallest to the largest: (a) n·log₂(n) (b) n + n² + n³ (c) 2⁴ (d) n^0.5

Answer with Explanation

Reorder the following efficiencies from the smallest to the largest: (a) n·log2(n) (b) n + n2 + n3 (c) 24 (d) n0.5

Answer with Explanation

Reorder the following efficiencies from the smallest to the largest: (a) n·log₂(n) (b) n + n² + n³ (c) 2⁴ (d) n^0.5