Running time of amstrong algorithm

Question

I have a problem how to find best, worst, average case in armstrong number algorithm?

Here the pseudo-code :

Declaration:
      n : integer
      Sum : integer
      Temp : integer
      Rem  : integer
Algorithm:
      Input(n)
      Temp <-- n
      Sum <-- 0
      Rem <--- 0
      While ( n != 0 ) do
               Rem <-- n mod 10
               Sum <-- Sum + Rem * Rem * Rem
               n <-- n / 10
      Endwhile

      if ( Sum = n ) then
         output (n ," is Armstrong number")
     else
         Output(n," is not armstrong number")
         Endif

I find

Best-case : Tmin(n) = 3,

Worst-case : Tmax(n) = 3n,

Average-case : Tavg(n) = 3 [ n(n+1)/2 ] / n = 3(n+1)/2 = 3n+3/2

Sn= n(n+1)/2 The sum of the natural numbers from 1 to n

Thanks for you help.

score 0 · Accepted Answer · answered Nov 05 '15 at 01:03

Essentially, the only thing that affects the run time of your algorithm is the loop. This will iterate for as many times as it takes to get from $n$ to 0 by dividing $n$ by 10. So we're asking, how many times will it be before $n, n/10, n/100, n/1000, \dotsc$ gets to 0? That is, when is $n/10^k = 0$? Clearly, when $k=\log_{10}n$. The loop is unaffected by any difference between the $n$ values, as long as they have the same number of decimal digits, so best-case time = worst-case time = average-case time = $3\log n$ plus some constant overhead for setting the variables and displaying the output.

Running time of amstrong algorithm

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