C Question - 1

Replies

• 

  Harshad Italiya

  unsigned char S1[]="MBCAE"
   
  unsigned char S2[]="XYZAC"
   
  unsigned char i,j,match;
   
  void main()
   
  {
   
  for(i=0;i<5;i++)
   
  {
   
  for(j=0;j<5;j++)
   
  {
   
  if(S1==S2[j])
   
  {
   
  match =1;
   
  break;
   
  }
   
  }
   
  if(match==1)
   
  break;
   
  }
   
  print(S1);
   
  }
   
  

  PS:- Haven't compile this code this is logic only It may be wrong.. 😉
• 

  Saandeep Sreerambatla

  Ok, the logic is correct.
  
  Here you are using two for loops, so the worst case scenario is you have the number of comparisions on the order of n[sup]2[sup].
• 

  Saandeep Sreerambatla

  Now, I need ideas on how will you reduce the order of the program.
  
  PS. THis is not my assignment or my project work 😛
• 

  Saandeep Sreerambatla

  TO add curiosity , this is a Microsoft Interview question! faced by my friend.
  
  Hink: Sorting technique will reduce the order to some extent.
  
  There are other ways as well to reduce the order to only "n". So think and post your answers.
• 

  simplycoder

  Sort and use binary search. Where, sort can be implemented in O(n) time and binary search takes up O(lgn)
• 

  Saandeep Sreerambatla

  Great! so we have reduced the order to nlogn now.
  
  There is a way where we can reduce the way only to "n" ways.
  
  That is using the hast table technique.
• 

  Saandeep Sreerambatla

  Next question on the same one, what all scenarios you can think in testing the above code?
  
  One is already done, we test the order and reduced it to N ways.
  
  Consider string 1 and string 2 are pointers and give the test scenarios... There are many interesting things to learn. let me know your thoughts first.