How many minimum number of comparisons required to find minimum and maximum of 100 numbers

• It requires 148 comparisons.
  To find maximum or minimum it take 3n/2 -2 comparissions
  So 150-2=148
• 10 years agoHelpfull: Yes(5) No(0)
• minimum comparisions will be 99
  min=arr[0];
  max=arr[0];
  for(i=1;i
• 10 years agoHelpfull: Yes(4) No(0)
• 98 comparisons only
• 10 years agoHelpfull: Yes(1) No(0)
• Ans: minimum number of comparison require to find minimum and maximum is: Approach is divide and conquer ....
  
  T(n) = T(floor(n/2)) + T(ceil(n/2)) + 2
  T(2) = 1 // if two element then compare both and return max and min
  T(1) = 0 // if one element then return both max and min same
  If n is a power of 2, then we can write T(n) as:
  
  T(n) = 2T(n/2) + 2
  After solving above recursion, we get
  
  T(n) = 3/2n -2
  Thus, the approach does 3/2n -2 comparisons if n is a power of 2. And it does more than 3/2n -2 comparisons if n is not a power of 2.
  
  So, here in this case put n=100 and we will get (3/2)(100) - 2 = 148 comparison .....
• 7 years agoHelpfull: Yes(1) No(0)
• How 98??
  Please explain.
• 10 years agoHelpfull: Yes(0) No(0)
• Using Merge sort or Heap sort, we can find min. and max. 100 nos. as worst case efficiency for both algorithms is o(nlog n)... i.e. 100*log 100 = 200.
• 10 years agoHelpfull: Yes(0) No(0)
• 98 comparisions
• 10 years agoHelpfull: Yes(0) No(0)
• n-1 => 99 comparisons given the data is unsorted.
  
• 10 years agoHelpfull: Yes(0) No(0)
• 1 + 2(n-2) in worst case and 1 + n – 2 in best case
  here consider best case.
  1+n-2
  =1+100-2
  =99
  
• 10 years agoHelpfull: Yes(0) No(0)