Write a C function to return minimum and maximum in an array. Your program should make the minimum number of comparisons. 

First of all, how do we return multiple values from a C function? We can do it either using structures or pointers. 
We have created a structure named pair (which contains min and max) to return multiple values. 

And the function declaration becomes: struct pair getMinMax(int arr[], int n) where arr[] is the array of size n whose minimum and maximum are needed. 

METHOD 1 (Simple Linear Search) 
Initialize values of min and max as minimum and maximum of the first two elements respectively. Starting from 3rd, compare each element with max and min, and change max and min accordingly (i.e., if the element is smaller than min then change min, else if the element is greater than max then change max, else ignore the element) 

Output: 

Minimum element is 1
Maximum element is 3000

Time Complexity: O(n)

In this method, the total number of comparisons is 1 + 2(n-2) in the worst case and 1 + n – 2 in the best case. 
In the above implementation, the worst case occurs when elements are sorted in descending order and the best case occurs when elements are sorted in ascending order.

METHOD 2 (Tournament Method) 
Divide the array into two parts and compare the maximums and minimums of the two parts to get the maximum and the minimum of the whole array.

Pair MaxMin(array, array_size)
   if array_size = 1
      return element as both max and min
   else if arry_size = 2
      one comparison to determine max and min
      return that pair
   else    /* array_size  > 2 */
      recur for max and min of left half
      recur for max and min of right half
      one comparison determines true max of the two candidates
      one comparison determines true min of the two candidates
      return the pair of max and min

Implementation 

Output: 

Minimum element is 1
Maximum element is 3000

Time Complexity: O(n) 

Total number of comparisons: let the number of comparisons be T(n). T(n) can be written as follows: 
Algorithmic Paradigm: Divide and Conquer 

             
  T(n) = T(floor(n/2)) + T(ceil(n/2)) + 2  
  T(2) = 1
  T(1) = 0

If n is a power of 2, then we can write T(n) as: 

   T(n) = 2T(n/2) + 2

After solving the above recursion, we get 

  T(n)  = 3n/2 -2

Thus, the approach does 3n/2 -2 comparisons if n is a power of 2. And it does more than 3n/2 -2 comparisons if n is not a power of 2.

METHOD 3 (Compare in Pairs) 
If n is odd then initialize min and max as first element. 
If n is even then initialize min and max as minimum and maximum of the first two elements respectively. 
For rest of the elements, pick them in pairs and compare their 
maximum and minimum with max and min respectively. 

Output: 

Minimum element is 1
Maximum element is 3000

Time Complexity: O(n)

Total number of comparisons: Different for even and odd n, see below: 

       If n is odd:    3*(n-1)/2  
       If n is even:   1 Initial comparison for initializing min and max, 
                           and 3(n-2)/2 comparisons for rest of the elements  
                      =  1 + 3*(n-2)/2 = 3n/2 -2

Second and third approaches make the equal number of comparisons when n is a power of 2. 
In general, method 3 seems to be the best.
Please write comments if you find any bug in the above programs/algorithms or a better way to solve the same problem.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.