How to find max. and min. in array using minimum comparisons?

Question

This is a interview question: given an array of integers find the max. and min. using minimum comparisons.

Obviously, I can loop over the array twice and use ~

Answer 1

Compare in Pairs will work best for minimum comparisons

# Initialization # 
- if len(arr) is even, min = min(arr[0], arr[1]), max = max(arr[0], arr[1])
- if len(arr) is odd, min = min = arr[0], max = arr[0]

# Loop over pairs # 
- Compare bigger of the element with the max, and smaller with min,
- if smaller element less than min, update min, similarly with max.

Total Number of comparisons -

• For size = odd, 3(n - 1) / 2 where n is size of array
• For size = even, 1 + 3*(n - 2)/2 = 3n/2 - 2

Below is the python code for the above pseudo-code

class Solution(object):
    def min_max(self, arr):
        size = len(arr)
        if size == 1:
            return arr[0], arr[0]

        if size == 2:
            return arr[0], arr[1]

        min_n = None
        max_n = None
        index = None
        if size % 2 == 0:       # One comparison
            min_n = min(arr[0], arr[1])
            max_n = max(arr[0], arr[1])
            st_index = 2

        else:
            min_n = arr[0]
            max_n = arr[0]
            st_index = 1
        for index in range(st_index, size, 2):
            if arr[index] < arr[index + 1]:
                min_n = min(arr[index], min_n)
                max_n = max(arr[index + 1], max_n)
            else:
                min_n = min(arr[index + 1], min_n)
                max_n = max(arr[index], max_n)

        return min_n, max_n