Min and Max of a List in Python (without using min/max function)

Question

I was wondering if there is a way to find min & max of a list without using min/max functions in Python. So i wrote a small code for the same using recursion. My logic is very naive: I make two stacks (min_stack and max_stack) which keep track of minimum and maximum during each recursive call. I have two questions:

1. Could somebody help me estimate the complexity of my code?
2. Is there a better way to do this? Will sorting the list using mergesort/quicksort and picking up first and last element give a better performance?

Thank you

Here is my attempt in Python:

minimum = []
maximum = []

# Defining Stack Class
class Stack:
    def __init__(self) :
        self.items = []

    def push(self, item) :
        self.items.append(item)

    def pop(self) :
        return self.items.pop()

    def access(self, index):
        return self.items[index]

    def isEmpty(self) :
        return (self.items == [])

    def length(self):
        return len(self.items)

def minmax(input_list):
    # make two stacks, one for min and one for max
    min_stack = Stack()
    max_stack = Stack()
    # comparing the first two elements of the list and putting them in appropriate stack
    if input_list[0]<input_list[1]:
        min_stack.push(input_list[0])
        max_stack.push(input_list[1])
    else:
        max_stack.push(input_list[0])
        min_stack.push(input_list[1])

    # Pushing remaining elements of the list into appropriate stacks. 
    for i in range(2, len(input_list)):
        if input_list[i] < min_stack.access(-1):
            min_stack.push(input_list[i])
        else:
            max_stack.push(input_list[i])

    # to find minimum
    minlist = []
    while min_stack.length() > 0:
        minlist.append(min_stack.pop())

    # to find maximum
    maxlist = []
    while max_stack.length() > 0:
        maxlist.append(max_stack.pop())

    if len(minlist) > 1:
        minmax(minlist)
    else:
        minimum.append(minlist)


    if len(maxlist) > 1:
        minmax(maxlist)
    else:
        maximum.append(maxlist)

def main():
    input_list = [2, 0, 2, 7, 5, -1, -2]
    print 'Input List is: ', input_list
    minmax(input_list)

print 'Global Minimum is: ', minimum[0]
print 'Global Maximum is: ', maximum[len(maximum)-1]

if __name__ == "__main__":
    main()

Answer 1

Using sorted() would, of course, be reliable, quick to write, and high performance for moderate-sized lists because it is built-in. For large lists, an O(n) algorithm would be faster e.g.:

def minmax1 (x):
    # this function fails if the list length is 0 
    minimum = maximum = x[0]
    for i in x[1:]:
        if i < minimum: 
            minimum = i 
        else: 
            if i > maximum: maximum = i
    return (minimum,maximum)

print(minmax1([9,8,7,6,5,4,3,2,1,11,12,13,14,15,16,17,18,19]))
print(minmax1([1]))
print(minmax1([2, 0, 2, 7, 5, -1, -2]))

... for which the output is:

(1, 19)
(1, 1)
(-2, 7)

I was interested to check the performance of the two alternatives. On my PC running Windows XP and Python 3.2.3, I found that the sorting approach is faster than the minmax1() function defined above for lists of fewer than 500 elements but, for longer lists, the O(n) minmax1() is faster. My timing test code was as follows:

def minmax_sort(x):
    x = sorted(x)
    return (x[0],x[-1])

import timeit

aa = list(range(0,100))
a = aa
while (1):
    stime = min(timeit.repeat('minmax_sort(a)', "from __main__ import minmax_sort,a",number=1000))
    mtime = min(timeit.repeat('minmax1(a)', "from __main__ import minmax,a",number=1000))
    if (stime > mtime):
        break
    else:
        a = a + aa
print(len(a))