Implement a function that returns the minimal and the maximal value of a list (in this order).

Difficulty: introductory

Starter code:

def get_min_max(seq):
	

Solutions:
def get_min_max(seq): 
    return min(seq), max(seq)

def get_min_max(L): 
    return (min(L),max(L))

get_min_max = lambda seq: (min(seq), max(seq))

def get_min_max(seq):
    max = seq[0]
    min = seq[0]
    for a in seq:
        if a > max: max = a
        if a < min: min = a
    return min,max


def get_min_max(seq): 
    return (min(seq),max(seq)) if len(seq)>1 else (seq[0],seq[0])

def get_min_max(seq): 
    s = sorted(seq)
    return (s[0], s[-1])


# Computational complexity of 1.5n
def get_min_max(seq):
    seq = iter(seq)
    seq_min = seq_max = next(seq)

    while True:
        try:
            a = next(seq)
        except StopIteration:
            break

        b = next(seq, seq_min)
        if a > b:
            seq_min = min(b, seq_min)
            seq_max = max(a, seq_max)
        else:
            seq_min = min(a, seq_min)
            seq_max = max(b, seq_max)

    return seq_min, seq_max

def get_min_max(seq):
    seq.sort()
    return (seq[0],seq[0]) if len(seq) is 1 else (seq[0],seq[len(seq)-1])


def get_min_max(seq): 
    return (min(seq),max(seq))# your code here
# print(min(seq),max(seq))pass


def get_min_max(seq): 
    min = seq[0]
    max = seq[0]
    for number in seq:
        if number > max:
            max = number
      
    for number in seq:
        if number < min:
            min = number
          
    return (min, max)