How to implement three stacks using a single array

Question

I came across this problem in an interview website. The problem asks for efficiently implement three stacks in a single array, such that no stack overflows until there is no

Answer 1

Another solution in PYTHON, please let me know if that works as what you think.

    class Stack(object):

    def __init__(self):
        self.stack = list()

        self.first_length = 0
        self.second_length = 0
        self.third_length = 0

        self.first_pointer = 0
        self.second_pointer = 1

    def push(self, stack_num, item):

        if stack_num == 1:
            self.first_pointer += 1
            self.second_pointer += 1
            self.first_length += 1
            self.stack.insert(0, item)

        elif stack_num == 2:
            self.second_length += 1
            self.second_pointer += 1
            self.stack.insert(self.first_pointer, item)
        elif stack_num == 3:
            self.third_length += 1
            self.stack.insert(self.second_pointer - 1, item)
        else:
            raise Exception('Push failed, stack number %d is not allowd' % stack_num)

    def pop(self, stack_num):
        if stack_num == 1:
            if self.first_length == 0:
                raise Exception('No more element in first stack')
            self.first_pointer -= 1
            self.first_length -= 1
            self.second_pointer -= 1
            return self.stack.pop(0)
        elif stack_num == 2:
            if self.second_length == 0:
                raise Exception('No more element in second stack')
            self.second_length -= 1
            self.second_pointer -= 1
            return self.stack.pop(self.first_pointer)
        elif stack_num == 3:
            if self.third_length == 0:
                raise Exception('No more element in third stack')
            self.third_length -= 1
            return self.stack.pop(self.second_pointer - 1)

    def peek(self, stack_num):
        if stack_num == 1:
            return self.stack[0]
        elif stack_num == 2:
            return self.stack[self.first_pointer]
        elif stack_num == 3:
            return self.stack[self.second_pointer]
        else:
            raise Exception('Peek failed, stack number %d is not allowd' % stack_num)

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

s = Stack()

# push item into stack 1
s.push(1, '1st_stack_1')
s.push(1, '2nd_stack_1')
s.push(1, '3rd_stack_1')
#
## push item into stack 2
s.push(2, 'first_stack_2')
s.push(2, 'second_stack_2')
s.push(2, 'third_stack_2')
#
## push item into stack 3
s.push(3, 'FIRST_stack_3')
s.push(3, 'SECOND_stack_3')
s.push(3, 'THIRD_stack_3')
#
print 'Before pop out: '
for i, elm in enumerate(s.stack):
    print '\t\t%d)' % i, elm

#
s.pop(1)
s.pop(1)
#s.pop(1)
s.pop(2)
s.pop(2)
#s.pop(2)
#s.pop(3)
s.pop(3)
s.pop(3)
#s.pop(3)
#
print 'After pop out: '
#
for i, elm in enumerate(s.stack):
    print '\t\t%d)' % i, elm