Given an integer array, find three numbers whose product is maximum and output the maximum product.

Example 1:

Input: [1,2,3]
Output: 6



Example 2:

Input: [1,2,3,4]
Output: 24



Note:

The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer.

Difficulty: EASY
Starter code: class Solution:
    def maximumProduct(self, nums: List[int]) -> int:
        

Solutions:
class Solution:

	def maximumProduct(self, nums):
		import heapq
		(a, b) = (heapq.nlargest(3, nums), heapq.nsmallest(2, nums))
		return max(a[0] * a[1] * a[2], a[0] * b[0] * b[1])


class Solution:

	def maximumProduct(self, nums):
		(h, m, l) = (-float('inf'), -float('inf'), -float('inf'))
		(lowest, second_lowest) = (float('inf'), float('inf'))
		for num in nums:
			if num > h:
				l = m
				m = h
				h = num
			elif num > m:
				l = m
				m = num
			elif num > l:
				l = num
			if num < lowest:
				second_lowest = lowest
				lowest = num
			elif num < second_lowest:
				second_lowest = num
		return max(h * m * l, h * lowest * second_lowest)


class Solution:

	def maximumProduct1(self, nums):
		nums.sort()
		top = nums[~0] * nums[~1] * nums[~2]
		bottom = nums[0] * nums[1] * nums[~0]
		return max(top, bottom)

	def maximumProduct(self, nums):
		min1 = min2 = float('inf')
		max1 = max2 = max3 = -float('inf')
		for n in nums:
			if n <= min1:
				(min1, min2) = (n, min1)
			elif n <= min2:
				min2 = n
			if n >= max1:
				(max1, max2, max3) = (n, max1, max2)
			elif n >= max2:
				(max2, max3) = (n, max2)
			elif n >= max3:
				max3 = n
		return max(min1 * min2 * max1, max1 * max2 * max3)


class Solution:

	def maximumProduct(self, nums):
		(max1, max2, max3, min1, min2) = (float('-Inf'), float('-Inf'), float('-Inf'), float('Inf'), float('Inf'))
		for num in nums:
			if num >= max1:
				(max3, max2, max1) = (max2, max1, num)
			elif num >= max2:
				(max3, max2) = (max2, num)
			elif num > max3:
				max3 = num
			if num <= min1:
				(min2, min1) = (min1, num)
			elif num < min2:
				min2 = num
		return max(max1 * max2 * max3, min1 * min2 * max1)


class Solution:

	def maximumProduct(self, nums):
		a = b = c = -1001
		d = e = 1001
		for num in nums:
			if num >= a:
				c = b
				b = a
				a = num
			elif num >= b:
				c = b
				b = num
			elif num >= c:
				c = num
			if num <= d:
				e = d
				d = num
			elif num <= e:
				e = num
		return max(a * b * c, a * d * e)


class Solution:

	def maximumProduct(self, nums):
		max_num = -1000
		s_max = -1000
		t_max = -1000
		min_num = 1000
		s_min = 1000
		for i in range(len(nums)):
			if nums[i] > max_num:
				t_max = s_max
				s_max = max_num
				max_num = nums[i]
			elif nums[i] > s_max:
				t_max = s_max
				s_max = nums[i]
			elif nums[i] > t_max:
				t_max = nums[i]
			if nums[i] < min_num:
				s_min = min_num
				min_num = nums[i]
			elif nums[i] < s_min:
				s_min = nums[i]
		print(max_num, s_max, t_max, min_num, s_min)
		a = max_num * s_max * t_max
		b = min_num * s_min * max_num
		if a > b:
			return a
		else:
			return b


class Solution:

	def maximumProduct(self, nums):
		new = [0 for i in range(2001)]
		for i in range(len(nums)):
			new[1000 + nums[i]] = new[1000 + nums[i]] + 1
		count = 0
		product = 1
		digit = []
		for i in range(2000, -1, -1):
			while new[i] > 0:
				digit.append(i - 1000)
				count = count + 1
				new[i] = new[i] - 1
				if count == 3:
					break
			if count == 3:
				break
		for i in range(2001):
			while new[i] > 0:
				digit.append(i - 1000)
				count = count + 1
				new[i] = new[i] - 1
				if count == 5:
					break
			if count == 5:
				break
		if len(digit) == 3:
			return digit[0] * digit[1] * digit[2]
		elif len(digit) == 4:
			return max(digit[0] * digit[1] * digit[2], digit[3] * digit[1] * digit[2])
		else:
			p1 = digit[0] * digit[3] * digit[4]
			p2 = digit[0] * digit[1] * digit[2]
			return max(p1, p2)