Given a list of positive integers, the adjacent integers will perform the float division. For example, [2,3,4] -> 2 / 3 / 4.

However, you can add any number of parenthesis at any position to change the priority of operations. You should find out how to add parenthesis to get the maximum result, and return the corresponding expression in string format. Your expression should NOT contain redundant parenthesis.

Example:

Input: [1000,100,10,2]
Output: "1000/(100/10/2)"
Explanation:
1000/(100/10/2) = 1000/((100/10)/2) = 200
However, the bold parenthesis in "1000/((100/10)/2)" are redundant, since they don't influence the operation priority. So you should return "1000/(100/10/2)". 

Other cases:
1000/(100/10)/2 = 50
1000/(100/(10/2)) = 50
1000/100/10/2 = 0.5
1000/100/(10/2) = 2



Note:

The length of the input array is [1, 10].
Elements in the given array will be in range [2, 1000].
There is only one optimal division for each test case.

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: Dynamic programming

Data structures

Mathematics

Starter code: class Solution:
    def optimalDivision(self, nums: List[int]) -> str:
        

Solutions:
class Solution:

	def optimalDivision(self, nums):
		if len(nums) == 0:
			return ''
		if len(nums) == 1:
			return str(nums[0])
		if len(nums) == 2:
			return str(nums[0]) + '/' + str(nums[1])
		res = str(nums[0]) + '/' + '('
		for i in range(1, len(nums) - 1):
			res += str(nums[i])
			res += '/'
		res += str(nums[-1])
		res += ')'
		return res


class Solution:

	def optimalDivision(self, nums):
		s = '/'.join(map(str, nums))
		if len(nums) <= 2:
			return s
		s = s.replace('/', '/(', 1) + ')'
		return s


class Solution:

	def optimalDivision(self, nums):
		n = len(nums)
		if n == 1:
			return str(nums[0])
		elif n == 2:
			return str(nums[0]) + '/' + str(nums[1])
		else:
			result = str(nums[0]) + '/('
			for i in range(1, n - 1):
				result += str(nums[i]) + '/'
			result += str(nums[-1]) + ')'
			return result


class Solution:

	def optimalDivision(self, nums):
		nums = list(map(str, nums))
		if len(nums) == 1:
			return nums[0]
		if len(nums) == 2:
			return '/'.join(nums)
		nums[1] = '(' + nums[1]
		nums[-1] = nums[-1] + ')'
		return '/'.join(nums)


class Solution:

	def optimalDivision(self, nums):
		table = [[None for i in range(len(nums))] for j in range(len(nums))]
		for i in range(len(nums)):
			table[i][i] = (nums[i], str(nums[i]))
		return self.findmax(table, 0, len(nums) - 1)[1]

	def findmax(self, table, i, j):
		if table[i][j] is not None:
			return table[i][j]
		if j - i == 1:
			(a, a_s) = table[i][i]
			(b, b_s) = table[j][j]
			ret = (a / b, a_s + '/' + b_s)
			table[i][j] = ret
			return ret
		m = -1
		ret = None
		for d in range(i, j):
			(ma, max_rep) = self.findmax(table, i, d)
			(mi, min_rep) = self.findmin(table, d + 1, j)
			div = ma / mi
			if div > m:
				m = div
				ret = (max_rep, min_rep)
		rep = self.assemble(ret[0], ret[1])
		print('max %d, %d: v, %f, %s' % (i, j, m, rep))
		ret = (m, rep)
		table[i][j] = ret
		return ret

	def findmin(self, table, i, j):
		if table[j][i] is not None:
			return table[j][i]
		if j - i == 1:
			(a, a_s) = table[i][i]
			(b, b_s) = table[j][j]
			ret = (a / b, a_s + '/' + b_s)
			table[j][i] = ret
			return ret
		m = float('inf')
		ret = None
		for d in range(i, j):
			(ma, max_rep) = self.findmin(table, i, d)
			(mi, min_rep) = self.findmax(table, d + 1, j)
			div = ma / mi
			if div < m:
				m = div
				ret = (max_rep, min_rep)
		rep = self.assemble(ret[0], ret[1])
		print('min %d, %d: v, %f, %s' % (i, j, m, rep))
		ret = (m, rep)
		table[j][i] = ret
		return ret

	def assemble(self, up, down):
		if '/' in down:
			return '%s/(%s)' % (up, down)
		else:
			return '%s/%s' % (up, down)


class Solution:

	def optimalDivision(self, nums):
		if len(nums) == 1:
			return str(nums[0]) + ''
		elif len(nums) == 2:
			return str(nums[0]) + '/' + str(nums[1])
		answer = str(nums[0]) + '/(' + str(nums[1])
		for i in nums[2:]:
			answer += '/' + str(i)
		answer += ')'
		return answer


class Solution:

	def optimalDivision(self, nums):
		result = ''
		numsCount = len(nums)
		for i in range(numsCount):
			thisNum = nums[i]
			strNum = str(thisNum)
			if i == 0:
				result = result + strNum
			elif i == 1 and numsCount > 2:
				result = result + '/(' + strNum
			elif i == numsCount - 1 and numsCount > 2:
				result = result + '/' + strNum + ')'
			else:
				result = result + '/' + strNum
		return result


class Solution:

	def optimalDivision(self, nums):
		if len(nums) < 2:
			return str(nums[0])
		if len(nums) == 2:
			return str(nums[0]) + '/' + str(nums[1])
		l = str(nums[0]) + '/('
		for x in nums[1:]:
			l = l + str(x) + '/'
		return l[:-1] + ')'