Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

 Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

Example 1:

Input: [1,1,2,2,2]
Output: true

Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.



Example 2:

Input: [3,3,3,3,4]
Output: false

Explanation: You cannot find a way to form a square with all the matchsticks.



Note:

The length sum of the given matchsticks is in the range of 0 to 10^9.
The length of the given matchstick array will not exceed 15.

Difficulty: MEDIUM_HARD
Tags: Dynamic programming

Bit manipulation

Data structures

Complete search
Skill types: Dynamic programming

Bit manipulation

Data structures

Complete search

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

Solutions:
class Solution:

	def makesquare(self, nums):
		if len(nums) < 4:
			return False
		length = sum(nums)
		if length % 4:
			return False
		length = int(length / 4)
		nums.sort(reverse=True)
		if length < nums[0]:
			return False
		elif length == nums[0]:
			stack = list([(set([0]), 1, length, 1)])
		else:
			stack = list([(set([0]), 1, length - nums[0], 2)])
		while stack:
			(usedSet, startIndex, target, remainRounds) = stack.pop()
			for i in range(len(nums) - 1, startIndex - 1, -1):
				if i in usedSet:
					continue
				num = nums[i]
				if num < target and i + 1 < len(nums):
					stack.append((usedSet | {i}, i + 1, target - num, remainRounds))
				elif num == target:
					if remainRounds == 0:
						return True
					else:
						stack.append((usedSet | {i}, 1, length, remainRounds - 1))
		return False


class Solution:

	def makesquare(self, nums):
		total = sum(nums)
		if total % 4 != 0 or len(nums) < 4:
			return False
		size = total / 4
		nums.sort(reverse=True)
		used = [False] * len(nums)

		def dfs(i, expect):
			if i >= len(nums):
				return expect % size == 0
			if used[i]:
				return dfs(i + 1, expect)
			used[i] = True
			if nums[i] == expect:
				return True
			if nums[i] < expect:
				expect -= nums[i]
				available = [j for j in range(i + 1, len(nums)) if not used[j]]
				for x in available:
					if dfs(x, expect):
						return True
			used[i] = False
			return False
		for i in range(len(nums)):
			if not dfs(i, size):
				return False
		return True


class Solution:

	def makesquare(self, nums):

		def checkv(t, c, nums, used):
			if t == 0:
				return True
			if t < 0:
				return False
			while c < len(nums) and used[c]:
				c = c + 1
			if c >= len(nums):
				return False
			if checkv(t - nums[c], c + 1, nums, used):
				used[c] = True
				return True
			if checkv(t, c + 1, nums, used):
				return True
			return False
		edgel = sum(nums)
		if edgel % 4 != 0 or edgel == 0:
			return False
		edgel = edgel / 4
		nums.sort(key=lambda x: -x)
		n = len(nums)
		used = [False] * n
		for p in range(3):
			t = edgel
			if not checkv(t, 0, nums, used):
				return False
		return True


class Solution:

	def makesquare(self, nums):
		s = sum(nums)
		if not s % 4 == 0:
			return False
		l = s // 4
		from collections import Counter
		self.c = Counter(nums)
		for _ in range(4):
			n = self.f(0, sorted(self.c.elements(), reverse=True), l, ())
			if not n:
				return False
			self.c.subtract(n)
		return True

	def f(self, index, keys, sum, nums):
		if sum == 0:
			return nums
		if sum < 0 or index >= len(keys):
			return None
		return self.f(index + 1, keys, sum - keys[index], (*nums, keys[index])) or self.f(index + 1, keys, sum, nums)