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: interview

Starter code:

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

Solutions:
class Solution:
     def makesquare(self, nums):
         """
         :type nums: List[int]
         :rtype: bool
         """
         if len(nums) < 4:
             return False
         
         length = sum(nums)
         if length % 4:
             return False
         length = (int) (length / 4)
         
         nums.sort(reverse=True)
         #print(nums)
         
         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)])  # (usedIndexSet, searchStartFromIndex, target, remainRounds)
         while stack:
             usedSet, startIndex, target, remainRounds = stack.pop()
             #print(usedSet, set(range(0, len(nums))) - usedSet, target, remainRounds)
             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))
                 # Else not valid path, continue
         return False
             
             


class Solution:
     def makesquare(self, nums):
         """
         :type nums: List[int]
         :rtype: bool
         """
 #         if sum(nums) % 4 != 0 or len(nums) < 4 or max(nums) > sum(nums) / 4 :
 #             return False
         
 #         nums.sort(reverse=True)      
         
 #         if nums[0] < sum(nums) / 4 and nums[0]  + nums[-1] >  sum(nums):
 #             return False
         
 #         def dfs(nums, pos, target):
 #             if pos == len(nums):
 #                 return True
             
 #             for i in range(4):
 #                 if target[i] >= nums[pos]:
 #                     target[i] -= nums[pos]
 #                     if dfs(nums, pos + 1, target):
 #                         return True
 #                     target[i] += nums[pos]
                     
                     
 #             return False
             
 #         return dfs(nums, 0,  [sum(nums) / 4] * 4)
     
     
         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):
         """
         :type nums: List[int]
         :rtype: bool
         """
         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)