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| 1 | +class Solution(object): |
| 2 | + |
| 3 | + def makesquare(self, nums): |
| 4 | + """ |
| 5 | + :type nums: List[int] |
| 6 | + :rtype: bool |
| 7 | + """ |
| 8 | + |
| 9 | + # If there are no matchsticks, then we can't form any square. |
| 10 | + if not nums: |
| 11 | + return False |
| 12 | + |
| 13 | + # Number of matchsticks |
| 14 | + L = len(nums) |
| 15 | + |
| 16 | + # Possible perimeter of our square |
| 17 | + perimeter = sum(nums) |
| 18 | + |
| 19 | + # Possible side of our square from the given matchsticks |
| 20 | + possible_side = perimeter // 4 |
| 21 | + |
| 22 | + # If the perimeter isn't equally divisible among 4 sides, return False. |
| 23 | + if possible_side * 4 != perimeter: |
| 24 | + return False |
| 25 | + |
| 26 | + # Memoization cache for the dynamic programming solution. |
| 27 | + memo = {} |
| 28 | + |
| 29 | + # mask and the sides_done define the state of our recursion. |
| 30 | + def recurse(mask, sides_done): |
| 31 | + |
| 32 | + # This will calculate the total sum of matchsticks used till now using the bits of the mask. |
| 33 | + total = 0 |
| 34 | + for i in range(L - 1, -1, -1): |
| 35 | + if not (mask & (1 << i)): |
| 36 | + total += nums[L - 1 - i] |
| 37 | + |
| 38 | + # If some of the matchsticks have been used and the sum is divisible by our square's side, then we increment the number of sides completed. |
| 39 | + if total > 0 and total % possible_side == 0: |
| 40 | + sides_done += 1 |
| 41 | + |
| 42 | + # If we were successfully able to form 3 sides, return True |
| 43 | + if sides_done == 3: |
| 44 | + return True |
| 45 | + |
| 46 | + # If this recursion state has already been calculated, just return the stored value. |
| 47 | + if (mask, sides_done) in memo: |
| 48 | + return memo[(mask, sides_done)] |
| 49 | + |
| 50 | + # Common variable to store answer from all possible further recursions from this step. |
| 51 | + ans = False |
| 52 | + |
| 53 | + # rem stores available space in the current side (incomplete). |
| 54 | + c = int(total / possible_side) |
| 55 | + rem = possible_side * (c + 1) - total |
| 56 | + |
| 57 | + # Iterate over all the matchsticks |
| 58 | + for i in range(L - 1, -1, -1): |
| 59 | + |
| 60 | + # If the current one can fit in the remaining space of the side and it hasn't already been taken, then try it out |
| 61 | + if nums[L - 1 - i] <= rem and mask & (1 << i): |
| 62 | + |
| 63 | + # If the recursion after considering this matchstick gives a True result, just break. No need to look any further. |
| 64 | + # mask ^ (1 << i) makes the i^th from the right, 0 making it unavailable in future recursions. |
| 65 | + if recurse(mask ^ (1 << i), sides_done): |
| 66 | + ans = True |
| 67 | + break |
| 68 | + # cache the result for the current recursion state. |
| 69 | + memo[(mask, sides_done)] = ans |
| 70 | + return ans |
| 71 | + |
| 72 | + # recurse with the initial mask with all matchsticks available. |
| 73 | + return recurse((1 << L) - 1, 0) |
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