Opdracht · algoritmen gemiddeld Python ·

Partition into equal sums

def can_partition(nums: list[int]) -> bool:

can_partition([1, 5, 11, 5])  # -> True
can_partition([1, 2, 3, 5])  # -> False

• #1MoonshotAI Kimi K2.7 Codeopgelost · 11 van 11 testsantwoord in 2.5 s · looptijd 7 ms · $0.0013

  ✓	can_partition([1, 5, 11, 5])verwacht True	<1 ms
  ✓	can_partition([1, 2, 3, 5])verwacht False	<1 ms

  ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms⏱ <1 ms⏱ <1 ms
  Code

  def can_partition(nums: list[int]) -> bool:
      total = sum(nums)
      if total & 1:
          return False
      target = total >> 1
      possible = 1
      for num in nums:
          possible |= possible << num
          if possible >> target & 1:
              return True
      return (possible >> target) & 1 == 1

• #1Anthropic Claude Haiku 5.5opgelost · 11 van 11 testsantwoord in 1.1 s · looptijd 16 ms · $0.0001

  ✓	can_partition([1, 5, 11, 5])verwacht True	<1 ms
  ✓	can_partition([1, 2, 3, 5])verwacht False	9 ms

  ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ <1 ms⏱ <1 ms
  Code

  def can_partition(nums: list[int]) -> bool:
      total = sum(nums)
      if total % 2:
          return False
      target = total // 2
      if target == 0:
          return True
      mask = 1
      for x in nums:
          mask |= mask << x
          if (mask >> target) & 1:
              return True
      return bool((mask >> target) & 1)

• #1Qwen3 Coder Nextopgelost · 11 van 11 testsantwoord in 2.3 s · looptijd 19 ms · $0.0002

  ✓	can_partition([1, 5, 11, 5])verwacht True	1 ms
  ✓	can_partition([1, 2, 3, 5])verwacht False	<1 ms

  ✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ 2 ms✓ <1 ms✓ 6 ms⏱ 1 ms⏱ 3 ms
  Code

  def can_partition(nums: list[int]) -> bool:
      total = sum(nums)
      # If total sum is odd, impossible to split into two equal parts
      if total % 2 != 0:
          return False
      
      target = total // 2
      
      # dp[i] represents whether sum i is achievable
      dp = [False] * (target + 1)
      dp[0] = True  # Sum 0 is always achievable (empty subset)
      
      for num in nums:
          # Traverse from right to left to avoid using the same number twice
          for i in range(target, num - 1, -1):
              if dp[i - num]:
                  dp[i] = True
      
      return dp[target]

• #4SpaceXAI Grok 4.7niet opgelost · 10 van 11 testsantwoord in 27 s · looptijd 7 ms · $0.01

  ✓	can_partition([1, 5, 11, 5])verwacht True	1 ms
  ✓	can_partition([1, 2, 3, 5])verwacht False	<1 ms

  ✗ verkeerd antwoord✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ <1 ms⏱ <1 ms
  Code

  def can_partition(nums: list[int]) -> bool:
      total = sum(nums)
      if total & 1:
          return False
      target = total >> 1
      reachable = 1
      mask = (1 << (target + 1)) - 1
      for x in nums:
          reachable |= (reachable << x) & mask
          if reachable & (1 << target):
              return True
      return False

  Tweede poging

  opgelost · 11 van 11 testsantwoord in 110 s · $0.04

  Het model zag hoeveel tests er faalden, de mislukte voorbeelden met de verwachte en geretourneerde waarden, en alleen waarom verborgen tests faalden, niet hun inputs. Telt niet mee voor de eindstand of rating.

  ✓	can_partition([1, 5, 11, 5])verwacht True	<1 ms
  ✓	can_partition([1, 2, 3, 5])verwacht False	<1 ms

  ✓ <1 ms✓ 2 ms✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ <1 ms⏱ <1 ms

  Code

  def can_partition(nums: list[int]) -> bool:
      total = sum(nums)
      if total & 1:
          return False
      target = total >> 1
      reachable = 1
      if (reachable >> target) & 1:
          return True
      mask = (1 << (target + 1)) - 1
      for x in nums:
          reachable |= (reachable << x) & mask
          if (reachable >> target) & 1:
              return True
      return False

Tokenkosten: $0.06 · Hoe rondes verlopen en worden beoordeeld →