Tugas · algoritma sedang 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 Codeselesai · 11 dari 11 pengujianrespons dalam 2.5 dtk · waktu eksekusi 7 md · $0.0013

  ✓	can_partition([1, 5, 11, 5])diharapkan True	<1 md
  ✓	can_partition([1, 2, 3, 5])diharapkan False	<1 md

  ✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ 1 md✓ <1 md✓ <1 md⏱ <1 md⏱ <1 md
  Kode

  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 5.5selesai · 11 dari 11 pengujianrespons dalam 1.1 dtk · waktu eksekusi 16 md · $0.0001

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

  ✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md⏱ <1 md⏱ <1 md
  Kode

  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 Nextselesai · 11 dari 11 pengujianrespons dalam 2.3 dtk · waktu eksekusi 19 md · $0.0002

  ✓	can_partition([1, 5, 11, 5])diharapkan True	1 md
  ✓	can_partition([1, 2, 3, 5])diharapkan False	<1 md

  ✓ <1 md✓ 1 md✓ <1 md✓ <1 md✓ 2 md✓ <1 md✓ 6 md⏱ 1 md⏱ 3 md
  Kode

  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 4.7tidak selesai · 10 dari 11 pengujianrespons dalam 27 dtk · waktu eksekusi 7 md · $0.01

  ✓	can_partition([1, 5, 11, 5])diharapkan True	1 md
  ✓	can_partition([1, 2, 3, 5])diharapkan False	<1 md

  ✗ jawaban salah✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md⏱ <1 md⏱ <1 md
  Kode

  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

  Percobaan kedua

  selesai · 11 dari 11 pengujianrespons dalam 110 dtk · $0.04

  Model melihat berapa banyak tes yang gagal, contoh yang gagal beserta nilai yang diharapkan dan yang dihasilkan, serta hanya alasan mengapa tes tersembunyi gagal, bukan inputnya. Tidak memengaruhi peringkat atau rating.

  ✓	can_partition([1, 5, 11, 5])diharapkan True	<1 md
  ✓	can_partition([1, 2, 3, 5])diharapkan False	<1 md

  ✓ <1 md✓ 2 md✓ <1 md✓ 1 md✓ <1 md✓ <1 md✓ <1 md⏱ <1 md⏱ <1 md

  Kode

  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

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