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60 changes: 60 additions & 0 deletions grind75_hard/4_Median of Two Sorted Arrays/level_1.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,60 @@
# 2つのリストをマージしたものを作成していく(マージソートの要領)
# O(m + n)
class Solution:
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
marged_list = []
i = 0
j = 0
while i < len(nums1) and j < len(nums2):
if (len(nums1) + len(nums2)) // 2 < i + j:
break
if nums1[i] <= nums2[j]:
marged_list.append(nums1[i])
i += 1
else:
marged_list.append(nums2[j])
j += 1
else:
if len(nums1) <= i:
marged_list += nums2[j : (len(nums1) + len(nums2)) // 2 - i + 1]
else:
marged_list += nums1[i : (len(nums1) + len(nums2)) // 2 - j + 1]

if (len(nums1) + len(nums2)) % 2 == 1:
return marged_list[-1]
else:
return (marged_list[-2] + marged_list[-1]) / 2


# 2分探索
# k番目に小さい要素を求める関数を作成
# 小さい方からk-1個を削除すればk番目に小さいものは求められる。
# nums1: [1, 4, 5, 7, 9] nums2: [2, 3, 6, 8]
# nums1[3]=5 > nums2[1]=3 なので、nums2[0], nums2[1]は中央値にならないから捨てる
# というように、小さい方からk-1個を削除していく
# O(log(min(m, n)))
class Solution:
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
# k=0のとき最小値を返す
def get_kth_small(k, a_start, a_end, b_start, b_end):
if a_start >= a_end:
return nums2[k - a_start]
if b_start >= b_end:
return nums1[k - b_start]
a_middle = (a_start + a_end) // 2
b_middle = (b_start + b_end) // 2
if a_middle + b_middle < k:
if nums1[a_middle] < nums2[b_middle]:
return get_kth_small(k, a_middle + 1, a_end, b_start, b_end)
return get_kth_small(k, a_start, a_end, b_middle + 1, b_end)
if nums1[a_middle] < nums2[b_middle]:
return get_kth_small(k, a_start, a_end, b_start, b_middle)
return get_kth_small(k, a_start, a_middle, b_start, b_end)

nums_length = len(nums1) + len(nums2)
if nums_length % 2 == 1:
return get_kth_small(nums_length // 2, 0, len(nums1), 0, len(nums2))
return (
get_kth_small(nums_length // 2 - 1, 0, len(nums1), 0, len(nums2))
+ get_kth_small(nums_length // 2, 0, len(nums1), 0, len(nums2))
) / 2
26 changes: 26 additions & 0 deletions grind75_hard/4_Median of Two Sorted Arrays/level_2.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,26 @@
# 2分探索
# インデックスではなく、配列を渡すように修正
class Solution:
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
def get_kth_small(k, nums1, nums2):
if not nums1:
return nums2[k]
if not nums2:
return nums1[k]
middle_1 = len(nums1) // 2
middle_2 = len(nums2) // 2
if middle_1 + middle_2 < k:
if nums1[middle_1] < nums2[middle_2]:
return get_kth_small(k - middle_1 - 1, nums1[middle_1 + 1 :], nums2)
return get_kth_small(k - middle_2 - 1, nums1, nums2[middle_2 + 1 :])
if nums1[middle_1] < nums2[middle_2]:
return get_kth_small(k, nums1, nums2[:middle_2])
return get_kth_small(k, nums1[:middle_1], nums2)

nums_length = len(nums1) + len(nums2)
if nums_length % 2 == 1:
return get_kth_small(nums_length // 2, nums1, nums2)
return (
get_kth_small(nums_length // 2 - 1, nums1, nums2)
+ get_kth_small(nums_length // 2, nums1, nums2)
) / 2
24 changes: 24 additions & 0 deletions grind75_hard/4_Median of Two Sorted Arrays/level_3.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
class Solution:
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
def get_kth_small(k, nums1, nums2):
if not nums1:
return nums2[k]
if not nums2:
return nums1[k]
middle_1 = len(nums1) // 2
middle_2 = len(nums2) // 2
if middle_1 + middle_2 < k:
if nums1[middle_1] < nums2[middle_2]:
return get_kth_small(k - middle_1 - 1, nums1[middle_1 + 1 :], nums2)
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これはここでスライスを取っているからコピーされていませんか。

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level1のようにindexで管理したほうが良さそうですね。
引数減らせて読みやすいかと思いましたが、遅くなるデメリットのほうが大きいですね。

return get_kth_small(k - middle_2 - 1, nums1, nums2[middle_2 + 1 :])
if nums1[middle_1] < nums2[middle_2]:
return get_kth_small(k, nums1, nums2[:middle_2])
return get_kth_small(k, nums1[:middle_1], nums2)

nums_length = len(nums1) + len(nums2)
if nums_length % 2 == 1:
return get_kth_small(nums_length // 2, nums1, nums2)
return (
get_kth_small(nums_length // 2 - 1, nums1, nums2)
+ get_kth_small(nums_length // 2, nums1, nums2)
) / 2