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[shinsj4653] Week 14 Solutions #1634

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15 changes: 15 additions & 0 deletions binary-tree-level-order-traversal/shinsj4653.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
"""
[문제풀이]
# Inputs

# Outputs

# Constraints

# Ideas

[회고]

"""


52 changes: 52 additions & 0 deletions counting-bits/shinsj4653.py
Original file line number Diff line number Diff line change
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"""
[문제풀이]
# Inputs
n: int
# Outputs

ans: arr
길이: n + 1

ans내 각 원소들 : ans[i]: i의 이진법에서 1의 개수

# Constraints
0 <= n <= 10^5

# Ideas
2중 for문 이면 안됨

[회고]
dp를 활용한 풀이도 같이 알아두자
"""

# 내 풀이
class Solution:
def countBits(self, n: int) -> List[int]:

ans = []

for i in range(n + 1):
if i == 0 or i == 1:
ans.append(i)
continue

num = i
cnt = 0
while num > 0:
num, n = num // 2, num % 2
if n == 1:
cnt += 1

ans.append(cnt)

return ans

# 해설
class Solution:
def countBits(self, n: int) -> List[int]:
dp = [0] * (n + 1)
for num in range(1, n + 1):
dp[num] = dp[num // 2] + (num % 2)
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가장 오른쪽 비트를 기준으로 하는 DP로 풀이하신 것 같네요!

저의 경우에는 // 2% 2 연산은 비트 연산으로 대체하는 편인데요, 비트 연산이 성능 측면에서 미세하게나마 더 효율적이라고 합니다! 따라서 dp[num] = dp[num >> 1] + (num & 1)로도 작성할 수 있을 것 같아요.

또한, 가장 왼쪽 비트를 기준으로 하는 DP로도 한 번 풀이해보시는 걸 추천드려요~!

return dp


15 changes: 15 additions & 0 deletions house-robber-ii/shinsj4653.py
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"""
[문제풀이]
# Inputs

# Outputs

# Constraints

# Ideas

[회고]

"""


15 changes: 15 additions & 0 deletions meeting-rooms-ii/shinsj4653.py
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"""
[문제풀이]
# Inputs

# Outputs

# Constraints

# Ideas

[회고]

"""


15 changes: 15 additions & 0 deletions word-search-ii/shinsj4653.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
"""
[문제풀이]
# Inputs

# Outputs

# Constraints

# Ideas

[회고]

"""