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60 changes: 60 additions & 0 deletions contains-duplicate/EGON.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,60 @@
from unittest import TestCase, main
from typing import List
from collections import Counter


class Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
return self.solve_3(nums=nums)

"""
Runtime: 412 ms (Beats 75.17%)
Analyze Complexity: O(n)
Memory: 31.92 MB (Beats 45.93%)
"""
def solve_1(self, nums: List[int]) -> bool:
return len(nums) != len(set(nums))

"""
Runtime: 423 ms (Beats 39.66%)
Analyze Complexity: O(n)
Memory: 34.54 MB (Beats 14.97%)
"""
def solve_2(self, nums: List[int]) -> bool:
counter = {}
for num in nums:
if num in counter:
return True
else:
counter[num] = True
else:
return False

"""
Runtime: 441 ms (Beats 16.59%)
Analyze Complexity: O(n)
Memory: 34.57 MB (Beats 14.97%)
"""
def solve_3(self, nums: List[int]) -> bool:
return Counter(nums).most_common(1)[0][1] > 1


class _LeetCodeTCs(TestCase):
def test_1(self):
nums = [1, 2, 3, 1]
output = True
self.assertEqual(Solution.containsDuplicate(Solution(), nums=nums), output)

def test_2(self):
nums = [1, 2, 3, 4]
output = False
self.assertEqual(Solution.containsDuplicate(Solution(), nums=nums), output)

def test_3(self):
nums = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]
output = True
self.assertEqual(Solution.containsDuplicate(Solution(), nums=nums), output)


if __name__ == '__main__':
main()
111 changes: 111 additions & 0 deletions kth-smallest-element-in-a-bst/EGON.py
Original file line number Diff line number Diff line change
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from typing import Optional, List
from unittest import TestCase, main
from heapq import heappush, heappop


class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right


class Solution:
def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
return self.solve_2(root, k)

"""
Runtime: 50 ms (Beats 25.03%)
Analyze Complexity: O(n)
Memory: 19.55 MB (Beats 15.91%)
"""
def solve_1(self, root: Optional[TreeNode], k: int) -> int:
visited = []
stack = [root]
while stack:
curr_node = stack.pop()
heappush(visited, curr_node.val)
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다양한 풀이로 하셔서 좋네요 :) 요부분 조금 궁금한게 있는데, bst의 heappush의 Time complexity가 제 기억엔 logn 이었던거로 기억하는데, 혹시 어떻게 n으로 도출하셨는지 알려주실수 있을까요?

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n개의 노드를 순회하며 각각 heap push를 하니 말씀하신 것 처럼 n * log n 이 맞습니다. 주석 복붙하다 빼먹은 것 같네요. 리뷰 감사합니다.

if curr_node.left is not None:
stack.append(curr_node.left)
if curr_node.right is not None:
stack.append(curr_node.right)

result = visited[0]
for _ in range(k):
result = heappop(visited)

return result

"""
Runtime: 43 ms (Beats 69.91%)
Analyze Complexity: O(n)
Memory: 19.46 MB (Beats 60.94%)
"""
def solve_2(self, root: Optional[TreeNode], k: int) -> int:
vals = []

def inorder_traverse(root: Optional[TreeNode]):
if root is None:
return

if len(vals) >= k:
return
Comment on lines +51 to +52
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이 조건으로 빠르게 탐색을 마칠 수 있어 정말 좋네요! 👍


inorder_traverse(root.left)
vals.append(root.val)
inorder_traverse(root.right)

inorder_traverse(root)
return vals[k - 1]


class _LeetCodeTCs(TestCase):
def test_1(self):
root = TreeNode(
val=3,
left=TreeNode(
val=1,
left=None,
right=TreeNode(
val=2,
left=None,
right=None,
)
),
right=TreeNode(
val=4,
left=None,
right=None
)
)

k = 1
output = 1
self.assertEqual(Solution.kthSmallest(Solution(), root, k), output)

def test_2(self):
root = TreeNode(
val=5,
left=TreeNode(
val=3,
left=TreeNode(
val=2,
left=TreeNode(
val=1
)
),
right=TreeNode(
val=4
)
),
right=TreeNode(
val=6
)
)
k = 3
output = [3]
self.assertEqual(Solution.kthSmallest(Solution(), root, k), output)


if __name__ == '__main__':
main()
72 changes: 72 additions & 0 deletions number-of-1-bits/EGON.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,72 @@
from unittest import TestCase, main


class Solution:
def hammingWeight(self, n: int) -> int:
return self.solve_4(n=n)

"""
Runtime: 26 ms (Beats 97.13%)
Analyze Complexity: O(log n), bin(int)가 O(log(n))
Memory: 16.56 MB (Beats 22.67%)
"""
def solve_1(self, n: int) -> int:
return bin(n).count('1')

"""
Runtime: 31 ms (Beats 85.00%)
Analyze Complexity: O(log n)
Memory: 16.60 MB (Beats 22.67%)
"""
def solve_2(self, n: int) -> int:
hamming_weight = 0
while n:
hamming_weight += n % 2
n = n >> 1
return hamming_weight

"""
Runtime: 30 ms (Beats 88.73%)
Analyze Complexity: O(k), k는 2진수의 1의 갯수 (== O(log(n)))
Memory: 16.56 MB (Beats 22.67%)
"""
# Brian Kernighan's Algorithm
def solve_3(self, n: int) -> int:
hamming_weight = 0
while n:
n &= (n - 1)
hamming_weight += 1
return hamming_weight

"""
Runtime: 36 ms (Beats 53.56%)
Analyze Complexity: O(k), k는 2진수의 1의 갯수 (== O(log(n)))
Memory: 16.55 MB (Beats 22.67%)
"""
def solve_4(self, n: int) -> int:
hamming_weight = 0
while n:
hamming_weight += n & 1
n >>= 1
return hamming_weight


class _LeetCodeTCs(TestCase):
def test_1(self):
n = 11
output = 3
self.assertEqual(Solution.hammingWeight(Solution(), n=n), output)

def test_2(self):
n = 128
output = 1
self.assertEqual(Solution.hammingWeight(Solution(), n=n), output)

def test_3(self):
n = 2147483645
output = 30
self.assertEqual(Solution.hammingWeight(Solution(), n=n), output)


if __name__ == '__main__':
main()
57 changes: 57 additions & 0 deletions palindromic-substrings/EGON.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,57 @@
from unittest import TestCase, main


class Solution:
def countSubstrings(self, s: str) -> int:
return self.solve_2(s)

"""
Runtime: 776 ms (Beats 10.15%)
Analyze Complexity: O(n ** 3), for * for * slicing
Memory: 16.43 MB (Beats 73.44%)
"""
def solve_1(self, s: str) -> int:
count = 0
for left in range(len(s)):
for right in range(left, len(s)):
substring = s[left: right + 1]
count += substring == substring[::-1]

return count

"""
Runtime: 373 ms (Beats 19.31%)
Analyze Complexity: O(n ** 2)
Memory: 25.15 MB (Beats 11.85%)
"""
def solve_2(self, s: str) -> int:
dp = [[left == right for left in range(len(s))] for right in range(len(s))]
count = len(s)

for i in range(len(s) - 1):
dp[i][i + 1] = s[i] == s[i + 1]
count += dp[i][i + 1]

for length in range(3, len(s) + 1):
for left in range(len(s) - length + 1):
right = left + length - 1
dp[left][right] = dp[left + 1][right - 1] and s[left] == s[right]
count += dp[left][right]

return count


class _LeetCodeTCs(TestCase):
def test_1(self):
s = "abc"
output = 3
self.assertEqual(Solution.countSubstrings(Solution(), s), output)

def test_2(self):
s = "aaa"
output = 6
self.assertEqual(Solution.countSubstrings(Solution(), s), output)


if __name__ == '__main__':
main()
74 changes: 74 additions & 0 deletions top-k-frequent-elements/EGON.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,74 @@
from collections import Counter
from typing import List
from unittest import TestCase, main


class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
return self.solve_3(nums, k)

"""
Runtime: 82 ms (Beats 87.87%)
Analyze Complexity: O(n log n)
most_common의 정렬이 O(n log n)
Memory: 21.13 MB (Beats 60.35%)
"""
def solve_1(self, nums: List[int], k: int) -> List[int]:
return [key for key, value in Counter(nums).most_common(k)]

"""
Runtime: 88 ms (Beats 62.46%)
Analyze Complexity: O(n log n)
counter 생성에 O(n), 정렬에 O(n log n), slicing에 O(k)
Memory: 21.17 MB (Beats 60.35%)
"""
def solve_2(self, nums: List[int], k: int) -> List[int]:
counter = {}
for num in nums:
counter[num] = 1 + counter.get(num, 0)

sorted_counter = sorted(counter.items(), key=lambda item: -item[1])
return [item[0] for item in sorted_counter[:k]]


"""
Runtime: 81 ms (Beats 90.60%)
Analyze Complexity: O(n)
counter 생성이 O(n), counter_matrix 생성이 O(n), reversed는 O(1), early-return으로 O(k)
Memory: 22.10 MB (Beats 12.57%)
"""
def solve_3(self, nums: List[int], k: int) -> List[int]:
counter = {}
for num in nums:
counter[num] = 1 + counter.get(num, 0)

counter_matrix = [[] for _ in range(len(nums) + 1)]
for key, val in counter.items():
counter_matrix[val].append(key)

result = []
for num_list in reversed(counter_matrix):
for num in num_list:
result.append(num)
if len(result) >= k:
return result
else:
return result


class _LeetCodeTCs(TestCase):
def test_1(self):
nums = [1, 1, 1, 2, 2, 3]
k = 2
output = [1, 2]
self.assertEqual(Solution.topKFrequent(Solution(), nums, k), output)

def test_2(self):
nums = [1]
k = 1
output = [1]
self.assertEqual(Solution.topKFrequent(Solution(), nums, k), output)


if __name__ == '__main__':
main()