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[bhyun-kim, bh_kim] Solution for week1 assignments #40

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May 3, 2024
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43 changes: 43 additions & 0 deletions best-time-to-buy-and-sell-stock/bhyun-kim.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,43 @@
"""
Solution

Algorithm:
1. Iterate through the list in reverse.
2. Keep track of the maximum value seen so far.
3. Calculate the profit by subtracting the current value from the maximum value.
4. Update the profit if it is greater than the current profit.

Time complexity: O(n)
Space complexity: O(1)
"""


from typing import List

class Solution:
def maxProfit(self, prices: List[int]) -> int:
profit = 0
prev_max = 0

for i in reversed(range(len(prices)-1)):
prev_max = max(prev_max, prices[i+1])
profit = max(profit, prev_max-prices[i])

return profit



def main():
test_cases = [
[[7,1,5,3,6,4], 5],
[[7,6,4,3,1], 0]
]
s = Solution()

for test_case in test_cases:
prices_input, expected = test_case
assert s.maxProfit(prices_input) == expected

if __name__ == '__main__':
main()

34 changes: 34 additions & 0 deletions contains-duplicate/bhyun-kim.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,34 @@
"""
Solution

Algorithm:
1. Create a set from the list.
2. If the length of the set is not equal to the length of the list, return True.
3. Otherwise, return False.

Time complexity: O(n)
Space complexity: O(n)
"""


from typing import List

class Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
return len(set(nums)) != len(nums)

def main():
test_cases = [
[[1,2,3,1], True,]
[[1,2,3,4], False],
[[1,1,1,3,3,4,3,2,4,2], True]
]
s = Solution()

for test_case in test_cases:
nums_input, expected = test_case
assert s.containsDuplicate(nums_input) == expected

if __name__ == '__main__':
main()

42 changes: 42 additions & 0 deletions two-sum/bhyun-kim.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,42 @@
"""
Solution

Algorithm:
1. Create a hashmap to store the index of each element.
2. Iterate through the list.
3. Check if the remaining value is in the hashmap.
4. If it is, return the current index and the index of the remaining value in the hashmap.

Time complexity: O(n)
Space complexity: O(n)
"""

from typing import List

class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:

hashmap = {}

for i in range(len(nums)):
remaining = target - nums[i]
if remaining in hashmap:
return [i, hashmap[remaining]]

hashmap[nums[i]] = i

def main():
test_cases = [
[2,7,11,15], 9, [0,1],
[3,2,4], 6, [1,2],
[3,3], 6, [0,1]
]
s = Solution()

for test_case in test_cases:
nums_input, target_input, expected = test_case
assert sorted(s.twoSum(nums_input, target_input)) == expected

if __name__ == '__main__':
main()

34 changes: 34 additions & 0 deletions valid-anagram/bhyun-kim.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,34 @@
"""
Solution

Algorithm:
1. Sort the strings and compare them.
2. If they are equal, return True. Otherwise, return False.

Time complexity: O(nlogn)
Space complexity: O(1)
"""

class Solution:
def isAnagram(self, s: str, t: str) -> bool:

if sorted(s) == sorted(t):
return True
else:
return False


def main():
test_cases = [
["anagram", "nagaram", True],
["rat", "car", False]
]
s = Solution()

for test_case in test_cases:
s_input, t_input, expected = test_case
assert s.isAnagram(s_input, t_input) == expected

if __name__ == '__main__':
main()

35 changes: 35 additions & 0 deletions valid-palindrome/bhyun-kim.py
Original file line number Diff line number Diff line change
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"""
Solution:

Algorithm:
1. Convert the string to lowercase.
2. Remove all non-alphanumeric characters.
3. Check if the string is equal to its reverse.

Time complexity: O(n)
Space complexity: O(n)
"""


class Solution:
def isPalindrome(self, s: str) -> bool:
s = s.lower()
s = list([c for c in s
if c.isalpha() or c.isnumeric()])
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이미 리스트인데, list() 함수로 감싸는 것은 불필요한 것 같습니다.

Suggested change
s = list([c for c in s
if c.isalpha() or c.isnumeric()])
s = [c for c in s
if c.isalpha() or c.isnumeric()]

s = "".join(s)
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굳이 문자열로 다시 재변환 할 필요가 있을까요? 이 줄을 지우면 그냥 리스트 간 비교가 될 것 같네요.

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문자열 변환과 리스트 변환 중복을 제거했습니다. 리트코드 홈페이지에서 실행하니 평균적으로 10% 정도 속도가 줄어들었네요!

return s == s[::-1]

def main():
test_cases = [
["A man, a plan, a canal: Panama", True],
["race a car", False],
[" ", True]
]
s = Solution()

for test_case in test_cases:
s_input, expected = test_case
assert s.isPalindrome(s_input) == expected

if __name__ == '__main__':
main()