The document contains code snippets and explanations for solving three LeetCode problems: Power of Two, Valid Parentheses, and Find Minimum in Rotated Sorted Array. For Power of Two, it provides an O(log n) solution that uses modulo and division to check if a number is a power of two. For Valid Parentheses, it provides an O(n) solution that uses a string to track opening and closing parentheses. For Find Minimum, it provides both an O(n) solution that finds the minimum by checking if each number is less than the previous, and an O(log n) solution that recursively searches halves of the array to find the minimum.

2. LeetCode – Power of Two (Easy)

3. 𝑂(log 𝑛) 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
class Solution:
def isPowerOfTwo(self, n: int) -> bool:
if n <= 0:
return False
while n>1:
if n%2 == 0:
n = n/2
else:
return False
return True

4. LeetCode – Valid Parentheses (Easy)

5. 𝑂(𝑛) 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
class Solution:
def isValid(self, s: str) -> bool:
new_s = ' '
for text in s:
if new_s[-1] == '{' and text == '}':
new_s = new_s[:-1]
continue
elif new_s[-1] == '(' and text == ')':
new_s = new_s[:-1]
continue
elif new_s[-1] == '[' and text == ']':
new_s = new_s[:-1]
continue
elif new_s[-1] == '{' and (text == ')' or text == ']'):
return False
elif new_s[-1] == '(' and (text == ']' or text == '}'):
return False
elif new_s[-1] == '[' and (text == ')' or text == '}'):
return False
new_s += text
if new_s[1:] == '':
return True
else:
return False

6. LeetCode – Find Minimum in Rotated Sorted Array (Medium)

7. 𝑂(𝑛) 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
class Solution:
def findMin(self, nums: List[int]) -> int:
prev = nums[0] - 1
for num in nums:
if prev > num:
return num
prev = num
return nums[0]

8. 𝑂 log 𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
class Solution:
def findMin(self, nums: List[int]) -> int:
cur = nums
while len(cur)>0:
leng = len(cur)
if leng == 1:
return cur[0]
mid = int(leng/2)
if len(cur[:mid]) > 1:
if cur[:mid][0] > cur[:mid][-1] and cur[:mid][-1] < cur[mid:][0]:
cur = cur[:mid]
elif cur[:mid][0] < cur[:mid][-1] and cur[:mid][0] < cur[mid:][-1]:
return cur[:mid][0]
else:
cur = cur[mid:]
elif len(cur[:mid]) == 0:
cur = cur[mid:]
elif len(cur[:mid]) == 1:
if cur[:mid][0] > cur[mid:][-1]:
cur = cur[mid:]
else:
return cur[:mid][0]

9. Thank You