Coding Test Review 3

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