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[i-mprovising] Week 02 solutions #1249
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""" | ||
Time complexity O(n) | ||
Space complexity O(n) | ||
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Dynamic programming | ||
""" | ||
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class Solution: | ||
def climbStairs(self, n: int) -> int: | ||
dp = [0, 1, 2] # distinct ways to reach i steps | ||
if n <= 2: | ||
return dp[n] | ||
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for i in range(3, n+1): | ||
dp.append(dp[i-1] + dp[i-2]) | ||
return dp[n] |
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Original file line number | Diff line number | Diff line change |
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""" | ||
Time complexity O(n) | ||
Space complexity O(n) | ||
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Prefix sum | ||
""" | ||
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class Solution: | ||
def productExceptSelf(self, nums: List[int]) -> List[int]: | ||
products = [1] | ||
reverse_products = [1] | ||
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tmp = 1 | ||
for n in nums[:-1]: | ||
tmp *= n | ||
products.append(tmp) | ||
tmp = 1 | ||
for n in nums[::-1][:-1]: | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 이렇게 표현사는 방법도 있네요! 👍 |
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tmp *= n | ||
reverse_products.append(tmp) | ||
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answer = [ | ||
products[i] * reverse_products[-i-1] | ||
for i in range(len(nums)) | ||
] | ||
return answer |
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Original file line number | Diff line number | Diff line change |
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""" | ||
Time complexity O(n) | ||
Space complexity O(n) | ||
""" | ||
from collections import defaultdict | ||
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class Solution: | ||
def isAnagram(self, s: str, t: str) -> bool: | ||
s_cnt = defaultdict(int) | ||
t_cnt = defaultdict(int) | ||
for char in s: | ||
s_cnt[char] += 1 | ||
for char in t: | ||
t_cnt[char] += 1 | ||
if s_cnt != t_cnt: | ||
return False | ||
return True |
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지금도 코드가 깔끔하고 좋네요! 👍
배열 대신 이전 값을 저장하는 변수 2개를 사용하는 방법도 생각해볼 수 있을 것 같아요.
공간복잡도가 O(1)이 될 것 같아서요!
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굳이 배열로 전부 저장할 필요가 없네요! 피드백 감사합니다 :)