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[parkhojeong] WEEK 11 Solutions #2851

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parkhojeong merged 1 commit into from
Sep 5, 2026
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[parkhojeong] WEEK 11 Solutions #2851

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15 changes: 15 additions & 0 deletions missing-number/parkhojeong.py
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@dalestudy dalestudy Bot Sep 5, 2026

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🏷️ 알고리즘 패턴 분석

missing-number/parkhojeong.py 
class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        N = max(nums)
        num_set = set()

        for i in range(N + 1):
            num_set.add(i)

        for num in nums:
            num_set.remove(num)

        if len(num_set) == 0:
            return N + 1
        else:
            return num_set.pop()
  • 패턴: Hash Map / Hash Set, Greedy, Binary Search, Dynamic Programming, Two Pointers, Sliding Window, Backtracking, Divide and Conquer, Union Find, Trie, Bit Manipulation, DFS, BFS, Monotonic Stack, Heap / Priority Queue
  • 설명: 집합을 이용해 존재 여부를 체크하는 방식으로 누락된 수를 찾는다. 전체 가능한 숫자 세트를 만든 뒤 주어진 nums에서 제거하는 해시 셋 활용 패턴에 해당한다.

📊 시간/공간 복잡도 분석

복잡도
Time O(N)
Space O(N)

피드백: 정수 집합을 만들어 가능한 모든 값을 저장한 뒤 남은 값을 찾는다.

개선 제안: 현재 구현은 최악의 경우 추가적 공간이 필요하므로, 가정된 수의 합이나 비트 연산 등을 이용한 상수 공간 방법을 고려해볼 수 있다.

💡 풀이에 시간/공간 복잡도를 주석으로 남겨보세요!

Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
class Solution:
def missingNumber(self, nums: List[int]) -> int:
N = max(nums)
num_set = set()
Comment on lines +3 to +4

@Yiseull Yiseull Sep 5, 2026

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정답 범위는 항상 0~n이니까 max 대신 len을 쓰면 더 효율적일 것 같아요!


for i in range(N + 1):
num_set.add(i)
Comment on lines +4 to +7

@Yiseull Yiseull Sep 5, 2026

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num_set = set(range(N + 1)) 로 한 줄에 됩니다! 그리고 크로드한테 물어보니까 저 코드는 파이썬 레벨 루프가 C 레벨로 내려가서 눈에 띄게 빨라진다고 합니다!!👍


for num in nums:
num_set.remove(num)

if len(num_set) == 0:
return N + 1
else:
return num_set.pop()
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