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[kayden] Week 07 Solutions #487

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Sep 27, 2024
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20 changes: 20 additions & 0 deletions longest-substring-without-repeating-characters/kayden.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
class Solution:
# 시간복잡도: O(N)
# 공간복잡도: O(N)
# set에서 제외하는 로직을 없애기 위해, map을 사용해서 idx값을 저장후, start와 비교해서 start보다 작은 idx를 가진 경우에는 중복이 아니라고 판단했습니다.
def lengthOfLongestSubstring(self, s: str) -> int:

last_idx = {}
answer = 0
start = 0

for idx, ch in enumerate(s):
# 중복 조회시 idx값과 start 값 비교를 통해, 삭제하는 로직을 없이 중복을 확인했습니다.
if ch in last_idx and last_idx[ch] >= start:
start = last_idx[ch] + 1
last_idx[ch] = idx
else:
answer = max(answer, idx - start + 1)
last_idx[ch] = idx

return answer
36 changes: 36 additions & 0 deletions number-of-islands/kayden.py
Original file line number Diff line number Diff line change
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# 시간복잡도: O(M*N)
# 공간복잡도: O(M*N)

from collections import deque


class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
dx = [0, 0, -1, 1]
dy = [-1, 1, 0, 0]
m = len(grid)
n = len(grid[0])
q = deque()

def bfs(a, b):
q.append((a, b))
while q:
x, y = q.popleft()

for i in range(4):
nx = x + dx[i]
ny = y + dy[i]
if not (0 <= nx < m and 0 <= ny < n): continue

if grid[nx][ny] == '1':
grid[nx][ny] = '0'
q.append((nx, ny))

count = 0
for i in range(m):
for j in range(n):
if grid[i][j] == '1':
count += 1
bfs(i, j)

return count
26 changes: 26 additions & 0 deletions reverse-linked-list/kayden.java
Original file line number Diff line number Diff line change
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
// 시간복잡도: O(N)
// 공간복잡도: O(1)
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반복문 내에서 새로운 ListNode를 만들어내기 때문에 공간 복잡도 또한 N에 비례하여 선형적으로 증가할 것 같습니다

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답변 감사합니다! 최적화해서 수정 했습니다!

class Solution {
public ListNode reverseList(ListNode head) {
ListNode prev = null;

while (head != null){
ListNode next = head.next;
head.next = prev;
prev = head;
head = next;
}

return prev;
}
}
23 changes: 23 additions & 0 deletions set-matrix-zeroes/kayden.py
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# 시간복잡도: O(m*n)
# 공간복잡도: O(m+n)
class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
m = len(matrix)
n = len(matrix[0])

rows = set()
cols = set()

for i in range(m):
for j in range(n):
if matrix[i][j] == 0:
rows.add(i)
cols.add(j)

for row in rows:
for j in range(n):
matrix[row][j] = 0

for col in cols:
for i in range(m):
matrix[i][col] = 0
17 changes: 17 additions & 0 deletions unique-paths/kayden.py
Original file line number Diff line number Diff line change
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from math import comb
class Solution:
# 시간복잡도: O(m+n)
# 공간복잡도: O(1)
def uniquePaths(self, m: int, n: int) -> int:
return comb(m+n-2, n-1) # m+n-2Cn-1

# 시간복잡도: O(m*n)
# 공간복잡도: O(n)
def uniquePaths2(self, m: int, n: int) -> int:
dp = [1] * n

for _ in range(1, m):
for j in range(1, n):
dp[j] = dp[j-1] + dp[j]
Comment on lines +11 to +15
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저는 배열 두 개를 관리 했는데, 배열 하나면 충분하군요 잘 배워갑니다 :)

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@bky373 bky373 Sep 26, 2024

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오 그렇네요 좋은 풀이 감사합니다!


return dp[n-1]