Skip to content
Merged
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
Original file line number Diff line number Diff line change
@@ -0,0 +1,30 @@
'''
1. 아이디어 :
모든 (i, j) 쌍을 완전탐색하면서, 두 구간이 포함하는 정수 set을 각각 만들어
교집합이 비어있지 않으면 교차로 카운트한다.
닫힌 구간이라 끝점(경계)만 공유해도 교차로 세어진다.
2. 시간복잡도 :
o(n^2 * m) # m = 구간 길이(end - start)
3. 자료구조/알고리즘 :
set(집합), 완전탐색
'''

class Solution:
def countIntersectingIntervals(self, intervals: list[list[int]]) -> int:
n = len(intervals)
ans = 0

for i in range(n-1):
for j in range(i+1,n):
st_a, st_b = set(), set()
a1,a2 = intervals[i]
b1,b2 = intervals[j]
for a in range(a1,a2+1):
st_a.add(a)
for b in range(b1,b2+1):
st_b.add(b)
# print(st_a, st_b)
if len(st_a & st_b) > 0:
ans +=1

return ans
Loading