https://school.programmers.co.kr/learn/courses/30/lessons/42890
1) 코드
uniqueList = []
def checkUnique(arr, relation):
selected = set()
for row in relation:
newRow = ""
for col in arr:
newRow += row[col]
if newRow in selected:
return False
else:
selected.add(newRow)
return True
## 2^8 = 256이다
def dfs(pos, total, arr, relation):
if pos == total:
if len(arr) != 0:
result = checkUnique(arr, relation)
if result == True:
arrNums = ""
for num in arr:
arrNums += str(num)
uniqueList.append(arrNums)
return
arr.append(pos)
dfs(pos+1, total, arr, relation)
arr.pop()
dfs(pos+1, total, arr, relation)
def solution(relation):
answer = 0
total = len(relation[0])
arr = []
dfs(0, total, arr, relation)
sorted_uniqueList = sorted(uniqueList, key=len)
answer = []
for i in range(len(sorted_uniqueList)):
curr = sorted_uniqueList[i]
addable = True
for j in range(len(answer)):
tmp = answer[j]
if all(char in curr for char in tmp):
addable = False
break
if addable == True:
answer.append(curr)
return len(answer)2) 해설
- Uniqueness와 Minimality를 잘 판단해야 하는 문제이다
- if all(char in curr for char in tmp) 문법을 잘 기억해도록 하자
- sorted(uniqueList, key=len) 문법도 잘 기억해도록 하자
hello