# 가장 중요한 것: 변수에 내가 원하는 값이 제대로 들어가고 있는지 먼저 테스트하는 것.
# 이 부분을 빨리 캐치하지 못하면 로직을 제대로 짜고도 '왜 안나오지'만 반복하게 된다.
from collections import deque
def solution(q1, q2):
#
if (sum(q1) + sum(q2)) % 2 != 0:
return -1
if sum(q1) == sum(q2) :
return 0
#
n = len(q1)
q1 = deque(q1)
q2 = deque(q2)
sum1 = sum(q1)
sum2 = sum(q2)
cnt = 0
while sum1 != sum2:
if cnt >= 4*n:
return -1
if sum1 < sum2:
v = q2.popleft()
q1.append(v)
sum1 += v
sum2 -= v
elif sum1 > sum2:
v = q1.popleft()
q2.append(v)
sum1 -= v
sum2 += v
cnt += 1
return cnt
# 내 풀이 (시간 초과)
def solution(seq, k):
n = len(seq)
s = n - 1
e = n
while True:
while sum(seq[s:e]) < k:
s -= 1
if sum(seq[s:e]) == k:
if seq[e - 1] * (e - s) == k:
s, e = seq.index(seq[e - 1]), seq.index(seq[e - 1]) + (e - s)
break
else:
e -= 1
return [s, e - 1]
# 내 풀이2 (시간 초과): sum(seq[s:e]) 때문에 불필요하게 리스트를 돌면서 부분합을 구하게 됨, 차이가 나는 원소 값만 계산하면 되는데..
def solution(seq, k):
n = len(seq)
s = n-1
e = n
sumSeq = seq[s]
while True:
while sumSeq < k:
s -= 1
sumSeq += seq[s]
if sumSeq == k:
if seq[e-1]*(e-s) == k:
while s>=1 and seq[s] == seq[s-1]:
s-=1
e-=1
break
else:
e -= 1
s = e-1
sumSeq = seq[s]
return [s, e-1]
# 내 풀이3 (통과): s는 초기화할 필요가 없었는데 이 때문에 불필요한 부분을 계속 체크해서 시간초과 발생
# 연속된 부분 수열의 합: 투 포인터 생각
def solution(seq, k):
n = len(seq)
s = n-1
e = n
sumSeq = seq[s]
while True:
while sumSeq < k:
s -= 1
sumSeq += seq[s]
if sumSeq == k:
if seq[e-1]*(e-s) == k:
while s>=1 and seq[s] == seq[s-1]:
s-=1
e-=1
break
else:
e -= 1
sumSeq -= seq[e]
return [s, e-1]
# 내 풀이 : 성공, Counter, combinations 사용
from itertools import combinations as comb
from collections import Counter
def solution(orders, course):
result = []
for n in course:
lst = []
for order in orders:
order = [x for x in order]
for c in comb(order,n):
s = ''.join(sorted(c))
lst.append(s)
dic = Counter(lst).most_common()
if dic and dic[0][1] != 1:
result += [d[0] for d in dic if d[1]==dic[0][1]]
return sorted(result)
# 참고풀이 : 형태는 다르지만 나와 같은 로직의 풀이였다.
import collections
import itertools
def solution(orders, course):
result = []
for course_size in course:
order_combinations = []
for order in orders:
order_combinations += itertools.combinations(sorted(order), course_size)
most_ordered = collections.Counter(order_combinations).most_common()
result += [ k for k, v in most_ordered if v > 1 and v == most_ordered[0][1] ]
return [ ''.join(v) for v in sorted(result) ]
# 내 풀이 (런타임 에러)
food = 0
def solution(maps):
global food
dx = [-1, 1, 0, 0]
dy = [0, 0, -1, 1]
def dfs(x, y):
global food
if not visited[x][y] and graph[x][y] != 'X':
food += int(graph[x][y])
visited[x][y] = 1
for i in range(4):
nx = x + dx[i]
ny = y + dy[i]
if 0 <= nx < n and 0 <= ny < m:
dfs(nx, ny)
answer = []
graph = [[x for x in m] for m in maps]
n = len(graph)
m = len(graph[0])
visited = [[0 for x in m] for m in maps]
# dfs
for i in range(len(graph)):
for j in range(len(graph[0])):
food = 0
dfs(i, j)
if food != 0:
answer.append(food)
return sorted(answer) or [-1]
# 내 풀이2: 재귀 문제는 최대 깊이 제한 꼭 풀어주자. 파이썬은 최대 깊이가 1000이라 10^6까지 풀어주는 게 좋다.
import sys
limit_number = 10000
sys.setrecursionlimit(limit_number)
food = 0
def solution(maps):
global food
dx = [-1, 1, 0, 0]
dy = [0, 0, -1, 1]
def dfs(x, y):
global food
if not visited[x][y] and graph[x][y] != 'X':
food += int(graph[x][y])
visited[x][y] = 1
for i in range(4):
nx = x + dx[i]
ny = y + dy[i]
if 0 <= nx < n and 0 <= ny < m:
dfs(nx, ny)
answer = []
graph = [[x for x in m] for m in maps]
n = len(graph)
m = len(graph[0])
visited = [[0 for x in m] for m in maps]
# dfs
for i in range(len(graph)):
for j in range(len(graph[0])):
food = 0
dfs(i, j)
if food != 0:
answer.append(food)
return sorted(answer) or [-1]
# 내 풀이 : 실패 (86.7) // return '(None)' 을 한 줄 때문에. 해당 하는 값이 없으면 (None)출력 하는 걸 깜박함
def change(music):
temp = ''
for i,x in enumerate(music):
if x == '#':
s = temp[-1]
temp = temp[:-1]
temp += s.lower()
else:
temp += x
return temp
def play(time, music):
music = change(music)
length = len(music)
if time>length:
answer = music*(time//length)
answer += music[:time%length]
else:
answer = music[:time]
return answer
def timeDiff(tA, tB):
return int(tB[0:2])*60+int(tB[3:5]) - (int(tA[0:2])*60+int(tA[3:5]))
def solution(m, musicinfos):
m = change(m)
musicinfos = [[i+1]+list(musicinfo.split(',')) for i, musicinfo in enumerate(musicinfos)]
musicinfos = [[x[0], timeDiff(x[1], x[2]), x[3], play(timeDiff(x[1],x[2]), x[4])] for x in musicinfos]
musicinfos = sorted(musicinfos, key = lambda x: (-x[1], x[0]))
for musicinfo in musicinfos:
if m in musicinfo[3]:
return musicinfo[2]
# 내 풀이2 : 성공 //
def change(music):
temp = ''
for i,x in enumerate(music):
if x == '#':
s = temp[-1]
temp = temp[:-1]
temp += s.lower()
else:
temp += x
return temp
def play(time, music):
music = change(music)
length = len(music)
if time>length:
answer = music*(time//length)
answer += music[:time%length]
else:
answer = music[:time]
return answer
def timeDiff(tA, tB):
return int(tB[0:2])*60+int(tB[3:5]) - (int(tA[0:2])*60+int(tA[3:5]))
def solution(m, musicinfos):
m = change(m)
musicinfos = [[i+1]+list(musicinfo.split(',')) for i, musicinfo in enumerate(musicinfos)]
musicinfos = [[x[0], timeDiff(x[1], x[2]), x[3], play(timeDiff(x[1],x[2]), x[4])] for x in musicinfos]
musicinfos = sorted(musicinfos, key = lambda x: (-x[1], x[0]))
for musicinfo in musicinfos:
if m in musicinfo[3]:
return musicinfo[2]
return '(None)'
# 참고 풀이
def shap_to_lower(s):
s = s.replace('C#','c').replace('D#','d').replace('F#','f').replace('G#','g').replace('A#','a')
return s
def solution(m,musicinfos):
answer=[0,'(None)'] # time_len, title
m = shap_to_lower(m)
for info in musicinfos:
split_info = info.split(',')
time_length = (int(split_info[1][:2])-int(split_info[0][:2]))*60+int(split_info[1][-2:])-int(split_info[0][-2:])
title = split_info[2]
part_notes = shap_to_lower(split_info[-1])
full_notes = part_notes*(time_length//len(part_notes))+part_notes[:time_length%len(part_notes)]
if m in full_notes and time_length>answer[0]:
answer=[time_length,title]
return answer[-1]
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