시나위

from itertools import permutations def solution(numbers): answer = 0 ## permutation of elements of numbers c_arr = list(set(map(int, [''.join(j) for i in range(len(numbers)) for j in permutations(numbers, i+1)]))) ## Sieve of Eratosthenes prime_num=[] max_arr=max(c_arr) chk=[False]*2+[True]*(max_arr-1) for i in range(2,max_arr+1): if chk[i]: prime_num.append(i) for j in range(2*i,max_arr+1,i): i..

from collections import deque def solution(heights): answer = [] heights=deque(heights) len_h=len(heights) for i in range(len_h): tmp=heights.pop() cnt=0 ind=0 for j in heights: cnt+=1 if tmp: ind=cnt answer.insert(0,ind) return answer

def solution(prices): len_p=len(prices) answer = [0]*len_p for i in range(len_p): tmp=prices[i+1:] for j in tmp: if prices[i]>j: answer[i]+=1 break else : answer[i]+=1 return answer def solution(prices): len_pr=len(prices) answer = [] for i in range(len_pr): tmp=prices[i+1:] cnt=0 for j in tmp: cnt+=1 if prices[i]>j: break answer.append(cnt) return answer from collections import deque def soluti..

def solution(n): answer = '' while 1: tmp=n%3 if tmp == 0: n-=1 answer+='4' else: answer+=str(tmp) n= n//3 if n==0: break return answer[::-1] 생각할때는 쉽겠는데 하고 들어간 문제였는데 생각보다 만만치 않았다. 한시간 이상 고민하다가 다른분이 하신 코드를 봤는데 나머지가 0일때 몫에서 1을 빼주는게 요점이었다.

def solution(priorities, location): answer = 0 while priorities!=[]: max_pr=max(priorities) ind_zr=priorities.pop(0) if max_pr==ind_zr: answer+=1 if location==0: break else: location -=1 else: priorities.append(ind_zr) if location==0: location=len(priorities)-1 else: location-=1 return answer

def solution(arrangement): answer = 0 iron = 0 arr=arrangement.replace("()","*") for i in arr: if i == '*': answer+=iron elif i =='(': iron+=1 elif i ==')': iron-=1 answer+=1 #print("i:{}\tiron:{}\tanswer:{}".format(i,iron,answer)) return answer 다른 사람풀이도 나랑 비슷하게 푼거 같으니 추가는 안하겠다

def solution(progresses, speeds): answer = [] while (progresses!=[]): ed=[] # complete len_p=len(progresses) for i in range(len_p): progresses[i]+=speeds[i] for i in range(len_p): if progresses[i]>=100: ed.append(progresses[i]) else: break if ed!=[]: answer.append(len(ed)) for i in ed: del progresses[0] del speeds[0] return answer

def solution(bridge_length, weight, truck_weights): answer = 0 ing=[] # 다리를 건너는 트럭 cnt=[] # 트럭당 지난 거리 ed=[] # 다리를 지난 트력 while 1: if (truck_weights == []) and (ing == []): break answer+=1 # count total time for i in range(len(cnt)): # decrease the times of trucks above the bridge cnt[i]-=1 if (cnt!=[])and(cnt[0]==0): ed.append(ing[0]) del cnt[0] del ing[0] if (sum(ing) max_weight: bridge.append(0..

def solution(x, n): answer = [] for i in range(n): answer.append((i+1)*x) return answer

def solution(arr1, arr2): answer = [] row= len(arr1) #row col= len(arr1[0]) #column for i in range(row): tmp=[] for j in range(col): tmp.append(arr1[i][j]+arr2[i][j]) answer.append(tmp) return answer 원소끼리는 더하는 방법은 알았는데 어떻게 배열을 만들지가 문제였다.