코딩테스트 스터디 풀이 업로드

Author: hoeen

solution_codes/programmers_스택, 큐_같은 숫자는 싫어_1회.py

@@ -0,0 +1,23 @@
+from collections import deque
+
+def solution(arr):
+    q = deque(arr)
+    p1 = q.popleft()
+    ans = deque([])
+    if not q:
+        return arr
+    while q:
+        
+        p2 = q.popleft() # 하나 더뽑음
+
+        if p1 == p2:
+            if not q:
+                ans.append(p1)
+            continue
+        else:
+            ans.append(p1)
+            p1 = p2
+            if not q:
+                ans.append(p2)
+            
+    return list(ans)

solution_codes/programmers_스택, 큐_기능개발_1회.py

@@ -0,0 +1,41 @@
+from collections import deque
+
+def solution(pro, spd):
+    answer = []
+    comp = deque([])
+    qpro = deque(pro)
+    qspd = deque(spd)
+    p = qpro.popleft()
+    s = qspd.popleft()
+    total_date = (100 - p) // s  # 며칠후에 배포가능
+    if (100 - p) % s > 0: 
+        total_date += 1 # 안맞아떨어지면 하루 추가
+    
+    comp.append(p)
+    
+    while qpro:
+        qp = qpro.popleft()
+        qs = qspd.popleft()
+        
+        due_date = (100 - qp) // qs
+        if (100 - qp) % qs > 0:
+            due_date += 1
+   
+        if total_date >= due_date: 
+            # 다음 작업 기간이 총진행일보다 같거나 일찍끝남
+            comp.append(qp)
+            if not qpro:
+                answer.append(len(comp))
+                break
+        else: # 다음작업이 더걸림 - 앞선것을 배포
+            answer.append(len(comp)) # 배포
+            comp = deque([])
+            total_date = due_date
+            if not qpro:
+                answer.append(1)
+                break
+            else:
+                qpro.appendleft(qp)
+                qspd.appendleft(qs)
+        
+    return answer

solution_codes/programmers_스택, 큐_올바른 괄호_1회.py

@@ -0,0 +1,20 @@
+from collections import deque
+
+def solution(s):
+    answer = True
+    oc = 0 
+    qs = deque(s)
+    while qs:
+        q = qs.popleft()
+        if q == '(':
+            oc += 1
+        else:
+            oc -= 1
+        if oc < 0:
+            return False
+        
+        
+    if oc != 0:
+        return False
+
+    return answer

solution_codes/programmers_스택, 큐_프린터_1회.py

@@ -0,0 +1,24 @@
+# ~ 10:30
+from collections import deque
+
+def solution(pri, loc):
+    answer = 0
+    q_pri = deque(tuple(zip(range(len(pri)), pri)))
+    # (위치정보 0~n-1, 중요도)
+    
+    count = 0 # 인쇄된 문서 카운트
+    
+    while q_pri:
+        # pri에서 하나빼고, 이게 뺀 큐의 max보다 큰지 판단
+        position, priority = q_pri.popleft()   # 위치정보, 중요도
+        if not q_pri: # 마지막까지 다뺸경우
+            return count+1
+        if priority >= max(q_pri, key = lambda x: x[1])[1]:
+            # 뺀다
+            count += 1
+            if position == loc:
+                return count
+        else:
+            q_pri.append((position, priority))
+
+    return count

solution_codes/programmers_완전탐색_모의고사_1회.py

@@ -0,0 +1,29 @@
+# ~ 11:30
+def solution(answers):
+    p1 = [1, 2, 3, 4, 5]
+    p2 = [2, 1, 2, 3, 2, 4, 2, 5]
+    p3 = [3, 3, 1, 1, 2, 2, 4, 4, 5, 5]
+    
+    pe1 = p1*(len(answers)//len(p1) + 1)
+    pe2 = p2*(len(answers)//len(p2) + 1)
+    pe3 = p3*(len(answers)//len(p3) + 1)
+    
+    pe1 = pe1[:len(answers)]
+    pe2 = pe2[:len(answers)]
+    pe3 = pe3[:len(answers)]
+    
+    cor = [0, 0, 0, 0]
+    for i in range(len(answers)):
+        if pe1[i] == answers[i]:
+            cor[1] += 1
+        if pe2[i] == answers[i]:
+            cor[2] += 1
+        if pe3[i] == answers[i]:
+            cor[3] += 1
+    result = []
+    max_c = max(cor)
+    for i in range(1, 4):
+        if cor[i] == max_c:
+            result.append(i)
+    
+    return result

solution_codes/programmers_완전탐색_소수 찾기_1회.py

@@ -0,0 +1,28 @@
+# ~ 11:00
+from itertools import permutations
+import numpy as np 
+
+def solution(numbers):
+    # 소수 판별? 2~sqrt(자기자신)까지 해서 나머지 계속 생기면 됨 
+    def is_sosu(number):
+        for i in range(2, int(np.sqrt(number))+1):
+            if number%i == 0:
+                return False
+        return True
+    
+    answer = []
+    
+    perm = []
+    for l in range(1, len(numbers)+1):
+        perm += list(set(permutations(numbers, l)))
+    
+    
+    # print(perm)
+    for i in perm:
+        num = int(''.join(i))
+        if num in [0, 1]: continue
+        # print(is_sosu(num))
+        if is_sosu(num):
+            answer.append(num)
+            
+    return len(set(answer))

solution_codes/programmers_완전탐색_최소직사각형_1회.py

@@ -0,0 +1,12 @@
+def solution(sizes):
+    max_h, max_v = 0, 0
+    # 가로 < 세로로 맞춤
+    for h, v in sizes:
+        if h < v:
+            h, v = v, h
+        if max_h < h:
+            max_h = h
+        if max_v < v:
+            max_v = v
+            
+    return max_h*max_v

solution_codes/programmers_힙_더 맵게_1회.py

@@ -0,0 +1,20 @@
+from heapq import heapify, heappop, heappush
+
+def solution(scoville, K):
+    
+    count = 0
+    # 불가능 경우 처리하기
+    heapify(scoville) # 최소 - 최대
+    # for _ in range(int(1e6)):
+    while True:
+        dish1 = heappop(scoville)
+        # K 체크
+        if dish1 >= K:
+            return count
+        # 큐 비어있으면 -1 리턴
+        if not scoville: return -1
+        dish2 = heappop(scoville)
+        mix = dish1 + (dish2*2)
+        heappush(scoville, mix)
+        count += 1
+    

solution_codes/programmers_힙_이중우선순위큐_1회답.py

@@ -0,0 +1,19 @@
+# 최댓값, 최솟값을 한번에 O(logn)으로 빼는 방법?
+# 1. 그때그때 정렬
+def solution(operations):
+    answer = []
+    queue = []
+    for op in operations:
+        com, num = op.split()
+        num = int(num)
+        if com == 'I':
+            queue.append(num)
+        elif com == 'D' and len(queue) > 0:
+            if num == 1:
+                queue.sort()
+                queue.pop() # 최댓값 삭제  
+            elif num == -1:
+                queue.sort()
+                queue.pop(0) # 최솟값 삭제
+        
+    return [max(queue), min(queue)] if queue else [0, 0]

temp/baekjoon_숨바꼭질 3.py

@@ -0,0 +1,32 @@
+import sys
+sys.setrecursionlimit = 5000
+n, k = map(int, input().split())
+
+# top-down
+def find(x, t, memo: dict):
+    # memoization - 확인해서 값 있으면 갱신하고 리턴
+    if x == k:
+        return t
+    # k가 아닐 경우 메모하고 나옴
+    if x-1 >= 0:
+        if x-1 in memo:
+            memo[x] = min(memo[x], memo[x-1]+1)
+            return
+        else:
+            find(x-1, t+1, memo)
+    if x+1 <= 100000:
+        if x+1 in memo:
+            memo[x] = min(memo[x], memo[x+1]+1)
+            return
+        else:
+            find(x+1, t+1, memo)
+    if 2*x <= 100000:
+        if 2*x in memo:
+            memo[x] = min(memo[x], memo[2*x])
+        else:
+            find(2*x, t, memo)
+
+print(find(n, 0, dict()))
+
+
+