주말코테

cmkim/2021 카카오 인턴쉽/프로그래밍1.py

@@ -0,0 +1,36 @@
+num = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine']
+
+def solution(s):
+    answer = 0
+    count = 0
+    arr = []
+    i = 0
+    while i < len(s):
+        #print('i = ', i)
+
+        if s[i].isdigit():
+            arr.append(int(s[i]))
+
+        else:
+            j = 0
+            #print('i = ', i)
+            # print(s[i:])
+            while j < 10:
+                if num[j] in s[i:i+5]:
+                    #print('num[j] =', num[j])
+                    arr.append(j)
+                    i += len(num[j]) - 1
+                    break
+                j += 1
+        i += 1
+
+    #print(arr)
+    answer = int("".join(map(str, arr)))
+    #print('+'.join(arr))
+    #answer = int("".join(arr))
+    print(answer)
+    return answer
+
+#solution("one4seveneight")
+solution('1zerotwozero3')
+

cmkim/2021 카카오 인턴쉽/프로그래밍2.py

@@ -0,0 +1,54 @@
+from collections import deque
+
+def solution(places):
+    answer = []
+
+    for i in range(5):
+        flag = 1
+        p_list = []
+        #p_ad_list = []
+        q = deque()
+
+        for x in range(5):
+            for y in range(5):
+                if places[i][x][y] == 'P':
+                    p_list.append((x, y))
+
+        for a in range(len(p_list)):
+            for b in range(a+1, len(p_list)):
+                if abs(p_list[a][0] - p_list[b][0]) + abs(p_list[a][1] - p_list[b][1]) < 3:
+                    q.append((p_list[a][0], p_list[a][1], p_list[b][0], p_list[b][1]))
+                    #p_ad_list.append((p_list[a][0], p_list[a][1], p_list[b][0], p_list[b][1]))
+                    #print('adjacent', p_list[a][0], p_list[a][1], p_list[b][0], p_list[b][1])
+
+#2칸 아래,
+#2칸 우측
+#1칸 좌측, 1칸 아래
+#1칸 우측, 1칸 아래
+
+        while q:
+            x1, y1, x2, y2 = q.popleft()
+            if x1 + 2 == x2:
+                if places[i][x1 + 1][y1] == 'O':
+                    flag -= 1
+            elif y1 + 2 == y2:
+                if places[i][x1][y1+1] == 'O':
+                    flag -= 1
+            elif y1 - 1 == y2 and x1 + 1 == x2: #왼쪽아래
+                if places[i][x1][y2] == 'O' or places[i][x2][y1] == 'O':
+                    flag -= 1
+            elif y1 + 1 == y2 and x1 + 1 == x2:
+                if places[i][x1][y2] == 'O' or places[i][x2][y1] == 'O':
+                    flag -= 1
+            elif x1 + 1 == x2 or y1 + 1 == y2:
+                flag -= 1
+
+
+        if flag > 0:
+            answer.append(1)
+        else:
+            answer.append(0)
+
+    return answer
+
+print(solution([["POOOP", "OXXOX", "OPXPX", "OOXOX", "POXXP"], ["POOPX", "OXPXP", "PXXXO", "OXXXO", "OOOPP"], ["PXOPX", "OXOXP", "OXPXX", "OXXXP", "POOXX"], ["OOOXX", "XOOOX", "OOOXX", "OXOOX", "OOOOO"], ["PXPXP", "XPXPX", "PXPXP", "XPXPX", "PXPXP"]]))

cmkim/2021 카카오 인턴쉽/프로그래밍4.py

@@ -0,0 +1,54 @@
+from collections import deque
+
+#INF = 1000000000
+
+
+def solution(n, start, end, roads, traps):
+    visited = [[0] * (n + 1) for _ in range(n+1)]
+    answer = 1000000
+    while q:
+        #print('roads = ', roads)
+        start, roads, traps, cost = q.popleft()
+
+        if start == end:
+            answer = min(answer, cost)
+            continue
+
+        for a, b, c in roads: # 출발, 도착, 비용
+            if start == a : # 이동할 다음노드가 있으면 이동
+                if b not in traps: #다음노드가 정상적으로 이동할 노드면
+                    if not visited[a][b]:
+                        visited[a][b] = 1
+                        q.append((b, roads, traps, cost+c))
+                    continue
+                else: # 다음노드가 함정노드면
+                    if not visited[a][b]:
+                        visited[a][b] = 1
+
+                        for i in range(len(roads)):
+                            if roads[i][0] == b or roads[i][1] == b:
+                                temp = roads[i][0]
+                                roads[i][0] = roads[i][1]
+                                roads[i][1] = temp
+                        q.append((b, roads, traps, cost+c))
+
+
+
+    print(answer)
+    return answer
+q = deque()
+
+# n = 3
+# start = 1
+# end = 3
+# roads = [[1, 2, 2], [3, 2, 3]]
+# traps = [2]
+
+n = 4
+start = 1
+end = 4
+roads = [[1, 2, 1], [3, 2, 1], [2, 4, 1]]
+traps = [2, 3]
+
+q.append((start, roads, traps, 0)) #마지막은 cost
+solution(n, start, end, roads, traps)

cmkim/2021 프로그래머스 써머코딩/프로그래밍1.py

@@ -0,0 +1,29 @@
+import sys
+
+
+# from collections import deque
+
+
+def solution(code, day, data):
+    answer = []
+    q = []
+
+    for i in data:
+        arr = i.split(' ')
+        time = []
+        if code in arr[1]:
+            if day in arr[2]:
+                time.append(int(arr[2][5:13]))
+                time.append(int(arr[2][13:15]))
+
+                num = arr[0].split('=')
+                # answer.append(int(num[1]))
+                price = int(num[1])
+                q.append((time[1], time[0], price))
+
+    q.sort()
+
+    for i in range(len(q)):
+        answer.append(q[i][2])
+
+    return answer

cmkim/2021 프로그래머스 써머코딩/프로그래밍2.py

@@ -0,0 +1,35 @@
+from collections import deque
+
+
+def solution(t, r):
+    q = []
+    wait = []
+    answer = []
+    flag = 0
+    for j in range(len(t)):
+        q.append((j, int(t[j]), int(r[j])))  # id, 나갈시간, 우선순위
+
+    for i in range(10000):
+        for j in range(len(q)):
+            if i == q[j][1]:
+                wait.append((q[j][0], q[j][2]))
+
+        index = 6
+        for k in range(len(wait)):
+            index = min(index, wait[k][1])
+
+        #print(wait)
+        for k in range(len(wait)):
+            if wait[k][1] == index:
+                pop_id = k
+                answer.append(wait[k][0])
+                wait.pop(pop_id)
+                break
+
+        flag = 0
+    print(answer)
+
+    return answer
+t = [0,1,3,0]
+r = [0,1,2,3]
+solution(t, r)