Merge pull request #173 from codesquad-backend-study/ayaan

Ayaan : 1월 셋째 주

Ayaan/programmers/완전탐색/모음사전.py

@@ -0,0 +1,21 @@
+def solution(word):
+    alphabets = ['A', 'E', 'I', 'O', 'U']
+    answer = []
+    order = 0
+
+    def dfs(target):
+        nonlocal order
+        if target == word:
+            answer.append(order)
+        if len(target) == 5:
+            return
+
+        for alphabet in alphabets:
+            order += 1
+            dfs(target + alphabet)
+
+    dfs('')
+    return answer[0]
+
+
+solution('I')

Ayaan/programmers/완전탐색/전력망을둘로나누기.py

@@ -0,0 +1,38 @@
+import collections
+
+
+def bfs(graph, start, visited):
+    q = collections.deque([start])
+    visited[start] = True
+    cnt = 1
+
+    while q:
+        v = q.popleft()
+        for n in graph[v]:
+            if not visited[n]:
+                q.append(n)
+                visited[n] = True
+                cnt += 1
+    return cnt
+
+
+def solution(n, wires):
+    answer = n
+    for i in range(len(wires)):
+        temp = wires[:]
+        graph = collections.defaultdict(list)
+        # i번째 연결을 뺀 graph를 만든다.
+        for idx, node in enumerate(temp):
+            if i == idx:
+                continue
+            graph[node[0]].append(node[1])
+            graph[node[1]].append(node[0])
+        # 한쪽의 개수를 구해서 차이를 계산한다.
+        cnt = bfs(graph, 1, [False] * (n + 1))
+        other = n - cnt
+        answer = min(answer, abs(cnt - other))
+
+    return answer
+
+
+solution(4, [[1, 2], [2, 3], [3, 4]])