[PGS] 단어변환 / lv3 /1hour

https://school.programmers.co.kr/learn/courses/30/lessons/43163

Author: zaqquum

Hongjoo/lv3/단어변환.py

@@ -0,0 +1,60 @@
+# 최소 편집 거리 =1 확인
+def check_edit1(a,b):
+    diff= 0 
+    for i in range(len(a)) :
+        if a[i] != b[i] :
+            diff+=1
+        if diff > 1 : 
+            return False
+    if diff == 1 : 
+        return True
+        
+from collections import deque
+def solution(begin, target, words):
+    answer = 0
+    #excep1 :words 에 target없으면 -> 변환 불가
+    if target not in words :
+        return 0 
+    
+    # 1. 인접 graph 생성
+    graph = [[] for _ in range(len(words)+2)] 
+    # begin -> words, target 단방향 
+    for k in range(len(words)) :
+        if check_edit1(begin, words[k]) :
+            graph[0].append(k+1)
+    if check_edit1(begin,target):
+        graph[0].append(len(words)+1)
+    # words -> wrods_1(쌍방향) , target(단방향)    
+    for k in range(len(words)):
+        for j in range(k ,len(words)):
+            if check_edit1(words[k], words[j]):
+                graph[k+1].append(j+1)
+                graph[j+1].append(k+1)
+        if check_edit1(words[k], target):
+            graph[k+1].append(len(words)+1)
+
+    #2.BFS 
+    visited = [ 0 for _ in range(len(words)+2)]
+    def bfs(visited):
+        visited[0] = 0
+        queue = deque()
+        queue.append(0)
+
+        while queue :
+            curr_idx = queue.popleft()
+            if len(graph[curr_idx]) <= 0 : # 이웃한 놈이 없는 경우 => return 0 
+                return 0
+
+            for next_idx in graph[curr_idx] : 
+                if not visited[next_idx] :  
+
+                    queue.append(next_idx)
+                    visited[next_idx] = visited[curr_idx] + 1 
+
+                if visited[-1] : #성공
+                    return visited[-1]
+        
+        return visited[-1] 
+    
+    
+    # print(graph)