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Author: wSangbae

week14/bsw/9935_문자열 폭발.py

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+string = input()
+bomb = input()
+
+answer = ''
+
+string = 's'*500000 + '4'*500000
+bomb = 'ssssssssssssssssss444444444444444444'
+for i in range(len(string)):
+    answer+=string[i]
+    if string[i] == bomb[-1]:
+        if answer[-len(bomb):] == bomb:
+            answer = answer[:-len(bomb)]
+
+if answer == '':
+    print('FRULA')
+else : print(answer)
+

week14/bsw/보석 쇼핑.py

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+# set 함수를 활용하려 했으나 시간복잡도가 O(N)인것을 고려해야 한다 -> 시간초과
+# dict의 길이를 활용하는 문제
+# dict를 활용하면 O(1)의 시간복잡도로 중복제거 기능을 활용 할 수 있음
+
+def solution(gems):
+    answer = []
+    
+    cnt_gems = len(set(gems))
+    cur_gems={}
+    l, r = 0, 0
+    
+    while l<len(gems) and r<len(gems):
+        
+        if gems[r] not in cur_gems:
+            cur_gems[gems[r]] = 1            
+        elif gems[r] in cur_gems:
+            cur_gems[gems[r]] += 1
+        
+        r+=1
+        
+        #print(cur_gems)
+        
+        if len(cur_gems) == cnt_gems:
+            
+            while l<r and len(cur_gems) == cnt_gems:
+                answer.append((l,r))    
+                if gems[l] not in cur_gems:
+                    break
+                elif gems[l] in cur_gems:
+                    cur_gems[gems[l]] -= 1
+                
+                if cur_gems[gems[l]] == 0:
+                    del cur_gems[gems[l]]
+                
+                l+=1      
+        
+    answer.sort(key = lambda x : (x[1]-x[0], x[0]))
+    return [answer[0][0]+1, answer[0][1]]

week14/bsw/합승 택시 요금.py

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+# 가중치 그래프 -> 다익스트라
+#
+# 다른사람 풀이 보고 작성
+# 다익스트라 연습이 더 필요
+
+def solution(n, s, a, b, fares):
+    answer = 1e9
+    INF = 1e9
+    from heapq import heappush, heappop
+    
+    graph=[[] for _ in range(n+1)]
+    
+    for fare in fares:
+        src, dst, cost = fare
+        graph[src].append((dst, cost))
+        graph[dst].append((src, cost))
+    
+    def dijk(s, d):
+        dists = [INF for i in range(n+1)]
+        dists[s] = 0
+        pq = []
+        heappush(pq, (dists[s], s))
+    
+        while pq:
+            cur_dist, cur_dst = heappop(pq)
+        
+            if dists[cur_dst] < cur_dist:
+                continue
+        
+            for new_dst, new_dist in graph[cur_dst]:
+
+                if cur_dist + new_dist < dists[new_dst]:
+                    dists[new_dst] = cur_dist + new_dist
+                    heappush(pq, (dists[new_dst], new_dst))
+        
+        return dists[d]
+    
+    for i in range(1, n+1):
+        answer = min(dijk(s, i) + dijk(i,a) + dijk(i,b), answer)
+    
+    return answer