From a27bbb31da2d861aa094bdb38c583dd8b3a5effc Mon Sep 17 00:00:00 2001
From: SOOOY <1401660@naver.com>
Date: Mon, 7 Oct 2019 17:52:50 +0900
Subject: [PATCH] update hash problem

---
 ...73\355\225\234\354\204\240\354\210\230.py" |  9 ++++++
 ...10\355\230\270\353\252\251\353\241\235.py" | 16 ++++++++++
 ...354\213\23403_\354\234\204\354\236\245.py" | 16 ++++++++++
 ...44\355\212\270\354\225\250\353\262\224.py" | 32 +++++++++++++++++++
 4 files changed, 73 insertions(+)
 create mode 100644 "SOOYEON/week1/\355\225\264\354\213\23401_\354\231\204\354\243\274\355\225\230\354\247\200\353\252\273\355\225\234\354\204\240\354\210\230.py"
 create mode 100644 "SOOYEON/week1/\355\225\264\354\213\23402_\354\240\204\355\231\224\353\262\210\355\230\270\353\252\251\353\241\235.py"
 create mode 100644 "SOOYEON/week1/\355\225\264\354\213\23403_\354\234\204\354\236\245.py"
 create mode 100644 "SOOYEON/week1/\355\225\264\354\213\23404_\353\262\240\354\212\244\355\212\270\354\225\250\353\262\224.py"

diff --git "a/SOOYEON/week1/\355\225\264\354\213\23401_\354\231\204\354\243\274\355\225\230\354\247\200\353\252\273\355\225\234\354\204\240\354\210\230.py" "b/SOOYEON/week1/\355\225\264\354\213\23401_\354\231\204\354\243\274\355\225\230\354\247\200\353\252\273\355\225\234\354\204\240\354\210\230.py"
new file mode 100644
index 0000000..e69252f
--- /dev/null
+++ "b/SOOYEON/week1/\355\225\264\354\213\23401_\354\231\204\354\243\274\355\225\230\354\247\200\353\252\273\355\225\234\354\204\240\354\210\230.py"
@@ -0,0 +1,9 @@
+# 풀이 1 - Counter 이용
+from collections import Counter
+
+def solution(participant, completion):
+    # Counter를 이용하면 리스트 내의 성분의 숫자를 딕셔너리 형태로 반환해준다
+    # ex) list1 =['a', 'b', 'c', 'a'] => Counter(list1) : {'a':2, 'b':1, 'c':1}
+    not_completion = Counter(participant) - Counter(completion) # Counter의 경우 사칙연산도 가능
+     
+    return list(not_completion.keys())[0]
diff --git "a/SOOYEON/week1/\355\225\264\354\213\23402_\354\240\204\355\231\224\353\262\210\355\230\270\353\252\251\353\241\235.py" "b/SOOYEON/week1/\355\225\264\354\213\23402_\354\240\204\355\231\224\353\262\210\355\230\270\353\252\251\353\241\235.py"
new file mode 100644
index 0000000..1a9d567
--- /dev/null
+++ "b/SOOYEON/week1/\355\225\264\354\213\23402_\354\240\204\355\231\224\353\262\210\355\230\270\353\252\251\353\241\235.py"
@@ -0,0 +1,16 @@
+# 시간복잡도 O(n^2)
+
+def solution(phone_book):
+    # phone_book 전체를 반복 => 기준 설정
+    for phone_num in phone_book:
+        # 반복문을 통해 나머지 번호들과 비교
+        for compare_num in phone_book:
+            if phone_num == compare_num:
+                continue
+            # 접두어 확인
+            # ex) '119','1195524421' 비교
+            # '1195524421'에서 '119'길이(3) 원소까지만 비교
+            if phone_num in compare_num[:len(phone_num)]:
+                return False
+    
+    return True 
diff --git "a/SOOYEON/week1/\355\225\264\354\213\23403_\354\234\204\354\236\245.py" "b/SOOYEON/week1/\355\225\264\354\213\23403_\354\234\204\354\236\245.py"
new file mode 100644
index 0000000..feaedcc
--- /dev/null
+++ "b/SOOYEON/week1/\355\225\264\354\213\23403_\354\234\204\354\236\245.py"
@@ -0,0 +1,16 @@
+def solution(clothes):
+    # 1. 주어진 정보를 바탕으로 자료 구성
+    clothes_info = {} # 의상 타입(key): 의상 수(value)
+    for c_name, c_type in clothes:
+        clothes_info[c_type] = clothes_info.get(c_type, 0) + 1
+    
+    # 2. 경우의 수 구하기
+    result = 1 # 조합 가능한 경우의 수
+    # a1, a2, a3, b1, b2 의상이 존재할 때 경우의 수
+    # (3 + 1) * (2 + 1) - 1 
+    # +1의 이유는 착용을 안하는 경우
+    # 마지막 -1의 경우는 문제에서 제시된 최소 하나는 착용한다
+    for value in clothes_info.values():
+        result *= value + 1
+    
+    return result - 1
diff --git "a/SOOYEON/week1/\355\225\264\354\213\23404_\353\262\240\354\212\244\355\212\270\354\225\250\353\262\224.py" "b/SOOYEON/week1/\355\225\264\354\213\23404_\353\262\240\354\212\244\355\212\270\354\225\250\353\262\224.py"
new file mode 100644
index 0000000..4af5e6a
--- /dev/null
+++ "b/SOOYEON/week1/\355\225\264\354\213\23404_\353\262\240\354\212\244\355\212\270\354\225\250\353\262\224.py"
@@ -0,0 +1,32 @@
+def solution(genres, plays):
+    GEN, PLAY = 0, 1 # 장르, 재생횟수
+    VALUE = 1        # 딕셔너리에서 value의 인덱스
+    
+    # 1. 주어진 정보 배열들을 알맞게 조작하기
+    song_info = [(gen, play) for gen, play in zip(genres, plays)]
+    
+    dic_songs = {} # 장르(key): 노래정보(고유번호, 재생횟수)
+    dic_plays = {} # 장르(key): 재생횟수 합
+    for idx, info in enumerate(song_info):
+        dic_songs[info[GEN]] = dic_songs.get(info[GEN], []) + [(idx, info[PLAY])]
+        dic_plays[info[GEN]] = dic_plays.get(info[GEN], 0) + info[PLAY]
+    
+    
+    # 2. 장르별 재생횟수 합이 높은 순서대로 정렬
+    dic_plays = sorted(dic_plays.items(), key=lambda x: x[VALUE], reverse=True)
+    
+    
+    result = [] # 정답 배열
+    
+    # 3. 2단계에서 구한 정보를 바탕으로 각 장르에서 재생횟수가 높은 2가지 음악 번호 배열에 담기
+    for gen, plays in dic_plays:
+        best_songs = dic_songs.get(gen) # best_songs = [(고유번호, 재생횟수), ...]
+        best_songs = sorted(best_songs, key=lambda x: x[PLAY], reverse=True)
+        if len(best_songs) >= 2 : # 해당 장르의 곡이 2개 이상일 경우
+            for idx in range(2):
+                result.append(best_songs[idx][0])
+        else: # 2개보다 작으면 전부 다 넣어
+            for idx in range(len(best_songs)):
+                result.append(best_songs[idx][0])
+        
+    return result