diff --git "a/leetcode3/\353\263\200\354\247\200\355\230\221/v2/3992. Rearrange String to Avoid Character Pair.py" "b/leetcode3/\353\263\200\354\247\200\355\230\221/v2/3992. Rearrange String to Avoid Character Pair.py"
new file mode 100644
index 00000000..e310a3b2
--- /dev/null
+++ "b/leetcode3/\353\263\200\354\247\200\355\230\221/v2/3992. Rearrange String to Avoid Character Pair.py"	
@@ -0,0 +1,24 @@
+
+'''
+1. 아이디어 :
+그냥 y를 먼저 붙이고, 나머지 문자를 붙인다.
+x 몰라도 그냥 풀림
+
+2. 시간복잡도 :
+o(n)
+
+3. 자료구조/알고리즘 :
+'''
+
+class Solution:
+    def rearrangeString(self, s: str, x: str, y: str) -> str:
+        ans = ''
+        tmp = ''
+        
+        for i in s:
+            if i == y:
+                tmp += i
+            else:
+                ans += i
+        
+        return tmp+ans
\ No newline at end of file
diff --git "a/leetcode3/\353\263\200\354\247\200\355\230\221/v2/948. Bag of Tokens.py" "b/leetcode3/\353\263\200\354\247\200\355\230\221/v2/948. Bag of Tokens.py"
new file mode 100644
index 00000000..22819446
--- /dev/null
+++ "b/leetcode3/\353\263\200\354\247\200\355\230\221/v2/948. Bag of Tokens.py"	
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+
+'''
+1. 아이디어 :
+점수를 잃을땐 최대 power를 얻고, 점수를 얻을 땐 최소 pwoer를 소모한다.
+
+2. 시간복잡도 :
+o(nlogn)
+
+3. 자료구조/알고리즘 :
+투포인터
+'''
+
+
+class Solution:
+    def bagOfTokensScore(self, tokens: List[int], power: int) -> int:
+        tokens.sort()
+        n = len(tokens)
+        
+        a = 0
+        b = n - 1
+        score = 0
+        ans = 0
+
+        while True:
+            print('a,b,power,score:', a,b,power)
+            if a > b:
+                break
+            
+            if power >= tokens[a]:
+                power -= tokens[a]
+                score +=1
+                ans = max([score, ans])
+                a += 1
+            else:
+                if score == 0:
+                    break
+                power += tokens[b]
+                score -= 1
+                b -= 1
+        
+        return ans
\ No newline at end of file