From cb6910126646069f0b6d1e94d07da7bf3873734c Mon Sep 17 00:00:00 2001
From: Won Joon Thomas Choi <113500771+724thomas@users.noreply.github.com>
Date: Thu, 20 Aug 2026 22:40:35 +0900
Subject: [PATCH 1/2] Create 3992. Rearrange String to Avoid Character Pair.py

---
 ...arrange String to Avoid Character Pair.py" | 27 +++++++++++++++++++
 1 file changed, 27 insertions(+)
 create mode 100644 "leetcode3/\354\265\234\354\233\220\354\244\200/3992. Rearrange String to Avoid Character Pair.py"

diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/3992. Rearrange String to Avoid Character Pair.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/3992. Rearrange String to Avoid Character Pair.py"
new file mode 100644
index 00000000..bdf21c5f
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/3992. Rearrange String to Avoid Character Pair.py"	
@@ -0,0 +1,27 @@
+#
+
+'''
+1. 아이디어 :
+
+
+2. 시간복잡도 :
+    O()
+
+3. 자료구조/알고리즘 :
+
+
+'''
+class Solution:
+    def rearrangeString(self, s: str, x: str, y: str) -> str:
+        counter = Counter(s)
+
+        ans = ""
+        if y in counter:
+            ans += y*counter[y]
+        if x in counter:
+            ans += x*counter[x]
+        for char, freq in counter.items():
+            if char == x or char == y:
+                continue
+            ans += char * freq
+        return ans

From 49abee49a2262b34c7153d18eb4ec7636adbf36b Mon Sep 17 00:00:00 2001
From: Won Joon Thomas Choi <113500771+724thomas@users.noreply.github.com>
Date: Thu, 20 Aug 2026 22:41:13 +0900
Subject: [PATCH 2/2] Create 948. Bag of Tokens.py

---
 .../948. Bag of Tokens.py"                    | 40 +++++++++++++++++++
 1 file changed, 40 insertions(+)
 create mode 100644 "leetcode3/\354\265\234\354\233\220\354\244\200/948. Bag of Tokens.py"

diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/948. Bag of Tokens.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/948. Bag of Tokens.py"
new file mode 100644
index 00000000..4d4a353e
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/948. Bag of Tokens.py"	
@@ -0,0 +1,40 @@
+#
+
+'''
+1. 아이디어 :
+더할때는 가장 큰것부터, 뺄때는 가장 작은것부터.
+
+2. 시간복잡도 :
+    O(nlogn)
+
+3. 자료구조/알고리즘 :
+two pointer
+
+'''
+class Solution:
+    def bagOfTokensScore(self, tokens: List[int], power: int) -> int:
+        n = len(tokens)
+        tokens.sort()
+        
+        left = 0
+        right = n-1
+        score = 0
+        ans = 0
+
+        while left<=right:
+            if power>=tokens[left]:
+                power -= tokens[left]
+                left+=1
+                score+=1
+                ans = max(ans, score)
+            elif score>0:
+                power += tokens[right]
+                right-=1
+                score-=1
+            else:
+                break
+        return ans
+
+        
+
+