[최원준] Day22

leetcode3/최원준/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

leetcode3/최원준/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
+
+
+
+