From 31d30f88ace15ec1dd0d41d46f43a6f3e82dea82 Mon Sep 17 00:00:00 2001
From: tom4649 <tkaaynakiyoshi@gmail.com>
Date: Fri, 29 May 2026 10:50:55 +0900
Subject: [PATCH] 767.Reorganize String

---
 0767.Reorganize-String/memo.md  | 15 +++++++++++++++
 0767.Reorganize-String/step1.py | 32 ++++++++++++++++++++++++++++++++
 0767.Reorganize-String/step2.py | 30 ++++++++++++++++++++++++++++++
 0767.Reorganize-String/step3.py | 29 +++++++++++++++++++++++++++++
 4 files changed, 106 insertions(+)
 create mode 100644 0767.Reorganize-String/memo.md
 create mode 100644 0767.Reorganize-String/step1.py
 create mode 100644 0767.Reorganize-String/step2.py
 create mode 100644 0767.Reorganize-String/step3.py

diff --git a/0767.Reorganize-String/memo.md b/0767.Reorganize-String/memo.md
new file mode 100644
index 0000000..ca9187a
--- /dev/null
+++ b/0767.Reorganize-String/memo.md
@@ -0,0 +1,15 @@
+# 767.Reorganize String
+
+## step1
+
+heapを使う解法を思いつく
+
+時間計算量は、文字列をK種類としたときO(Nlog K)。今K=26なのでほぼ線形。
+
+## step2
+
+https://leetcode.com/problems/reorganize-string/solutions/3947780/100-2-approaches-priority-queue-sort-by-66wlu/?envType=problem-list-v2&envId=7p5x763
+
+計算量は O(N)。こちらの方が速い。
+
+面接で聞かれたとしたらこちらまで聞かれるのだろうか。
diff --git a/0767.Reorganize-String/step1.py b/0767.Reorganize-String/step1.py
new file mode 100644
index 0000000..5b14d87
--- /dev/null
+++ b/0767.Reorganize-String/step1.py
@@ -0,0 +1,32 @@
+import heapq
+import collections
+
+
+class Solution:
+    def reorganizeString(self, s: str) -> str:
+        if len(s) == 1:
+            return s
+
+        counter = collections.Counter(s)
+        most_common = counter.most_common()
+
+        if most_common[0][1] > (len(s) - 1) // 2 + 1:
+            return ""
+
+        count_and_letter = [(count, c) for (c, count) in most_common]
+        heapq.heapify_max(count_and_letter)
+        result = []
+
+        while len(result) < len(s):
+            count, c = heapq.heappop_max(count_and_letter)
+            result.append(c)
+            count_next = 0
+            if len(count_and_letter) > 0:
+                count_next, c_next = heapq.heappop_max(count_and_letter)
+                result.append(c_next)
+            if count > 1:
+                heapq.heappush_max(count_and_letter, (count - 1, c))
+            if count_next > 1:
+                heapq.heappush_max(count_and_letter, (count_next - 1, c_next))
+
+        return "".join(result)
diff --git a/0767.Reorganize-String/step2.py b/0767.Reorganize-String/step2.py
new file mode 100644
index 0000000..ae4f1df
--- /dev/null
+++ b/0767.Reorganize-String/step2.py
@@ -0,0 +1,30 @@
+import collections
+
+
+class Solution:
+    def reorganizeString(self, s: str) -> str:
+        counter = collections.Counter(s)
+        n = len(s)
+
+        most_common_char, max_count = counter.most_common(1)[0]
+
+        if max_count > (n + 1) // 2:
+            return ""
+
+        result = [""] * n
+        index = 0
+
+        for _ in range(max_count):
+            result[index] = most_common_char
+            index += 2
+
+        del counter[most_common_char]
+
+        for char, count in counter.items():
+            for _ in range(count):
+                if index >= n:
+                    index = 1
+                result[index] = char
+                index += 2
+
+        return "".join(result)
diff --git a/0767.Reorganize-String/step3.py b/0767.Reorganize-String/step3.py
new file mode 100644
index 0000000..13f8987
--- /dev/null
+++ b/0767.Reorganize-String/step3.py
@@ -0,0 +1,29 @@
+import collections
+
+
+class Solution:
+    def reorganizeString(self, s: str) -> str:
+        counter = collections.Counter(s)
+        most_common_letter, most_common_count = counter.most_common()[0]
+
+        if most_common_count > (len(s) + 1) // 2:
+            return ""
+
+        result = [""] * len(s)
+        index = 0
+        while most_common_count > 0:
+            result[index] = most_common_letter
+            index += 2
+            most_common_count -= 1
+
+        del counter[most_common_letter]
+
+        for letter, count in counter.items():
+            while count > 0:
+                if index >= len(s):
+                    index = 1
+                result[index] = letter
+                index += 2
+                count -= 1
+
+        return "".join(result)