Kth symbol in grammar

arai60/kth_symbol_in_grammar/phase1.py

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+class Solution:
+    def kthGrammar(self, n: int, k: int) -> int:
+        moves = deque([]) # 0: move left 1: move right
+        # store the order from the leaf
+        for i in range(n-1):
+            if (k - 1) % 2 == 0:
+                moves.appendleft(0)
+            else:
+                moves.appendleft(1)
+            k = (k - 1) // 2 # 0-indexed 
+            k += 1 # 1-indexed 
+        kth_symbol = 0
+        for move in moves:
+            # move 0 kth_symbol 0 -> 0
+            # move 1 kth_symbol 0 -> 1
+            # move 0 kth_symbol 1 -> 1
+            # move 1 kth_symbol 1 -> 0
+            kth_symbol = kth_symbol ^ move
+
+        return kth_symbol

arai60/kth_symbol_in_grammar/phase2.py

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+"""
+Reference:
+Mike0121: https://github.com/Mike0121/LeetCode/pull/18/files
+再帰で書く書き方, 僕がphase1で書いたやり方は移動順を取り出してきて根から辿る方法だが, こっちだと明示的にfor文を書かなくてもかける
+
+Exzrgs: https://github.com/Exzrgs/LeetCode/pull/12/files
+shining-ai: https://github.com/shining-ai/leetcode/pull/46/files
+kを1で表した時の数の1の数と反転回数に注目, bitcountを用いて解く
+hayashi-ay: https://github.com/hayashi-ay/leetcode/pull/46/files
+親を表す単語が問題文にないことに注意して命名を行う
+"""
+
+# k = (k-1) // 2 + 1がk = (k+1) // 2でまとめられることに気づきまとめる
+class Solution:
+    def kthGrammar(self, n: int, k: int) -> int:
+        moves = deque([]) # 0: move left 1: move right
+        # store the order from the leaf
+        for i in range(n-1):
+            if (k - 1) % 2 == 0:
+                moves.appendleft(0)
+            else:
+                moves.appendleft(1)
+            k = (k + 1) // 2 
+        kth_symbol = 0
+        for move in moves:
+            # move 0 kth_symbol 0 -> 0
+            # move 1 kth_symbol 0 -> 1
+            # move 0 kth_symbol 1 -> 1
+            # move 1 kth_symbol 1 -> 0
+            kth_symbol = kth_symbol ^ move
+
+        return kth_symbol
+
+# 再帰
+class Solution:
+    def kthGrammar(self, n: int, k: int) -> int:
+        if k == 1:
+            return 0
+
+        previous_value = self.kthGrammar(n - 1, (k + 1) // 2)
+
+        if k % 2 == 0:
+            return 1 - previous_value
+        else:
+            return previous_value
+
+# bit count
+class Solution:
+    def kthGrammar(self, n: int, k: int) -> int:
+        return (k - 1).bit_count() % 2

arai60/kth_symbol_in_grammar/phase3.py

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+# 再帰で解く方法を選択
+
+class Solution:
+    def kthGrammar(self, n: int, k: int) -> int:
+        if k == 1:
+            return 0
+
+        previous_value = self.kthGrammar(n - 1, (k + 1) // 2)
+        if k % 2 == 0:
+            return 1 - previous_value
+        else:
+            return previous_value