Counting Bits

0338.Counting-Bits/memo.md

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+# 338. Counting Bits
+
+## step1
+4m ぐらい。hamming weightの問題の配列前計算の方法をまねた。計算量 O(n)。動的計画法である。
+
+動的計画法を使わないと計算量は O(nlog n)になる。
+
+しかし、実行速度自体は組み込み関数を使えばこの解法の方が速い。
+
+Follow up もその旨が書かれている。
+
+> It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
+
+## 他の人のコード
+https://github.com/rihib/leetcode/pull/44
+
+動的計画法だが考え方が異なる。自分のものは下一桁とそれ以外に分けていたが、これは上一桁とそれ以外に分けている。
+
+> Rightmost set bitをunsetする方法
+- x & (x - 1)
+- x - (x & -x)
+
+https://github.com/rihib/leetcode/blob/main/go/counting_bits.go

0338.Counting-Bits/step1_dp.py

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+class Solution:
+    def countBits(self, n: int) -> list[int]:
+        if n == 0:
+            return [0]
+
+        result = [0] * (n + 1)
+
+        for i in range(n + 1):
+            result[i] = result[i >> 1] + (i & 1)
+
+        return result
+

0338.Counting-Bits/step1_naive.py

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+class Solution:
+    def countBits(self, n: int) -> list[int]:
+        result = [0] * (n + 1)
+
+        for i in range(n + 1):
+            result[i] = i.bit_count()
+
+        return result
+

0338.Counting-Bits/step2_dpv2.py

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+class Solution:
+    def countBits(self, n: int) -> list[int]:
+        if n == 0:
+            return [0]
+
+        result = [0] * (n + 1)
+        power_of_two = 1
+
+        for i in range(1, n + 1):
+            if i == power_of_two << 1:
+                result[i] = 1
+                power_of_two = i
+            else:
+                result[i] = result[i - power_of_two] + 1
+
+        return result
+

0338.Counting-Bits/step2_naive.py

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+class Solution:
+    def countBits(self, n: int) -> list[int]:
+        return [i.bit_count() for i in range(n + 1)]
+

0338.Counting-Bits/step2_unset.py

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+class Solution:
+    def countBits(self, n: int) -> list[int]:
+        if n == 0:
+            return [0]
+
+        result = [0] * (n + 1)
+
+        for i in range(1, n + 1):
+            result[i] = result[i & (i - 1)] + 1
+
+        return result
+

0338.Counting-Bits/step2_unsetv2.py

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+class Solution:
+    def countBits(self, n: int) -> list[int]:
+        if n == 0:
+            return [0]
+
+        result = [0] * (n + 1)
+
+        for i in range(1, n + 1):
+            result[i] = result[i - (i & -i)] + 1
+
+        return result
+