Missing Number

0268.Missing-Number/memo.md

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+# 268. Missing Number
+
+## step1
+sortする解法: 時間 O(nlogn) 空間 O(n)
+setを用いる: 時間 O(n) 空間 O(n)
+和を用いる: 時間 O(n) 空間 O(1)
+
+> Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
+
+最後のが答えか？
+
+## 他の人のコード
+
+https://leetcode.com/problems/missing-number/solutions/8334452/video-gauss-sum-and-bitwise-xor-by-abstr-pi85/
+
+xorを使った解法。なるほど。これは覚えておきたい。
+
+https://github.com/Kitaken0107/GrindEasy/pull/20
+
+https://github.com/Nbotter/leetcode/pull/3/changes
+
+## step2
+sort解法の改善とxorの解法
+

0268.Missing-Number/step1_set.py

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+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        num_set = set(nums)
+        for i in range(len(nums)+1):
+            if i not in num_set:
+                return i
+
+

0268.Missing-Number/step1_sort.py

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+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        sorted_nums = sorted(nums)
+        expected = 0
+        for i in range(len(sorted_nums)):
+            if sorted_nums[i] != expected:
+                break
+            expected += 1
+
+        return expected

0268.Missing-Number/step1_sum.py

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+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        return (len(nums)+1) * len(nums) // 2 - sum(nums)

0268.Missing-Number/step2_sort.py

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+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        sorted_nums = sorted(nums)
+        expected = 0
+        for expected, actual in enumerate(sorted_nums):
+            if expected != actual:
+                return expected
+
+        return len(nums)

0268.Missing-Number/step2_xor.py

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+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        xor_sum = 0
+        for actual in nums:
+            xor_sum ^= actual
+        for expected in range(len(nums)+1):
+            xor_sum ^= expected
+
+        return xor_sum