Create set-mismatch.py

Python/set-mismatch.py

@@ -0,0 +1,50 @@
+# Time:  O(n)
+# Space: O(1)
+
+# The set S originally contains numbers from 1 to n.
+# But unfortunately, due to the data error, one of the numbers
+# in the set got duplicated to another number in the set, which results
+# in repetition of one number and loss of another number.
+#
+# Given an array nums representing the data status of this set after the error.
+# Your task is to firstly find the number occurs twice and then find the number
+# that is missing. Return them in the form of an array.
+#
+# Example 1:
+# Input: nums = [1,2,2,4]
+# Output: [2,3]
+# Note:
+# The given array size will in the range [2, 10000].
+# The given array's numbers won't have any order.
+
+class Solution(object):
+    def findErrorNums(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        result = [0] * 2
+        for i in nums:
+            if nums[abs(i)-1] < 0:
+                result[0] = abs(i)
+            else:
+                nums[abs(i)-1] *= -1
+        for i in xrange(len(nums)):
+            if nums[i] > 0:
+                result[1] = i+1
+            else:
+                nums[i] *= -1
+        return result
+
+# Time:  O(n)
+# Space: O(1)  
+class Solution2(object):
+    def findErrorNums(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        N = len(nums)
+        x_minus_y = sum(nums) - N*(N+1)//2
+        x_plus_y = (sum(x*x for x in nums) - N*(N+1)*(2*N+1)/6) // x_minus_y
+        return (x_plus_y+x_minus_y) // 2, (x_plus_y-x_minus_y) // 2