Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in $O(n)$ runtime? You may assume the returned list does not count as extra space.

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]

2. Analysis

Mark the according index with negative flag. Similar to hash solution.

3. Solution(s)

One line solution:

class (object):
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        return list(set([x for x in range(1,len(nums)+1)]) - set(nums))

Without extra space and in $O(n)$ runtime:

class (object):
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        res = []
        for i in range(len(nums)):
            index = abs(nums[i]) - 1
            if nums[index] > 0:
                nums[index] *= -1
        for i in range(len(nums)):
            if nums[i] > 0:
                res.append(i+1)
        return res

leetcode448: find all number disappeared in an array 2. Analysis 3. Solution(s)

2. Analysis

3. Solution(s)

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