find-all-duplicates-in-an-array.py

"""

442. Find All Duplicates in an Array
Medium

Given an integer array nums of length n where all the integers of nums are in the range [1, n] and each integer appears once or twice, return an array of all the integers that appears twice.

You must write an algorithm that runs in O(n) time and uses only constant extra space.

 

Example 1:

Input: nums = [4,3,2,7,8,2,3,1]
Output: [2,3]
Example 2:

Input: nums = [1,1,2]
Output: [1]
Example 3:

Input: nums = [1]
Output: []
 

Constraints:

n == nums.length
1 <= n <= 105
1 <= nums[i] <= n
Each element in nums appears once or twice.

"""

# V0
# IDEA : Counter
from collections import Counter
class Solution(object):
    def findDuplicates(self, nums):
        return [elem for elem, count in Counter(nums).items() if count == 2]

# V1
# http://bookshadow.com/weblog/2016/10/25/leetcode-find-all-duplicates-in-an-array/
# https://blog.csdn.net/fuxuemingzhu/article/details/79275549
# IDEA : FIND ALL ELEMENTS EXIST EXACTLY 2 TIMES IN THE GIVEN ARRAY 
class Solution(object):
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        ans = []
        for n in nums:
            if nums[abs(n) - 1] < 0:
                ans.append(abs(n))
            else:
                nums[abs(n) - 1] *= -1
        return ans

# V1'
# http://bookshadow.com/weblog/2016/10/25/leetcode-find-all-duplicates-in-an-array/
# https://blog.csdn.net/fuxuemingzhu/article/details/79275549
class Solution(object):
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        ans = []
        for i in range(len(nums)):
            while nums[i] and nums[i] != i + 1:
                n = nums[i]
                if nums[i] == nums[n - 1]:
                    ans.append(n)
                    nums[i] = 0
                else:
                    nums[i], nums[n - 1] = nums[n - 1], nums[i]
        return ans

# V2 
# Time:  O(n)
# Space: O(1)
class Solution(object):
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        result = []
        for i in nums:
            if nums[abs(i)-1] < 0:
                result.append(abs(i))
            else:
                nums[abs(i)-1] *= -1
        return result


# Time:  O(n)
# Space: O(1)
class Solution2(object):
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        result = []
        i = 0
        while i < len(nums):
            if nums[i] != nums[nums[i]-1]:
                nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]
            else:
                i += 1

        for i in range(len(nums)):
            if i != nums[i]-1:
                result.append(nums[i])
        return result


# Time:  O(n)
# Space: O(n), this doesn't satisfy the question
from collections import Counter
class Solution3(object):
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        return [elem for elem, count in Counter(nums).items() if count == 2]