leetcode - 442. Find All Duplicates in an Array

Description

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 <= 10^5
1 <= nums[i] <= n
Each element in nums appears once or twice.

Solution

Use value as the index, since all the values are in [1,n], we can use value - 1 as the index. And make the value under that index to be negative. Then if we use the index, and the number is already negative, then we find our answer.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

class Solution:
    def findDuplicates(self, nums: List[int]) -> List[int]:
        res = []
        for each_num in nums:
            if each_num < 0:
                each_num *= -1
            index = each_num - 1
            if nums[index] < 0:
                res.append(index + 1)
            else:
                nums[index] *= -1
        return res