LeetCode 442. Find All Duplicates in an Array (Medium)

Description:

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

Find all the elements that appear twice in this array.

Could you do it without extra space and in O(n) runtime?

Example:

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

Output:
[2,3]

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Analysis:

Traverse the array, and in each iteration, we multiply nums[abs(nums[i])-1]  by -1. After each multiplication, we check the value of nums[abs(nums[i])-1], if the value is above 0, it means that abs(nums[i])  occurs twice otherwise it only occurs once.

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Code:

class Solution {
    public List<Integer> findDuplicates(int[] nums) {
        List<Integer> rs = new ArrayList<>();
        
        for(int i = 0; i < nums.length; i++) {
            nums[Math.abs(nums[i])-1] *= -1;
            if(nums[Math.abs(nums[i])-1] > 0) {
                rs.add(Math.abs(nums[i]));
            }
        }
        
        return rs;
    }
}

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Complexity:

Time complexity: $O(n)$O(n) where $n$n represents the length of array $nums$nums.
Space complexity: $O(n)$O(n), but the program doesn’t use extra space except the $List$List that stores the results.