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]

 

 

Solution:

When finding number i, flip the number at position (i-1) to negative.

If the number at position (i-1) is already negative, i is the number that occurs twice.

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

 

Time complexity: O(n).

Space complexity: O(n).