Problem
Given an array nums
of n
integers where nums[i]
is in the range [1, n]
, return an array of all the integers in the range [1, n]
that do not appear in nums
.
Example 1:
Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]
Example 2:
Input: nums = [1,1]
Output: [2]
Constraints:
n == nums.length
1 <= n <= 10^5
1 <= nums[i] <= n
Follow up: Could you do it without extra space and in O(n)
runtime? You may assume the returned list does not count as extra space.
Solution (Java)
class Solution {
public List<Integer> findDisappearedNumbers(int[] nums) {
int[] arr = new int[nums.length];
int counter = 0;
for (int i = 0; i < nums.length; i++) {
arr[nums[counter] - 1] = 1;
counter++;
}
List<Integer> res = new ArrayList<>();
for (int i = 0; i < arr.length; i++) {
if (arr[i] == 0) {
res.add(i + 1);
}
}
return res;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).