# Java O(n) Time O(1) Space

• ``````public static int[] findErrorNums(int[] nums) {
int[] res = new int[2];
for (int i : nums) {
if (nums[Math.abs(i) - 1] < 0) res[0] = Math.abs(i);
else nums[Math.abs(i) - 1] *= -1;
}
for (int i=0;i<nums.length;i++) {
if (nums[i] > 0) res[1] = i+1;
}
return res;
}
``````

• Similar Solution. But maybe it's better to restore the input values.

``````public class Solution {
public int[] findErrorNums(int[] nums) {
int[] res = new int[2]; // duplicate, missing
//For each number we found, set nums[number-1] to its negative value (<0)
for(int i=0; i<nums.length; i++) {
int idx = Math.abs(nums[i])-1; //since index starts from 0, and the set starts from 1
if(nums[idx]> 0) nums[idx] = -nums[idx];
else res[0] = idx+1; //have already been found
}
// At this point, only nums[missingNumber-1] > 0
for(int i=0; i<nums.length; i++) {
if(nums[i] <0) nums[i] = -nums[i]; //restore the original values
else res[1]=i+1;//since index starts from 0, and the set starts from 1
}
return res;
}
}
``````

• smart idea! thanks!

• Great idea!
Little modification with only one loop. We can get sum of arithmetic progression by the formula and calculate actual sum.

``````class Solution {
public:
vector<int> findErrorNums(vector<int>& nums) {
int correctSum = (nums.size()*(2 + nums.size()-1))/2;
int actualSum = 0;
for (auto n : nums) {
nums[abs(n)-1] *= -1;
if (nums[abs(n)-1] > 0) {