## Tuesday, June 6, 2017

### Problem Statement

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements of [1, n] inclusive that do not appear in this array.
Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

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

Output:
[5,6]```

### Solution

• Iterate through the input array and mark elements as negative using `nums[nums[i] -1] = -nums[nums[i]-1]`. In this way all the numbers that we have seen will be marked as negative.
• In the second iteration, if a value is not marked as negative, it implies we have never seen that index before, so just add it to the return list.

```public class  AllNosDissappearedInArray
{
public static List<Integer> get(int[] nums) {
List<Integer> ret = new ArrayList<Integer>();

for(int i = 0; i < nums.length; i++) {
int val = Math.abs(nums[i]) - 1;
if(nums[val] > 0) {
nums[val] = -nums[val];
}
}

for(int i = 0; i < nums.length; i++) {
if(nums[i] > 0) {
}
}
return ret;
}

}
```

```{4,3,2,7,8,2,3,1} 1) 4-1 =3 //Third position is made negative 6) num=-num -7 2) 3-1 =2 //Second position is made negative num=-num -22:55 PM 5/8/2017 2:44 PM 5/8/2017 3) 2-1=1 //First position is made negative num=-num -3 4)7-1=6 //Sixth position is made negative num=-num -3 5)8-1=72:47 PM 5/8/2017 num=-num //Seventh position is made negative -1 6) 2-1=1 num=-num //Note it would not satisy the if condition.Hence{4,3,2,7,8,2,3,1}

1) 4-1 =3
//Third position is made negative     6)
num=-num
-7

2) 3-1 =2
num=-num
-22:55 PM 5/8/2017
2:44 PM 5/8/2017
3) 2-1=1
num=-num
-3

4)7-1=6
num=-num
-3

5)8-1=72:47 PM 5/8/2017
num=-num
-1
6) 2-1=1
num=-num
//Note it would not satisy the if condition.Hence it would not enter the loop
7)
-3

it would not enter the loop 7) -3
```