# [Solved] Find Two Unique Numbers from Array in O(n) Time

**Problem Statement: Find Two Unique Numbers from Array in O(n) Time**

Let's assume all numbers except two occur twice in an array. How do you get those two numbers to occur only once? For example, only two numbers, 4 and 6, in the array {2, 4, 3, 6, 3, 2, 5, 5} occur once, and the other numbers occur twice. Therefore, the output should be 4 and 6.

This question was asked in the NVIDIA interview coding round.

**Explanation:**

Let’s solve this problem into two steps:

Find the unique number in the array,

- when there is exactly one unique number
- when there are exactly two unique numbers

See one by one.

#### Find the unique number in the array when there is exactly one unique number.

If there is a single unique element in the array, you can easily find the duplicates using the XOR operation.

Things you should understand to solve this problem.

- XOR operation
- Bitwise operation

You can learn bitwise operations in Python.

XOR is a logical operation that returns 0 if both numbers are the same. When you XORed a number with zero, it returns the same number.

Suppose we have the following array of the numbers [2, 4, 3, 3, 2, 5, 5] which has exactly one unique number (4).

Performing XOR operation:

= 2^4^3^3^2^5^5 = 2^2^3^3^5^5^4 =0^0^0^4 =4

**Python Program:**

```
def findUnique(arr):
out = 0
for i in arr:
out = i ^ out
return out
print(findUnique([2, 4, 3, 3, 2, 5, 5]))
```

**Output:**

4

This way, using XOR operation, you can find the unique number in the array if there is exactly one unique number.

#### Find the unique numbers in the array where there are exactly two unique numbers.

What if you have two unique numbers in the array?

You have to split the given array into sub-arrays where each unique number is present in one sub-array.

**How to partition the array?**

- Let’s say this is our
`[2, 4, 3, 6, 3, 2, 5, 5]`

given array where two numbers are unique. - After performing XOR operation on each element in the array, it results in an output
`4(100)^6(110)=2 (binary 10)`

. - This means the second last bit of two unique numbers is different.
- We can split the array into two sub-arrays. All the numbers having second last bit one forms one sub-array. Keep the rest of the numbers in another sub-array.

Two sub-arrays after splitting:`[4, 5, 5]`

and`[2, 3, 6, 3, 2]`

- Find the unique number for each sub-array using the XOR operation. This will give us the desired output as 4 and 6.

Let’s write a program for it.

**Python Program:**

```
def findUnique(arr):
out=0
for i in arr:
out=i^out
return out
def createSubArray(arr, num):
num=(-1)*num
arr1=[]
arr2=[]
for i in arr:
if "{0:b}".format(i)[num]=="0":
arr1.append(i)
else:
arr2.append(i)
return arr1, arr2
def findFirstBit1FromLast(num):
strNum="{0:b}".format(num)
for i in range(0, len(strNum)):
if strNum[len(strNum)-1-i]=="1":
return i+1
arr=[2, 4, 3, 6, 3, 2, 5, 5]
temp=findUnique(arr)
bitLoc=findFirstBit1FromLast(temp)
arr1, arr2=createSubArray(arr, bitLoc)
print(findUnique(arr1))
print(findUnique(arr2))
```

**Output:**

4 6

**Complexity:**

We are calling `findUnique()`

method/function thrice in the above program- one is for the main array and one for each sub-array. For the main function, it will take time `O(n)`

. For sub-arrays, it will take `O(n/2)`

on average.

The total complexity will be `O(n)+O(n/2)+O(n/2)`

which is equivalent of `O(n)`

.

Using the XOR operation, it is possible to find the unique numbers in linear time.

Can you find two unique numbers from an array in O(n) time in any other programming language like C/C++ or Java? Share your code in the comment.

**Another similar kind of challenge:**[Solved] Find Duplicate in Array in O(n) Linear Time