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 in two steps:
Find the unique number in the array,
See them one by one.
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.
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 the XOR operation, you can find the unique number in the array if there is exactly one unique number.
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?
[2, 4, 3, 6, 3, 2, 5, 5]
given array where two numbers are unique.[4, 5, 5]
and [2, 3, 6, 3, 2]
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