# Find two non repeating elements of repeated array elements

The problem is to find two non-repeating elements in an array where all other elements are repeated exactly twice. In other words, there are two unique elements in the array, and all other elements occur twice. The task is to find those two unique elements.

## Example

Let's consider an array with repeated elements and two unique elements: [5, 7, 9, 3, 1, 7, 5, 4, 1, 3]

The two unique elements in this array are 9 and 4.

## Pseudocode

Procedure displayArray(num[], n)
// Display array elements
For i = 0 to n-1
Print num[i] + " "
End For
Print new line
End Procedure

Procedure twoNonRepeatingNo(num[], n)
// Resultant x and y
x = 0
y = 0

// Calculate xor of all array elements
xorValue = num[0]
For i = 1 to n-1
xorValue = xorValue XOR num[i]
End For

// Find the value of rightmost set (active) bit
rightActiveBit = xorValue AND ~(xorValue - 1)

// Find resultant x and y non-repeated element
For i = 0 to n-1
If (num[i] AND rightActiveBit) != 0 Then
// When num[i] contains the rightmost set bit
x = x XOR num[i]
Else
// Otherwise
y = y XOR num[i]
End If
End For

// Display array elements
displayArray(num, n)

// Display calculated result
Print "Result: x =", x, ", y =", y, new line
End Procedure

// Main Function
Function main
num[] = {5, 7, 9, 3, 1, 7, 5, 4, 1, 3}
// Get number of elements in the array
n = size of num[] / size of num[0]
// Test the function
twoNonRepeatingNo(num, n)
Return 0
End Function
1. Initialize two variables x and y to store the two unique elements.
2. Calculate the XOR of all elements in the array and store it in a variable xorValue.
3. Find the rightmost set bit (active bit) in xorValue and store it in rightActiveBit.
4. Iterate through the array again: a. If the rightActiveBit is set in the current element, XOR it with x. b. If the rightActiveBit is not set in the current element, XOR it with y.
5. After the loop, the variables x and y will contain the two non-repeating elements.

## Algorithm Explanation

>
1. The XOR operation is used because it is commutative and associative. XOR of a number with itself results in 0. Therefore, when we XOR all elements in the array, the repeated elements will cancel out, and only x and y will remain in the result.
2. After XORing all elements, the result will be a number that contains the XOR of x and y. Since x and y are distinct, there will be at least one set bit (1) in the result, indicating the positions where x and y differ.
3. We find the rightmost set bit (active bit) in the result using the bitwise AND operation with its two's complement. This operation isolates the rightmost set bit, and all other bits become zero. This is stored in the rightActiveBit.
4. In the final loop, we separate the elements based on the rightActiveBit. All elements that have a set bit in the same position as the rightActiveBit will contribute to the calculation of x, and the others will contribute to y.

## Code Solution

Here given code implementation process.

/*
C program for
Find two non repeating elements of repeated array elements
*/
// Include header file
#include <stdio.h>

void displayArray(int num[], int n)
{
// Display array elements
for (int i = 0; i < n; ++i)
{
printf("  %d", num[i]);
}
}
void twoNonRepeatingNo(int num[], int n)
{
// Resultant x and y
int x = 0;
int y = 0;
// get first array element
int xorValue = num[0];
// Calculate xor of all array elements
for (int i = 1; i < n; ++i)
{
xorValue = xorValue ^ num[i];
}
// Find the value of rightmost set (active) bit
int rightActiveBit = xorValue & ~(xorValue - 1);
// Find resultant x and y non repeated element
for (int i = 0; i < n; ++i)
{
if ((num[i] & rightActiveBit) != 0)
{
// When num[i] contain rightmost set bit
x = x ^ num[i];
}
else
{
// otherwise
y = y ^ num[i];
}
}
// Display array elements
displayArray(num, n);
// Display calculated result
printf("\n Result : x = %d, y = %d \n", x, y);
}
int main()
{
int num[] = {
5 , 7 , 9 , 3 , 1 , 7 , 5 , 4 , 1 , 3
};
// Get number of elements in number
int n = sizeof(num) / sizeof(num[0]);
// Test
twoNonRepeatingNo(num, n);
return 0;
}

#### Output

5  7  9  3  1  7  5  4  1  3
Result : x = 9, y = 4
// Java Program
// Find two non repeating elements of repeated array elements
class Finding
{
public void displayArray(int[] num, int n)
{
// Display array elements
for (int i = 0; i < n; ++i)
{
System.out.print(" " + num[i]);
}
}
public void twoNonRepeatingNo(int[] num, int n)
{
// Resultant x and y
int x = 0;
int y = 0;
// get first array element
int xorValue = num[0];
// Calculate xor of all array elements
for (int i = 1; i < n; ++i)
{
xorValue = xorValue ^ num[i];
}
// Find the value of rightmost set (active) bit
int rightActiveBit = xorValue & ~(xorValue - 1);
// Find resultant x and y non repeated element
for (int i = 0; i < n; ++i)
{
if ((num[i] & rightActiveBit) != 0)
{
// When num[i] contain rightmost set bit
x = x ^ num[i];
}
else
{
// otherwise
y = y ^ num[i];
}
}
// Display array elements
this.displayArray(num, n);
// Display calculated result
System.out.print("\n Result : x = " + x + ", y = " + y + " \n");
}
public static void main(String[] args)
{
Finding task = new Finding();
// Array which contains duplicate and two non repeated element
int[] num = {
5 , 7 , 9 , 3 , 1 , 7 , 5 , 4 , 1 , 3
};
// Get number of elements in number
int n = num.length;
// Test
}
}

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4
// Include header file
#include <iostream>
using namespace std;

// C++ Program
// Find two non repeating elements of repeated array elements

class Finding
{
public: void displayArray(int num[], int n)
{
// Display array elements
for (int i = 0; i < n; ++i)
{
cout << " " << num[i];
}
}
void twoNonRepeatingNo(int num[], int n)
{
// Resultant x and y
int x = 0;
int y = 0;
// get first array element
int xorValue = num[0];
// Calculate xor of all array elements
for (int i = 1; i < n; ++i)
{
xorValue = xorValue ^ num[i];
}
// Find the value of rightmost set (active) bit
int rightActiveBit = xorValue &~(xorValue - 1);
// Find resultant x and y non repeated element
for (int i = 0; i < n; ++i)
{
if ((num[i] &rightActiveBit) != 0)
{
// When num[i] contain rightmost set bit
x = x ^ num[i];
}
else
{
// otherwise
y = y ^ num[i];
}
}
// Display array elements
this->displayArray(num, n);
// Display calculated result
cout << "\n Result : x = " << x << ", y = " << y << " \n";
}
};
int main()
{
Finding *task = new Finding();
// Array which contains duplicate and two non repeated element
int num[] = {
5 , 7 , 9 , 3 , 1 , 7 , 5 , 4 , 1 , 3
};
// Get number of elements in number
int n = sizeof(num) / sizeof(num[0]);
// Test
return 0;
}

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4
// Include namespace system
using System;
// Csharp Program
// Find two non repeating elements of repeated array elements

public class Finding
{
public void displayArray(int[] num, int n)
{
// Display array elements
for (int i = 0; i < n; ++i)
{
Console.Write(" " + num[i]);
}
}
public void twoNonRepeatingNo(int[] num, int n)
{
// Resultant x and y
int x = 0;
int y = 0;
// get first array element
int xorValue = num[0];
// Calculate xor of all array elements
for (int i = 1; i < n; ++i)
{
xorValue = xorValue ^ num[i];
}
// Find the value of rightmost set (active) bit
int rightActiveBit = xorValue & ~(xorValue - 1);
// Find resultant x and y non repeated element
for (int i = 0; i < n; ++i)
{
if ((num[i] & rightActiveBit) != 0)
{
// When num[i] contain rightmost set bit
x = x ^ num[i];
}
else
{
// otherwise
y = y ^ num[i];
}
}
// Display array elements
this.displayArray(num, n);
// Display calculated result
Console.Write("\n Result : x = " + x + ", y = " + y + " \n");
}
public static void Main(String[] args)
{
Finding task = new Finding();
// Array which contains duplicate and two non repeated element
int[] num = {
5 , 7 , 9 , 3 , 1 , 7 , 5 , 4 , 1 , 3
};
// Get number of elements in number
int n = num.Length;
// Test
}
}

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4
<?php
// Php Program
// Find two non repeating elements of repeated array elements
class Finding
{
public function displayArray(\$num, \$n)
{
// Display array elements
for (\$i = 0; \$i < \$n; ++\$i)
{
echo(" ".\$num[\$i]);
}
}
public function twoNonRepeatingNo(\$num, \$n)
{
// Resultant x and y
\$x = 0;
\$y = 0;
// get first array element
\$xorValue = \$num[0];
// Calculate xor of all array elements
for (\$i = 1; \$i < \$n; ++\$i)
{
\$xorValue = \$xorValue ^ \$num[\$i];
}
// Find the value of rightmost set (active) bit
\$rightActiveBit = \$xorValue & ~(\$xorValue - 1);
// Find resultant x and y non repeated element
for (\$i = 0; \$i < \$n; ++\$i)
{
if ((\$num[\$i] & \$rightActiveBit) != 0)
{
// When num[i] contain rightmost set bit
\$x = \$x ^ \$num[\$i];
}
else
{
// otherwise
\$y = \$y ^ \$num[\$i];
}
}
// Display array elements
\$this->displayArray(\$num, \$n);
// Display calculated result
echo("\n Result : x = ".\$x.
", y = ".\$y.
" \n");
}
}

function main()
{
\$task = new Finding();
// Array which contains duplicate and two non repeated element
\$num = array(5, 7, 9, 3, 1, 7, 5, 4, 1, 3);
// Get number of elements in number
\$n = count(\$num);
// Test
}
main();

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4
package main
import "fmt"
// Go Program
// Find two non repeating elements of repeated array elements

func displayArray(num[] int, n int) {
// Display array elements
for i := 0 ; i < n ; i++ {
fmt.Print(" ", num[i])
}
}
func twoNonRepeatingNo(num[] int, n int) {
// Resultant x and y
var x int = 0
var y int = 0
// get first array element
var xorValue int = num[0]
// Calculate xor of all array elements
for i := 1 ; i < n ; i++ {
xorValue = xorValue ^ num[i]
}
// Find the value of rightmost set (active) bit
var rightActiveBit int = xorValue & ^(xorValue - 1)
// Find resultant x and y non repeated element
for i := 0 ; i < n ; i++ {
if (num[i] & rightActiveBit) != 0 {
// When num[i] contain rightmost set bit
x = x ^ num[i]
} else {
// otherwise
y = y ^ num[i]
}
}
// Display array elements
displayArray(num, n)
// Display calculated result
fmt.Print("\n Result : x = ", x, ", y = ", y, " \n")
}
func main() {

// Array which contains duplicate and two non repeated element
var num = [] int {5, 7, 9, 3, 1, 7, 5, 4, 1, 3}
// Get number of elements in number
var n int = len(num)
// Test
twoNonRepeatingNo(num, n)
}

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4
// Node JS Program
// Find two non repeating elements of repeated array elements
class Finding
{
displayArray(num, n)
{
// Display array elements
for (var i = 0; i < n; ++i)
{
process.stdout.write(" " + num[i]);
}
}
twoNonRepeatingNo(num, n)
{
// Resultant x and y
var x = 0;
var y = 0;
// get first array element
var xorValue = num[0];
// Calculate xor of all array elements
for (var i = 1; i < n; ++i)
{
xorValue = xorValue ^ num[i];
}
// Find the value of rightmost set (active) bit
var rightActiveBit = xorValue & ~(xorValue - 1);
// Find resultant x and y non repeated element
for (var i = 0; i < n; ++i)
{
if ((num[i] & rightActiveBit) != 0)
{
// When num[i] contain rightmost set bit
x = x ^ num[i];
}
else
{
// otherwise
y = y ^ num[i];
}
}
// Display array elements
this.displayArray(num, n);
// Display calculated result
process.stdout.write("\n Result : x = " + x + ", y = " + y + " \n");
}
}

function main()
{
var task = new Finding();
// Array which contains duplicate and two non repeated element
var num = [5, 7, 9, 3, 1, 7, 5, 4, 1, 3];
// Get number of elements in number
var n = num.length;
// Test
}
main();

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4
#  Python 3 Program
#  Find two non repeating elements of repeated array elements
class Finding :
def displayArray(self, num, n) :
i = 0
#  Display list elements
while (i < n) :
print(" ", num[i], end = "")
i += 1

def twoNonRepeatingNo(self, num, n) :
#  Resultant x and y
x = 0
y = 0
#  get first list element
xorValue = num[0]
i = 1
#  Calculate xor of all list elements
while (i < n) :
xorValue = xorValue ^ num[i]
i += 1

#  Find the value of rightmost set (active) bit
rightActiveBit = xorValue & ~(xorValue - 1)
i = 0
#  Find resultant x and y non repeated element
while (i < n) :
if ((num[i] & rightActiveBit) != 0) :
#  When num[i] contain rightmost set bit
x = x ^ num[i]
else :
#  otherwise
y = y ^ num[i]

i += 1

#  Display list elements
self.displayArray(num, n)
#  Display calculated result
print("\n Result : x = ", x ,", y = ", y ," ")

def main() :
#  Array which contains duplicate and two non repeated element
num = [5, 7, 9, 3, 1, 7, 5, 4, 1, 3]
#  Get number of elements in number
n = len(num)
#  Test

if __name__ == "__main__": main()

#### Output

5  7  9  3  1  7  5  4  1  3
Result : x =  9 , y =  4
#  Ruby Program
#  Find two non repeating elements of repeated array elements
class Finding
def displayArray(num, n)
i = 0
#  Display array elements
while (i < n)
print(" ", num[i])
i += 1
end

end

def twoNonRepeatingNo(num, n)
#  Resultant x and y
x = 0
y = 0
#  get first array element
xorValue = num[0]
i = 1
#  Calculate xor of all array elements
while (i < n)
xorValue = xorValue ^ num[i]
i += 1
end

#  Find the value of rightmost set (active) bit
rightActiveBit = xorValue & ~(xorValue - 1)
i = 0
#  Find resultant x and y non repeated element
while (i < n)
if ((num[i] & rightActiveBit) != 0)
#  When num[i] contain rightmost set bit
x = x ^ num[i]
else

#  otherwise
y = y ^ num[i]
end

i += 1
end

#  Display array elements
self.displayArray(num, n)
#  Display calculated result
print("\n Result : x = ", x ,", y = ", y ," \n")
end

end

def main()
#  Array which contains duplicate and two non repeated element
num = [5, 7, 9, 3, 1, 7, 5, 4, 1, 3]
#  Get number of elements in number
n = num.length
#  Test
end

main()

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4
// Scala Program
// Find two non repeating elements of repeated array elements
class Finding()
{
def displayArray(num: Array[Int], n: Int): Unit = {
var i: Int = 0;
// Display array elements
while (i < n)
{
print(" " + num(i));
i += 1;
}
}
def twoNonRepeatingNo(num: Array[Int], n: Int): Unit = {
// Resultant x and y
var x: Int = 0;
var y: Int = 0;
// get first array element
var xorValue: Int = num(0);
var i: Int = 1;
// Calculate xor of all array elements
while (i < n)
{
xorValue = xorValue ^ num(i);
i += 1;
}
// Find the value of rightmost set (active) bit
var rightActiveBit: Int = xorValue & ~(xorValue - 1);
i = 0;
// Find resultant x and y non repeated element
while (i < n)
{
if ((num(i) & rightActiveBit) != 0)
{
// When num[i] contain rightmost set bit
x = x ^ num(i);
}
else
{
// otherwise
y = y ^ num(i);
}
i += 1;
}
// Display array elements
this.displayArray(num, n);
// Display calculated result
print("\n Result : x = " + x + ", y = " + y + " \n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: Finding = new Finding();
// Array which contains duplicate and two non repeated element
var num: Array[Int] = Array(5, 7, 9, 3, 1, 7, 5, 4, 1, 3);
// Get number of elements in number
var n: Int = num.length;
// Test
}
}

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4
// Swift 4 Program
// Find two non repeating elements of repeated array elements
class Finding
{
func displayArray(_ num: [Int], _ n: Int)
{
var i: Int = 0;
// Display array elements
while (i < n)
{
print(" ", num[i], terminator: "");
i += 1;
}
}
func twoNonRepeatingNo(_ num: [Int], _ n: Int)
{
// Resultant x and y
var x: Int = 0;
var y: Int = 0;
// get first array element
var xorValue: Int = num[0];
var i: Int = 1;
// Calculate xor of all array elements
while (i < n)
{
xorValue = xorValue ^ num[i];
i += 1;
}
// Find the value of rightmost set (active) bit
let rightActiveBit: Int = xorValue & ~(xorValue - 1);
i = 0;
// Find resultant x and y non repeated element
while (i < n)
{
if ((num[i] & rightActiveBit)  != 0)
{
// When num[i] contain rightmost set bit
x = x ^ num[i];
}
else
{
// otherwise
y = y ^ num[i];
}
i += 1;
}
// Display array elements
self.displayArray(num, n);
// Display calculated result
print("\n Result : x = ", x ,", y = ", y ," ");
}
}
func main()
{
let task: Finding = Finding();
// Array which contains duplicate and two non repeated element
let num: [Int] = [5, 7, 9, 3, 1, 7, 5, 4, 1, 3];
// Get number of elements in number
let n: Int = num.count;
// Test
}
main();

#### Output

5  7  9  3  1  7  5  4  1  3
Result : x =  9 , y =  4
// Kotlin Program
// Find two non repeating elements of repeated array elements
class Finding
{
fun displayArray(num: Array < Int > , n: Int): Unit
{
var i: Int = 0;
// Display array elements
while (i < n)
{
print(" " + num[i]);
i += 1;
}
}
fun twoNonRepeatingNo(num: Array < Int > , n: Int): Unit
{
// Resultant x and y
var x: Int = 0;
var y: Int = 0;
// get first array element
var xorValue: Int = num[0];
var i: Int = 1;
// Calculate xor of all array elements
while (i < n)
{
xorValue = xorValue xor num[i];
i += 1;
}
// Find the value of rightmost set (active) bit
val rightActiveBit: Int = xorValue and(xorValue - 1).inv();
i = 0;
// Find resultant x and y non repeated element
while (i < n)
{
if ((num[i] and rightActiveBit) != 0)
{
// When num[i] contain rightmost set bit
x = x xor num[i];
}
else
{
// otherwise
y = y xor num[i];
}
i += 1;
}
// Display array elements
this.displayArray(num, n);
// Display calculated result
print("\n Result : x = " + x + ", y = " + y + " \n");
}
}
fun main(args: Array < String > ): Unit
{
val task: Finding = Finding();
// Array which contains duplicate and two non repeated element
val num: Array < Int > = arrayOf(5, 7, 9, 3, 1, 7, 5, 4, 1, 3);
// Get number of elements in number
val n: Int = num.count();
// Test
}

#### Output

5 7 9 3 1 7 5 4 1 3
Result : x = 9, y = 4

## Time Complexity

The time complexity of this algorithm is O(n) because we iterate through the array twice: once to calculate the XOR of all elements and then to separate the unique elements based on the rightmost set bit. Both iterations take linear time, and thus the overall complexity is O(n).

## Resultant Output Explanation

In the given example array [5, 7, 9, 3, 1, 7, 5, 4, 1, 3], the xorValue of all elements is 9. The rightmost set bit in 9 is at position 0 (from the right), which corresponds to the binary 1. Therefore, the rightActiveBit is 1.

After separating elements based on the rightActiveBit, we find x = 9 and y = 4, which are the two non-repeating elements in the array. The other elements, which are repeated twice, have canceled out during the XOR operations.

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