Two Non Repeating Elements in Array Using XOR in DSA Java - Build from Scratch

Two Non Repeating Elements in Array Using XOR

📖 Scenario: Imagine you have a list of numbers where every number appears twice except for two unique numbers that appear only once. You want to find those two unique numbers efficiently.

🎯 Goal: Build a Java program that finds the two non-repeating elements in an array using the XOR operation.

📋 What You'll Learn

Create an integer array called arr with the exact values: 2, 3, 7, 9, 2, 3

Create an integer variable called xorAll and initialize it to 0

Use a for loop with variable num to XOR all elements in arr and store the result in xorAll

Create two integer variables num1 and num2 initialized to 0

Find the rightmost set bit in xorAll and store it in setBit

Use a for loop with variable num to separate elements in arr into two groups based on setBit and XOR them into num1 and num2

Print the two unique numbers num1 and num2 separated by a space

💡 Why This Matters

🌍 Real World

Finding unique error codes or IDs in logs where most entries are duplicated.

💼 Career

Useful for software engineers working on debugging tools, data analysis, and optimization problems.

1

Create the array with given values

Create an integer array called arr with these exact values: 2, 3, 7, 9, 2, 3

DSA Java

# Create the array called arr with values 2, 3, 7, 9, 2, 3
// Your code here

Hint

Use int[] arr = {2, 3, 7, 9, 2, 3}; to create the array.

2

Initialize XOR accumulator variable

Create an integer variable called xorAll and set it to 0

DSA Java

int[] arr = {2, 3, 7, 9, 2, 3};
// Create an integer variable xorAll and set it to 0
// Your code here

Hint

Use int xorAll = 0; to start the XOR accumulator.

3

XOR all elements to find combined XOR

Use a for loop with variable num to XOR all elements in arr and store the result in xorAll

DSA Java

int[] arr = {2, 3, 7, 9, 2, 3};
int xorAll = 0;
// Use a for loop with variable num to XOR all elements in arr into xorAll
// Your code here

Hint

Use for (int num : arr) { xorAll ^= num; } to XOR all elements.

4

Find the two unique numbers and print them

Create two integer variables num1 and num2 initialized to 0. Find the rightmost set bit in xorAll and store it in setBit. Use a for loop with variable num to separate elements in arr into two groups based on setBit and XOR them into num1 and num2. Finally, print num1 and num2 separated by a space.

DSA Java

int[] arr = {2, 3, 7, 9, 2, 3};
int xorAll = 0;
for (int num : arr) {
    xorAll ^= num;
}
// Create num1 and num2 as 0
// Find rightmost set bit in xorAll and store in setBit
// Use for loop with num to separate arr elements by setBit and XOR into num1 and num2
// Print num1 and num2 separated by space
// Your code here

Hint

Use int setBit = xorAll & (-xorAll); to find the rightmost set bit. Then separate elements by checking (num & setBit) != 0. XOR into num1 and num2. Finally, print them.

Practice

1. What is the main idea behind using XOR to find two non-repeating elements in an array where all other elements repeat twice?

easy

A. XOR removes duplicates by summing elements

B. XOR counts the frequency of each element

C. XOR sorts the array to separate unique elements

D. XOR of all elements gives XOR of the two unique elements

Solution

1. Step 1: XOR all elements in the array

   When we XOR all elements, pairs cancel out because x ^ x = 0, leaving XOR of the two unique elements.

2. Step 2: Use the XOR result to separate unique elements

   The XOR result has set bits where the two unique elements differ, helping to divide elements into two groups.

3. Final Answer:

   XOR of all elements gives XOR of the two unique elements -> Option D

4. Quick Check:

   XOR all = unique1 ^ unique2 [OK]

Hint: XOR all elements to isolate combined unique XOR [OK]

Common Mistakes:

• Thinking XOR counts frequency
• Assuming XOR sorts elements
• Using sum instead of XOR

2. Which of the following Java code snippets correctly initializes the XOR variable to find two unique elements in an integer array arr?

easy

A. int xor = 1; for(int num : arr) xor += num;

B. int xor = 0; for(int num : arr) xor ^= num;

C. int xor = arr[0]; for(int i = 0; i < arr.length; i++) xor ^= arr[i];

D. int xor = 0; for(int i = 0; i < arr.length; i++) xor = xor + arr[i];

Solution

1. Step 1: Initialize XOR to zero and XOR all elements

   Starting with zero ensures XOR accumulates correctly. Using a for-each loop with ^= operator applies XOR to each element.

2. Step 2: Check correctness of other options

   int xor = 1; for(int num : arr) xor += num; uses addition, not XOR. int xor = arr[0]; for(int i = 0; i < arr.length; i++) xor ^= arr[i]; incorrectly loops from i=0 after initializing with arr[0], XORing arr[0] twice which cancels it out. int xor = 0; for(int i = 0; i < arr.length; i++) xor = xor + arr[i]; uses addition, not XOR.

3. Final Answer:

   int xor = 0; for(int num : arr) xor ^= num; -> Option B

4. Quick Check:

   Initialize xor=0 and XOR all elements [OK]

Hint: Start XOR with 0 and XOR all elements [OK]

Common Mistakes:

• Using addition instead of XOR
• Starting XOR with non-zero value
• Using wrong loop syntax

3. What will be the output of the following Java code snippet?

int[] arr = {2, 3, 7, 9, 2, 3};
int xor = 0;
for (int num : arr) xor ^= num;
int setBit = xor & (-xor);
int x = 0, y = 0;
for (int num : arr) {
  if ((num & setBit) != 0) x ^= num;
  else y ^= num;
}
System.out.println(x + ", " + y);

medium

A. 7, 9

B. 2, 3

C. 0, 0

D. 9, 7

Solution

1. Step 1: XOR all elements to get xor = 7 ^ 9 = 14

   Pairs (2,2) and (3,3) cancel out, leaving xor = 7 ^ 9 = 14.

2. Step 2: Find rightmost set bit and separate elements

   setBit = 14 & (-14) = 2. Elements with bit 2 set: 2, 2, 3, 3, 7, 9 split into groups. XOR group 1: 7, XOR group 2: 9.

3. Final Answer:

   7, 9 -> Option A

4. Quick Check:

   Unique elements are 7 and 9 [OK]

Hint: Use rightmost set bit to split and XOR groups [OK]

Common Mistakes:

• Mixing order of output
• Not using rightmost set bit correctly
• Confusing pairs with unique elements

4. Identify the error in the following Java code snippet that attempts to find two non-repeating elements using XOR:

int[] arr = {4, 1, 2, 1, 2, 5};
int xor = 0;
for (int i = 0; i <= arr.length; i++) {
  xor ^= arr[i];
}
int setBit = xor & (-xor);
int x = 0, y = 0;
for (int num : arr) {
  if ((num & setBit) != 0) x ^= num;
  else y ^= num;
}
System.out.println(x + ", " + y);

medium

A. Loop condition should be i < arr.length, not i <= arr.length

B. XOR initialization should be arr[0], not 0

C. setBit calculation is incorrect

D. Second loop should use index instead of for-each

Solution

1. Step 1: Check loop boundary in XOR calculation

   The for loop uses i <= arr.length, which causes ArrayIndexOutOfBoundsException because valid indices are 0 to arr.length-1.

2. Step 2: Correct loop condition to i < arr.length

   Changing to i < arr.length fixes the error and allows proper XOR of all elements.

3. Final Answer:

   Loop condition should be i < arr.length, not i <= arr.length -> Option A

4. Quick Check:

   Array index must be within bounds [OK]

Hint: Use i < length, not i <= length in loops [OK]

Common Mistakes:

• Using wrong loop boundary causing runtime error
• Incorrect XOR initialization
• Misunderstanding setBit calculation

5. Given an array arr = {10, 20, 10, 30, 40, 20, 50, 40}, which Java code snippet correctly finds the two non-repeating elements using XOR and prints them in ascending order?

hard

A. int xor = 0;
for(int num : arr) xor ^= num;
int setBit = xor & (-xor);
int x = 0, y = 0;
for(int num : arr) {
if((num & setBit) != 0) x ^= num;
else y ^= num;
}
System.out.println(x + y);

B. int xor = arr[0];
for(int i=1; i

C. int xor = 0;
for(int num : arr) xor ^= num;
int setBit = xor & (-xor);
int x = 0, y = 0;
for(int num : arr) {
if((num & setBit) != 0) x ^= num;
else y ^= num;
}
System.out.println(Math.min(x,y) + ", " + Math.max(x,y));

D. int xor = 0;
for(int num : arr) xor ^= num;
int setBit = xor & (-xor);
int x = 0, y = 0;
for(int num : arr) {
if((num & setBit) == 0) x ^= num;
else y ^= num;
}
System.out.println(x + ", " + y);

Solution

1. Step 1: XOR all elements and find rightmost set bit

   XOR of all elements gives xor = 30 ^ 50 = 44. setBit = 44 & (-44) = 4.

2. Step 2: Separate elements into two groups and XOR each group

   Elements with bit 4 set XOR to 50, others XOR to 30.

3. Step 3: Print the two unique elements in ascending order

   Using Math.min and Math.max prints "30, 50" as required.

4. Final Answer:

   Prints 30, 50 in ascending order -> Option C

5. Quick Check:

   Use XOR groups and sort output [OK]

Hint: Use XOR groups and Math.min/Math.max to order output [OK]

Common Mistakes:

• Using bitwise OR instead of AND for setBit
• Swapping group conditions incorrectly
• Printing sum instead of elements