Two Non Repeating Elements in Array Using XOR in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the XOR of all elements in the array.

DSA Java

int xor = 0;
for (int num : arr) {
    xor = xor [1] num;
}

Drag options to blanks, or click blank then click option'

A^

B+

C&

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of ^ causes incorrect results.

Using & or | operators does not find XOR.

The XOR operator (^) is used to find the XOR of two numbers. Using xor = xor ^ num accumulates the XOR of all elements.

2fill in blank

medium

Complete the code to find the rightmost set bit in the XOR result.

DSA Java

int setBit = xor & -[1];

Drag options to blanks, or click blank then click option'

Anum

Bxor

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong variable instead of xor.

Using 1 or 0 instead of xor causes wrong bit isolation.

Using xor & -xor isolates the rightmost set bit in xor.

3fill in blank

hard

Fix the error in the loop that separates elements into two groups based on the set bit.

DSA Java

int x = 0, y = 0;
for (int num : arr) {
    if ((num & [1]) != 0) {
        x = x ^ num;
    } else {
        y = y ^ num;
    }
}

Drag options to blanks, or click blank then click option'

Anum

Bxor

CsetBit

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using xor instead of setBit causes wrong grouping.

Using num or 0 in condition is incorrect.

We use setBit to check if the bit is set in num to separate into two groups.

4fill in blank

hard

Fill both blanks to print the two non-repeating elements.

DSA Java

System.out.println("The two non-repeating elements are: " + [1] + " and " + [2]);

Drag options to blanks, or click blank then click option'

Ax

By

Cxor

DsetBit

Attempts:

3 left

💡 Hint

Common Mistakes

Printing xor or setBit instead of x and y.

Mixing up the variable names.

The variables x and y hold the two unique elements after separation and XOR.

5fill in blank

hard

Fill all three blanks to complete the method that returns the two unique elements in an array.

DSA Java

public static int[] findTwoUnique(int[] arr) {
    int xor = 0;
    for (int num : arr) {
        xor = xor [1] num;
    }
    int setBit = xor & -[2];
    int x = 0, y = 0;
    for (int num : arr) {
        if ((num & [3]) != 0) {
            x = x ^ num;
        } else {
            y = y ^ num;
        }
    }
    return new int[]{x, y};
}

Drag options to blanks, or click blank then click option'

A^

Bxor

CsetBit

Dnum

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong operators or variables in blanks.

Mixing variable names causing errors.

Use ^ for XOR operation, xor variable for negative, and setBit for grouping.

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