Two Non Repeating Elements in Array Using XOR in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using XOR to find two non-repeating elements in an array?

XOR of all elements gives XOR of the two unique elements because pairs cancel out. Then, use a set bit in this XOR to separate elements into two groups to find the unique elements.

beginner

Why does XOR of a number with itself result in zero?

Because XOR compares bits and returns 1 only if bits differ. Same bits XORed give 0, so number XOR itself is 0.

intermediate

How do you find a set bit to separate two unique elements after XORing all elements?

Find the rightmost set bit (bit with value 1) in the XOR result using 'xor & (-xor)'. This bit differs between the two unique numbers.

beginner

What is the time complexity of finding two non-repeating elements using XOR?

It is O(n) because we scan the array a constant number of times without extra space.

intermediate

Explain why dividing elements into two groups based on a set bit helps find the unique elements.

Because the two unique elements differ at that bit, grouping elements by that bit separates them into different groups. XORing each group cancels duplicates, leaving unique elements.

What does XOR of all elements in an array with exactly two unique elements and others repeating twice give?

A Sum of all elements

B Zero

C XOR of the two unique elements

D Product of the two unique elements

Pairs cancel out in XOR, leaving XOR of the two unique elements.

How do you find the rightmost set bit in a number 'xor'?

A ~xor

B xor & (-xor)

C xor ^ (-xor)

D xor | (-xor)

The expression 'xor & (-xor)' isolates the rightmost set bit.

Why do we divide elements into two groups based on a set bit in XOR result?

A To count duplicates

B To sort the array

C To find the maximum element

D To separate the two unique elements

The two unique elements differ at that bit, so grouping separates them.

What is the space complexity of the XOR method to find two unique elements?

A O(1)

B O(n)

C O(log n)

D O(n^2)

Only a few variables are used, no extra data structures.

If an array has elements [2, 3, 7, 9, 2, 3], what are the two unique elements?

A 7 and 9

B 2 and 3

C 2 and 7

D 3 and 9

2 and 3 repeat twice; 7 and 9 appear once each.

Describe the step-by-step process to find two non-repeating elements in an array using XOR.

Think about how XOR cancels pairs and how bit differences help separate unique numbers.

Explain why XOR operation is useful for finding unique elements in an array where others repeat twice.

Recall how XOR behaves with pairs and unique values.

Practice

1. What is the main idea behind using XOR to find two non-repeating elements in an array where every other element repeats?

easy

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

B. XOR sorts the array to separate unique elements

C. XOR counts the frequency of each element

D. XOR removes duplicates by summing elements

Solution

1. Step 1: Understand XOR properties

   XOR of a number with itself is 0, and XOR with 0 is the number itself.

2. Step 2: Apply XOR to all elements

   XORing all elements cancels out duplicates, leaving XOR of two unique elements.

3. Final Answer:

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

4. Quick Check:

   XOR cancels duplicates, leaving unique pair XOR [OK]

Hint: XOR all elements to isolate unique pair XOR quickly [OK]

Common Mistakes:

• Thinking XOR sorts or counts elements
• Assuming XOR gives unique elements directly
• Confusing XOR with addition or subtraction

2. Which of the following Go code snippets correctly initializes the XOR variable to start finding two non-repeating elements?

easy

A. xor := nil

B. xor := 0

C. xor := 1

D. xor := []int{}

Solution

1. Step 1: Identify correct XOR initialization

   XOR operation starts with 0 because XOR with 0 returns the number itself.

2. Step 2: Check Go syntax for variable initialization

   "xor := 0" correctly initializes an integer variable for XOR operations.

3. Final Answer:

   xor := 0 -> Option B

4. Quick Check:

   Start XOR with 0 for correct accumulation [OK]

Hint: Initialize XOR with 0 to start XOR operations correctly [OK]

Common Mistakes:

• Using 1 instead of 0 for XOR start
• Trying to initialize XOR with nil or slice
• Incorrect Go syntax for variable declaration

3. What is the output of the following Go code snippet? arr := []int{2, 3, 7, 9, 2, 3, 9, 11} xor := 0 for _, v := range arr { xor ^= v } fmt.Println(xor)

medium

A. 12

B. 18

C. 0

D. 11

Solution

1. Step 1: XOR all elements in the array

   Calculate XOR stepwise: 0^2=2, 2^3=1, 1^7=6, 6^9=15, 15^2=13, 13^3=14, 14^9=7, 7^11=12.

2. Step 2: Verify final XOR value

   The final XOR value after processing all elements is 12.

3. Final Answer:

   12 -> Option A

4. Quick Check:

   XOR of all elements = 12 [OK]

Hint: XOR duplicates cancel; XOR unique pair remains [OK]

Common Mistakes:

• Miscomputing XOR stepwise
• Assuming XOR sums elements
• Ignoring duplicates canceling out

4. Identify the bug in this Go code snippet that tries to find two non-repeating elements using XOR: arr := []int{4, 1, 2, 1, 2, 5} xor := 0 for i := 0; i <= len(arr); i++ { xor ^= arr[i] } fmt.Println(xor)

medium

A. Use a map instead of XOR for this problem

B. XOR variable should be initialized to 1

C. Loop condition should be i < len(arr), not i <= len(arr)

D. Array elements should be converted to strings before XOR

Solution

1. Step 1: Check loop boundary

   Loop uses i <= len(arr), which causes out-of-range error because valid indices are 0 to len(arr)-1.

2. Step 2: Correct loop condition

   Change to i < len(arr) to avoid index out of range and correctly XOR all elements.

3. Final Answer:

   Loop condition should be i < len(arr), not i <= len(arr) -> Option C

4. Quick Check:

   Index out of range fixed by correct loop boundary [OK]

Hint: Use i < len(arr) to avoid index errors in loops [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Initializing XOR with wrong value
• Misunderstanding XOR usage for strings

5. Given the array arr := []int{6, 3, 5, 4, 5, 3, 4, 7}, which two numbers are the non-repeating elements found using the XOR method?

hard

A. [5, 6]

B. [3, 5]

C. [4, 7]

D. [6, 7]

Solution

1. Step 1: XOR all elements to get xorVal

   Calculate XOR: 6^3=5, 5^5=0, 0^4=4, 4^5=1, 1^3=2, 2^4=6, 6^7=1. Final xorVal = 1.

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

   Rightmost set bit in xorVal=1. Partition array into two groups based on this bit and XOR separately to find unique elements 6 and 7.

3. Final Answer:

   [6, 7] -> Option D

4. Quick Check:

   Unique elements are 6 and 7 after XOR partition [OK]

Hint: Use rightmost set bit to split and find unique elements [OK]

Common Mistakes:

• Not partitioning array by set bit
• Mixing up duplicates with unique elements
• Incorrect XOR calculations