Find the Only Non Repeating Element Using XOR in DSA Go - Time & Space Complexity

Time Complexity: Find the Only Non Repeating Element Using XOR

O(n)

Understanding Time Complexity

We want to understand how fast we can find the single element that does not repeat in a list using XOR.

How does the time needed grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func findNonRepeating(arr []int) int {
    result := 0
    for _, num := range arr {
        result ^= num
    }
    return result
}
    

This code finds the only number that appears once while all others appear twice by using XOR.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: XOR operation inside a single loop over the array.
• How many times: Once for each element in the array (n times).

How Execution Grows With Input

Each element is processed once, so the total work grows directly with the number of elements.

Input Size (n)	Approx. Operations
10	10 XOR operations
100	100 XOR operations
1000	1000 XOR operations

Pattern observation: The operations increase linearly as the input size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the unique element grows in direct proportion to the number of elements.

Common Mistake

[X] Wrong: "Using XOR means the time complexity is constant because XOR is a simple operation."

[OK] Correct: Even though XOR itself is fast, it must be done for every element, so the total time depends on the number of elements.

Interview Connect

Understanding this linear time approach shows you can solve problems efficiently by using bitwise operations and simple loops.

Self-Check

"What if the array contains multiple elements that appear only once? How would the time complexity change?"

Practice

1. What does the XOR operation do when applied between two identical numbers?

easy

A. It returns the sum of the numbers

B. It returns 0

C. It returns 1

D. It returns the number itself

Solution

1. Step 1: Understand XOR property for identical numbers

   XOR of a number with itself always results in 0 because bits cancel out.

2. Step 2: Apply this property to the question

   Since the two identical numbers cancel out, the result is 0.

3. Final Answer:

   It returns 0 -> Option B

4. Quick Check:

   n XOR n = 0 [OK]

Hint: XOR of same numbers always zeroes out [OK]

Common Mistakes:

• Thinking XOR returns the number itself
• Confusing XOR with OR operation
• Assuming XOR sums numbers

2. Which of the following Go code snippets correctly initializes a variable to store XOR results for an integer slice named nums?

easy

A. var result int = 0

B. result := []int{}

C. var result string = ""

D. result := make(map[int]int)

Solution

1. Step 1: Identify the correct type for XOR accumulation

   XOR operates on integers, so the variable must be an integer initialized to 0.

2. Step 2: Check each option

   var result int = 0 initializes an integer variable to 0, suitable for XOR operations. Others use wrong types or data structures.

3. Final Answer:

   var result int = 0 -> Option A

4. Quick Check:

   Integer zero initialization for XOR [OK]

Hint: XOR needs an integer variable initialized to zero [OK]

Common Mistakes:

• Using slice or map instead of int
• Initializing with empty string
• Not initializing variable

3. What is the output of the following Go code?

package main
import "fmt"
func main() {
 nums := []int{2, 3, 2, 4, 3}
 result := 0
 for _, num := range nums {
  result ^= num
 }
 fmt.Println(result)
}

medium

A. 2

B. 3

C. 4

D. 0

Solution

1. Step 1: XOR all elements in the slice

   Calculate XOR stepwise: 0^2=2, 2^3=1, 1^2=3, 3^4=7, 7^3=4.

2. Step 2: Identify the non-repeating element

   The final XOR result is 4, which is the only number appearing once.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   XOR cancels duplicates, leaves 4 [OK]

Hint: XOR duplicates cancel, leftover is unique [OK]

Common Mistakes:

• Adding numbers instead of XOR
• Miscomputing XOR steps
• Ignoring duplicates

4. Identify the error in this Go code snippet that tries to find the only non-repeating element using XOR:

package main
import "fmt"
func main() {
 nums := []int{5, 1, 5, 2, 1}
 var result int
 for i := 0; i <= len(nums); i++ {
  result ^= nums[i]
 }
 fmt.Println(result)
}

medium

A. Loop condition should be i < len(nums), not i <= len(nums)

B. Variable 'result' should be initialized to 1 instead of 0

C. The XOR operator '^=' is invalid in Go

D. The slice 'nums' should be declared as an array

Solution

1. Step 1: Check loop boundary condition

   The loop runs from i=0 to i<=len(nums), which causes out-of-range access at i=len(nums).

2. Step 2: Correct the loop condition

   Change condition to i < len(nums) to avoid index out of range error.

3. Final Answer:

   Loop condition should be i < len(nums), not i <= len(nums) -> Option A

4. Quick Check:

   Index out of range fixed by loop condition [OK]

Hint: Loop index must be less than length, not equal [OK]

Common Mistakes:

• Using <= instead of <
• Initializing result incorrectly
• Misunderstanding XOR operator

5. Given an array where every element appears twice except one element which appears once, how can you find the unique element using XOR in Go? Consider the array nums := []int{7, 9, 7, 8, 9, 10, 10}. What is the unique element?

hard

A. 7

B. 9

C. 10

D. 8

Solution

1. Step 1: XOR all elements in the array

   Calculate XOR stepwise: 0^7=7, 7^9=14, 14^7=9, 9^8=1, 1^9=8, 8^10=2, 2^10=8.

2. Step 2: Identify the unique element

   The final XOR result is 8, which appears only once in the array.

3. Final Answer:

   8 -> Option D

4. Quick Check:

   Duplicates cancel, unique is 8 [OK]

Hint: XOR all elements; duplicates vanish, unique remains [OK]

Common Mistakes:

• Adding numbers instead of XOR
• Forgetting to initialize result to zero
• Misreading array elements