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

Time Complexity: Find the Only Non Repeating Element Using XOR

O(n)

Understanding Time Complexity

We want to know how the time needed to find the unique element changes as the list grows.

How does the number of steps grow when the input list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function findUnique(arr) {
  let result = 0;
  for (let num of arr) {
    result ^= num;
  }
  return result;
}
    

This code finds the only number that does not repeat in the array by using XOR on all elements.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: One loop that goes through each number once.
• How many times: Exactly once for each element in the array.

How Execution Grows With Input

As the list gets bigger, the number of steps 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 steps increase in a straight line as input size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the unique element grows directly with the number of elements in the list.

Common Mistake

[X] Wrong: "Using XOR means the time is constant no matter the list size."

[OK] Correct: Even though XOR is fast, we still must apply it to every element, so time grows with list size.

Interview Connect

Understanding this helps you quickly explain how simple operations on all items lead to efficient solutions.

Self-Check

"What if the array was sorted? Would the time complexity change if we used a different method?"

Practice

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

easy

A. It returns 0

B. It returns the number itself

C. It returns 1

D. It returns the sum of the numbers

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 XOR to example

   For example, 5 ^ 5 = 0 as each bit is the same and cancels.

3. Final Answer:

   It returns 0 -> Option A

4. Quick Check:

   X ^ X = 0 [OK]

Hint: XOR 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 is the correct syntax to XOR two variables a and b in JavaScript?

easy

A. a && b

B. a ^ b

C. a xor b

D. a | b

Solution

1. Step 1: Recall JavaScript bitwise XOR operator

   In JavaScript, the XOR operator is represented by the caret symbol ^.

2. Step 2: Check each option

   a ^ b uses ^ correctly. a xor b uses 'xor' which is not valid syntax. Options A and D are logical AND and bitwise OR respectively.

3. Final Answer:

   a ^ b -> Option B

4. Quick Check:

   Use ^ for XOR in JavaScript [OK]

Hint: Use ^ symbol for XOR in JavaScript [OK]

Common Mistakes:

• Using 'xor' keyword
• Confusing XOR with && or ||
• Using | instead of ^

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

const arr = [2, 3, 2, 4, 3];
let result = 0;
for (const num of arr) {
  result ^= num;
}
console.log(result);

medium

A. 2

B. 0

C. 4

D. 3

Solution

1. Step 1: XOR all elements in the array

   Start with result = 0, then XOR with each element: 0 ^ 2 = 2, 2 ^ 3 = 1, 1 ^ 2 = 3, 3 ^ 4 = 7, 7 ^ 3 = 4.

2. Step 2: Final XOR result is the non-repeating element

   The only element that does not repeat is 4, which is the final XOR result.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   Final XOR = 4 [OK]

Hint: XOR all numbers; result is unique element [OK]

Common Mistakes:

• Not starting XOR from 0
• XORing incorrectly in loop
• Assuming sum instead of XOR

4. Identify the error in the following code that tries to find the non-repeating element using XOR:

const arr = [1, 2, 1, 3, 2];
let result;
for (let i = 0; i < arr.length; i++) {
  result ^= arr[i];
}
console.log(result);

medium

A. Variable 'result' is not initialized to 0

B. Loop condition should be i <= arr.length

C. XOR operator is incorrect, should be &&

D. Array should be sorted before XOR

Solution

1. Step 1: Check initialization of 'result'

   The variable 'result' is declared but not initialized, so it is undefined initially.

2. Step 2: Effect of uninitialized 'result' on XOR

   XOR with undefined results in NaN, so the final output will be incorrect. Initializing 'result' to 0 fixes this.

3. Final Answer:

   Variable 'result' is not initialized to 0 -> Option A

4. Quick Check:

   Initialize result = 0 before XOR loop [OK]

Hint: Always initialize XOR accumulator to 0 [OK]

Common Mistakes:

• Not initializing XOR variable
• Using wrong loop condition
• Misusing XOR operator

5. Given an array where every element appears three times except one element which appears once, can XOR be used directly to find the unique element? Why or why not?

hard

A. Yes, but only if the array length is odd

B. Yes, XOR works for any number of repetitions

C. No, because XOR requires sorted arrays

D. No, because XOR only cancels pairs of identical numbers

Solution

1. Step 1: Understand XOR behavior with repeated elements

   XOR cancels out pairs of identical numbers because x ^ x = 0.

2. Step 2: Analyze effect of triply repeated elements

   Elements repeated thrice do not cancel out with XOR because XOR of three identical numbers is the number itself, not zero.

3. Step 3: Conclusion on XOR usage

   Therefore, XOR alone cannot find the unique element if others repeat thrice; other methods are needed.

4. Final Answer:

   No, because XOR only cancels pairs of identical numbers -> Option D

5. Quick Check:

   XOR cancels pairs, not triples [OK]

Hint: XOR cancels pairs only, not triples [OK]

Common Mistakes:

• Assuming XOR works for any repetition count
• Thinking sorting helps XOR
• Confusing XOR with sum or count