Why Two Non Repeating Elements in Array Using XOR in DSA Python?

The Big Idea

Discover how a simple math trick can save you hours of tedious searching!

The Scenario

Imagine you have a big box of socks where every sock has a matching pair except for two unique socks. You want to find those two unique socks quickly without checking every single sock against all others.

The Problem

Checking each sock one by one against all others is slow and tiring. You might miss pairs or check the same socks many times. This manual way wastes time and can cause mistakes.

The Solution

Using XOR, a special math trick, you can quickly find the two unique socks by combining all socks in a way that pairs cancel out, leaving only the unique ones. This method is fast and simple.

Before vs After

✗ Before

for i in range(len(arr)):
    count = 0
    for j in range(len(arr)):
        if arr[i] == arr[j]:
            count += 1
    if count == 1:
        print(arr[i])

✓ After

xor_all = 0
for number in arr:
    xor_all ^= number
# Further steps to separate two unique numbers

What It Enables

This method lets you find two unique items in a large group quickly and without extra memory.

Real Life Example

Finding two unique defective products in a batch where all others are identical pairs, without checking each product twice.

Key Takeaways

Manual pair checking is slow and error-prone.

XOR helps cancel out pairs and isolate unique elements.

Efficient for large data with two unique items.

Practice

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

easy

A. To sort the array in ascending order

B. To count the frequency of each element

C. To cancel out all repeating elements and isolate the XOR of the two unique elements

D. To reverse the array elements

Solution

1. Step 1: Understand XOR properties

   XOR of a number with itself is 0, and XOR with 0 is the number itself. This helps cancel out repeating elements.

2. Step 2: Apply XOR on all elements

   When XOR is applied on the entire array, all repeating elements cancel out, leaving XOR of the two unique elements.

3. Final Answer:

   To cancel out all repeating elements and isolate the XOR of the two unique elements -> Option C

4. Quick Check:

   XOR cancels duplicates = isolate unique elements [OK]

Hint: XOR duplicates become zero, leaving unique elements XOR [OK]

Common Mistakes:

• Thinking XOR sorts the array
• Using XOR to count frequencies
• Assuming XOR reverses elements

2. Which of the following Python code snippets correctly initializes the XOR of all elements in a list named arr to zero before processing?

easy

A. xor = 1

B. xor = 0

C. xor = arr[0]

D. xor = None

Solution

1. Step 1: Initialize XOR variable

   We start XOR with 0 because XOR with 0 returns the other number unchanged.

2. Step 2: Confirm initialization correctness

   Starting with 0 ensures the first XOR operation correctly sets the value to the first element.

3. Final Answer:

   xor = 0 -> Option B

4. Quick Check:

   Initialize XOR with 0 for correct accumulation [OK]

Hint: Start XOR with 0 to accumulate correctly [OK]

Common Mistakes:

• Initializing XOR with 1 causes wrong results
• Using first element without XOR 0 initialization
• Setting XOR to None causes errors

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

arr = [2, 3, 7, 9, 2, 3]
xor = 0
for num in arr:
    xor ^= num
print(xor)

medium

A. 0

B. 9

C. 7

D. 14

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.

2. Step 2: Final XOR value

   The final XOR after processing all elements is 14.

3. Final Answer:

   14 -> Option D

4. Quick Check:

   Stepwise XOR = 14 [OK]

Hint: XOR all elements stepwise to find combined XOR [OK]

Common Mistakes:

• Forgetting XOR is cumulative and order matters
• Mistaking XOR with addition
• Stopping XOR early

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

arr = [4, 1, 2, 1, 2, 5]
xor = 0
for num in arr:
    xor ^= num
set_bit = xor & ~(xor - 1)
num1 = 0
num2 = 0
for num in arr:
    if (num & set_bit) > 0:
        num1 ^= num
    else:
        num2 ^= num
print(num1, num2)

medium

A. No error, code works correctly

B. XOR initialization should be arr[0] instead of 0

C. Incorrect calculation of set_bit using bitwise NOT

D. Using > 0 instead of != 0 in if condition

Solution

1. Step 1: Check set_bit calculation

   set_bit = xor & ~(xor - 1) correctly isolates rightmost set bit.

2. Step 2: Verify partition and XOR groups

   Elements are split by set_bit and XORed separately to find unique elements.

3. Step 3: Confirm output correctness

   Code correctly prints the two unique numbers without errors.

4. Final Answer:

   No error, code works correctly -> Option A

5. Quick Check:

   Correct set_bit and partitioning = code works [OK]

Hint: Use xor & ~(xor-1) to find rightmost set bit correctly [OK]

Common Mistakes:

• Misusing bitwise NOT causing wrong set_bit
• Initializing XOR incorrectly
• Using wrong condition in if statement

5. Given an array arr = [10, 20, 10, 30, 40, 20], which of the following Python code snippets correctly finds the two non-repeating elements using XOR and prints them in ascending order?

hard

A. xor = 0
for num in arr:
xor ^= num
set_bit = xor & (-xor)
num1 = 0
num2 = 0
for num in arr:
if num & set_bit:
num1 ^= num
else:
num2 ^= num
print(sorted([num1, num2]))

B. xor = 0
for num in arr:
xor ^= num
set_bit = xor | (xor - 1)
num1 = 0
num2 = 0
for num in arr:
if num & set_bit:
num1 ^= num
else:
num2 ^= num
print(sorted([num1, num2]))

C. xor = arr[0]
for num in arr[1:]:
xor ^= num
set_bit = xor & ~(xor - 1)
num1 = 0
num2 = 0
for num in arr:
if (num & set_bit) == 0:
num1 ^= num
else:
num2 ^= num
print(sorted([num1, num2]))

D. xor = 0
for num in arr:
xor ^= num
set_bit = xor & ~(xor + 1)
num1 = 0
num2 = 0
for num in arr:
if (num & set_bit) != 0:
num1 ^= num
else:
num2 ^= num
print(sorted([num1, num2]))

Solution

1. Step 1: Calculate XOR of all elements

   xor accumulates XOR of all elements, isolating XOR of two unique numbers.

2. Step 2: Find rightmost set bit correctly

   set_bit = xor & (-xor) correctly isolates the rightmost set bit for partitioning.

3. Step 3: Partition elements and XOR separately

   Elements are split by set_bit and XORed to find the two unique numbers.

4. Step 4: Sort and print unique elements

   Sorting ensures output is in ascending order as required.

5. Final Answer:

   Code in option A correctly finds and prints [30, 40] -> Option A

6. Quick Check:

   Rightmost set bit with xor & (-xor) + partition = correct [OK]

Hint: Use xor & (-xor) to find set bit and partition array [OK]

Common Mistakes:

• Using incorrect set_bit calculation like xor | (xor - 1)
• Wrong initialization of xor variable
• Incorrect condition in partition if statement