Data Structures and Algorithms

Identifying the Unpaired Element in an Array

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Find the Odd One Out: Identifying the Unpaired Element in an Array


Finding an unpaired element in an array is a common coding challenge. Imagine a room full of paired socks except for one odd sock.

The goal is to identify that odd one.

This problem becomes more interesting when the input size grows significantly, making efficiency a key consideration.

As an expert in algorithm optimization, I’ll show you a streamlined and efficient approach to solving this problem. Using Python, we’ll create a robust solution that efficiently finds the unpaired element without unnecessary computations.

The Problem

Given a non-empty array A of N integers, every element appears an even number of times except for one unpaired element. The challenge is to identify that unique unpaired value.

Example

# For the array

A = [9, 3, 9, 3, 9, 7, 9]

# Pairs: 9 (appears four times), 3 (appears twice).
# Unpaired: 7 (appears once).
# Output: 7.

Constraints

  • N is odd and within [1..1,000,000].
  • Array values are integers within [1..1,000,000,000].
  • Exactly one unpaired element exists.

Optimized Solution

Here’s the optimized approach using XOR (exclusive OR). XOR has a unique property:

  • x⊕x=0 (a number XORed with itself cancels out).
  • x⊕0=x (a number XORed with 0 remains unchanged).

Thus, applying XOR to all elements in the array eliminates paired elements, leaving only the unpaired value.

Optimized Python Code

Here’s the implementation:
 

def solution(A):
    """
    Finds the unpaired element in an array.

    Args:
        A (list): A non-empty array of integers with exactly one unpaired element.

    Returns:
        int: The value of the unpaired element.
    """
    # Initialize result to 0
    result = 0
    
    # Use XOR to cancel out paired elements
    for number in A:
        result ^= number  # XOR current number with the result
    
    return result


Step-by-Step Explanation

Step 1: Initialize Result

  • Start with result = 0. It will store the cumulative XOR of all numbers in the array.

Step 2: Iterate Through the Array

  • For each number in the array:
    • XOR the current result with the number
    • The XOR operation cancels out numbers that appear in pairs.

Step 3: Return the Result

  • Once all numbers are processed, the result will contain the unpaired value.

Example Walkthrough

Let’s break down the example step by step:

# Input: 

A = [9, 3, 9, 3, 9, 7, 9]

Execution


Step            Current Number         Current Result (⊕)               Explanation
1                            9                                      0⊕9=9                        Initialize with the first number.    
2                            3                                      9⊕3=10                      XOR with the next number.    
3                            9                                      10⊕9=3                      Cancel out one 9.    
4                            3                                      3⊕3=0                        Cancel out one 3.    
5                            9                                      0⊕9=9                        Add remaining 9.    
6                            7                                      9⊕7=14                      Add the unpaired 7.    
7                            9                                      14⊕9=7                      Cancel out final 9.    

Output

The unpaired value is 7.

Why Is This Solution Optimal?

Time Complexity

The algorithm runs in O(N), where N is the size of the array. Each element is processed once.

Space Complexity

The solution uses O(1) extra space. No additional data structures are required.

Scalability

Handles large arrays efficiently without performance degradation.

Make Your Algorithms Work for You

Now that you understand the optimized solution for finding unpaired elements try applying XOR to other problems involving pairs.

Practice with variations of this problem to deepen your understanding.

If you enjoyed this tutorial, share it with your peers or leave a comment. Let’s solve problems together, one algorithm at a time!

 

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