Efficient Algorithm for Finding Unique Integer in an Array

The problem of identifying the unique integer in an unsorted array of integers where every other integer appears exactly twice can be elegantly solved using the XOR bitwise operation. This method is efficient and straightforward, making it an ideal solution for this specific problem.

Key Property of XOR Bitwise Operation

The XOR bitwise operation is the key to solving this problem. One of the fundamental properties of XOR is that the XOR of a number with itself results in 0, and the XOR of a number with 0 is the number itself. Mathematically, for any integer x, the expression x ^ x equals 0, and x ^ 0 equals x. This property makes the XOR operation extremely useful for canceling out pairs of numbers in the array.

Step-by-Step Algorithm

Here is a detailed guide on how to implement the XOR bitwise operation for finding the unique integer:

Initialize a variable named result to 0.

Iterate through each number in the array and XOR it with the result.

After processing all elements in the array, the result will hold the unique integer that appears only once.

Python Implementation

The Python code for this algorithm is as follows:

def find_unique_integer(nums):    result  0    for num in nums:        result ^ num    return result

Example Usage

array  [4, 1, 2, 1, 2]single_number  find_unique_integer(array)print(single_number)  # Output: 4

Explanation of the Code

Initialization:

result is initialized to 0.

Loop through the array:

For each number in the array, perform the XOR operation with result.

Return the result:

After processing all numbers, result will hold the unique number.

Complexity Analysis

Time Complexity: O(n), where n is the number of elements in the array. We make a single pass through the array.

Space Complexity: O(1) as we use only a constant amount of extra space.

Self-Inverse Property of XOR

Another interesting approach involves leveraging the self-inverse property of the XOR operation. If there was a self-inverse operation on integers, we could calculate a[1] ⊕ a[2] ⊕ ... ⊕ a[n]. All the pairs would cancel each other out, leaving only the single integer.

Alternative Method: Sorting the Array

While the XOR method is efficient, an alternative approach is to sort the array and then compare successive pairs of elements for equality. However, if we wish to avoid sorting the array, the XOR method remains the most optimal solution.

Source Code Example:

module m;
  bit[7:0] array[6:0];
  bit[7:0] queue [6:0];
  initial
    begin
      array  7'b1231234;
      foreach array[i]
        begin
          queue [i]  1'b1;
        end
      foreach array[i]
        begin
          foreach queue[j]
            if (queue[j]  1'b1  i 1  7  array[i]  array[i 1])
              begin
                queue[j]  1'b0;
                queue[j 1]  1'b1;
              end
        end
    end
  endmodule

The provided Verilog code attempts a similar approach but it is not an optimal solution for the problem. Instead, it focuses on manually comparing adjacent elements, which is less efficient compared to the XOR method.