Let's write code to find a single number from the array, we use the XOR technique to solve this problem in C#. public static int SingleNumber(int[] nums) { // Initialize result to 0. This will hold the unique number. int result = 0; // Loop through each number in the array foreach (int num in nums) { // XOR the current number with the result // This will cancel out numbers that appear twice and leave the unique number result ^= num; } // Return the unique number found return result; } If I take the array [2, 2, 1]. Step-by-Step Execution Initialization: Initial State: result = 0We start with result initialized to 0. First Iteration (num = 2): Operation: result ^= 2 which means result = result ^ 2Calculation: 0 ^ 2 = 2 (since 0 XOR anything equals that thing)State after iteration: result = 2 Second Iteration (num = 2): Operation: result ^= 2 which means result = result ^ 2Calculation: 2 ^ 2 = 0 (since a XOR a = 0)State after iteration: result = 0 Third Iteration (num = 1): Operation: result ^= 1 which means result = result ^ 1Calculation: 0 ^ 1 = 1 (since 0 XOR anything equals that thing)State after iteration: result = 1 Final Result return result; Return Value: 1 and

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Single Number find

Steilla Mitchel11312-Jun-2024

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

You must implement a solution with a linear runtime complexity and use only constant extra space.

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Updated on 12-Jun-2024

Steilla Mitchel

I completed my post-graduation in 2013 in the engineering field. Engineering is the application of science and math to solve problems. Engineers figure out how things work and find practical uses for scientific discoveries. Scientists and inventors often get the credit for innovations that advance the human condition, but it is engineers who are instrumental in making those innovations available to the world. I love pet animals such as dogs, cats, etc.

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1 Answers

Ravi Vishwakarma

12-Jun-2024

Let's write code to find a single number from the array, we use the XOR technique to solve this problem in C#.

public static int SingleNumber(int[] nums)
{
    // Initialize result to 0. This will hold the unique number.
    int result = 0;

    // Loop through each number in the array
    foreach (int num in nums)
    {
        // XOR the current number with the result
        // This will cancel out numbers that appear twice and leave the unique number
        result ^= num;
    }

    // Return the unique number found
    return result;
}

If I take the array [2, 2, 1].

Step-by-Step Execution

Initialization:

Initial State: result = 0
We start with result initialized to 0.

First Iteration (num = 2):

Operation: result ^= 2 which means result = result ^ 2
Calculation: 0 ^ 2 = 2 (since 0 XOR anything equals that thing)
State after iteration: result = 2

Second Iteration (num = 2):

Operation: result ^= 2 which means result = result ^ 2
Calculation: 2 ^ 2 = 0 (since a XOR a = 0)
State after iteration: result = 0

Third Iteration (num = 1):

Operation: result ^= 1 which means result = result ^ 1
Calculation: 0 ^ 1 = 1 (since 0 XOR anything equals that thing)
State after iteration: result = 1

Final Result

return result;

Return Value: 1

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Single Number find

Steilla Mitchel

1 Answers

Step-by-Step Execution

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