Single Number
Given an array where every element appears twice except one, find the single element. Must run in O(n) time and O(1) space.
Example:
2,2,1
1
- We initialize a variable to 0, which will hold the result of the bitwise XOR operation.
- We iterate over the input array, applying the XOR operation to each element: result=result⊕2, result=result⊕2, result=result⊕1.
- Since a⊕a=0 and a⊕0=a, the XOR operation cancels out the duplicate elements, leaving only the single element: 0⊕2⊕2⊕1=0⊕0⊕1=1.
- The final output is the result of the XOR operation, which is the single element in the array.
Constraints:
- 1 <= len(nums) <= 3 * 10^4
- -3 * 10^4 <= nums[i] <= 3 * 10^4
Background Knowledge
The "Single Number" problem is a classic example of a bit manipulation problem. To understand this problem, you need to have a basic understanding of how bits work in computer science. In binary representation, each digit (or bit) can have a value of either 0 or 1. Bitwise operations are used to manipulate these bits. The key concept here is the XOR (exclusive OR) operation, which returns 1 if the two bits are different, and 0 if they are the same. The XOR operation has several useful properties, including a⊕0=a, a⊕a=0, and a⊕b⊕a=b.
In the context of this problem, we can use the properties of bitwise operations to find the single element in the array. Since every element appears twice except one, we can use the XOR operation to eliminate the elements that appear twice. This is because a⊕a=0, so when we XOR all the elements in the array, the elements that appear twice will cancel each other out, leaving only the single element.
The requirement of O(n) time and O(1) space complexity means that we need to find a solution that only iterates through the array once and uses a constant amount of space. This rules out solutions that involve sorting the array or using additional data structures that scale with the size of the input.
Algorithm/Approach
The general approach to solve this type of problem is to use a bit manipulation algorithm that takes advantage of the properties of bitwise operations. Specifically, we can use the XOR operation to eliminate the elements that appear twice and find the single element. This approach is efficient because it only requires a single pass through the array and uses a constant amount of space.
Step-by-Step Strategy
To implement the solution, follow these steps:
- Initialize a variable to store the result of the XOR operation. This variable will be used to accumulate the XOR of all the elements in the array.
- Iterate through the array, XORing each element with the result variable.
- After iterating through the entire array, the result variable will hold the value of the single element.
Common Pitfalls
When implementing the solution, watch out for the following common pitfalls:
- Forgetting to initialize the result variable to 0, which is the identity element for the XOR operation.
- Using the wrong bitwise operation, such as AND or OR instead of XOR.
- Not iterating through the entire array, which can cause some elements to be missed.
Time & Space Complexity
The expected time complexity of the solution is O(n), where n is the number of elements in the array. This is because we only need to iterate through the array once to find the single element. The expected space complexity is O(1), which means the space required does not change with the size of the input array, making it very efficient for large inputs.