Problem Description

The Leetcode problem "XOR Operation in an Array" asks you to perform a series of bitwise XOR operations on elements generated by a simple mathematical formula. Given two integers, n and start, you are to create an array where the i-th element is start + 2 * i for i from 0 to n-1. You must then compute the bitwise XOR of all elements in this array and return the result.

Constraints:

• 1 ≤ n ≤ 1000
• 0 ≤ start ≤ 1000
• Each element in the array is unique (since i increases and is multiplied by 2).
• There is exactly one valid answer for each input.

Thought Process

At first glance, the problem seems straightforward: generate the array, then XOR all its elements together. The brute-force approach is to loop through all indices, compute each value, and maintain a running XOR. However, since the array is not explicitly given and is generated via a formula, there may be opportunities to optimize further.

The XOR operation is associative and commutative, which means the order doesn't matter, and we can combine partial results as we go. Also, since the array elements are generated by a predictable formula, we might be able to compute the result without actually storing the array.

For this problem, the brute-force approach is efficient enough due to the small constraints, but it's always good to think about whether you can optimize further by leveraging properties of XOR and the formula used to generate the array.

Solution Approach

Let's break down the steps to solve the problem:

1. Initialize the Result:
   • Start with a variable (e.g., result) set to 0. This will hold the cumulative XOR.
2. Iterate Over the Array Indices:
   • Loop from 0 to n-1 (inclusive).
3. Generate Each Element:
   • For each index i, compute the value as start + 2 * i.
4. Apply XOR:
   • Update result by XOR-ing it with the current value.
5. Return the Result:
   • After the loop, result contains the answer. Return it.

This approach is efficient (linear time), uses constant extra space, and takes full advantage of the XOR operation's properties.

Example Walkthrough

Let's use n = 5 and start = 0 as an example.

• Step 1: Generate the array:
  • For i = 0: 0 + 2*0 = 0
  • For i = 1: 0 + 2*1 = 2
  • For i = 2: 0 + 2*2 = 4
  • For i = 3: 0 + 2*3 = 6
  • For i = 4: 0 + 2*4 = 8
  So the array is [0, 2, 4, 6, 8].
• Step 2: Compute the XOR:
  • Start with result = 0
  • result ^= 0 → 0
  • result ^= 2 → 2
  • result ^= 4 → 6
  • result ^= 6 → 0
  • result ^= 8 → 8
  Final result: 8

Each step applies the XOR operation, and we can see how the result changes as we process each element.

Time and Space Complexity

• Brute-force Approach:
  • Time Complexity: O(n), since we loop through n elements and perform a constant-time operation for each.
  • Space Complexity: O(1), as we don't need to store the array; we only keep a running result.
• Optimized Approach:
  • In this specific case, the brute-force method is already optimal due to the small constraints and the nature of the operation.
  • For much larger n, mathematical shortcuts using properties of XOR and the sequence could be explored, but they're not required here.

Summary

The "XOR Operation in an Array" problem is a great example of applying bitwise operations and recognizing when a direct, formula-based approach is both simple and efficient. By generating each array element on the fly and maintaining a running XOR, we achieve an optimal solution with minimal code and resource usage. The key insight is leveraging the properties of XOR and the predictable structure of the array to avoid unnecessary computation or storage.

Code Implementation

class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        result = 0
        for i in range(n):
            result ^= start + 2 * i
        return result
      

class Solution {
public:
    int xorOperation(int n, int start) {
        int result = 0;
        for (int i = 0; i < n; ++i) {
            result ^= start + 2 * i;
        }
        return result;
    }
};
      

class Solution {
    public int xorOperation(int n, int start) {
        int result = 0;
        for (int i = 0; i < n; i++) {
            result ^= start + 2 * i;
        }
        return result;
    }
}
      

var xorOperation = function(n, start) {
    let result = 0;
    for (let i = 0; i < n; i++) {
        result ^= start + 2 * i;
    }
    return result;
};