Xor Operation In An Array - Leetcode Solution

Problem Statement:

Given an integer n and an integer start. Define an array nums where nums[i] = start + 2*i (0-indexed) and n == nums.length.

Return the bitwise XOR of all elements of nums.

Example: Input: n = 5, start = 0 Output: 8 Explanation: Array nums is equal to [0, 2, 4, 6, 8] where (0^2^4^6^8) = 8.

Solution:

To solve this problem, we just need to calculate the bitwise XOR of all elements in the array. We can calculate the array elements by adding (start + 2*i) to the array for each index i from 0 to n-1. At the same time, we can also keep calculating the XOR of the array elements using a variable called xorVal.

Here is the Python code to solve this problem:

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

Time Complexity:

The time complexity of this algorithm is O(n) as we need to iterate over all n elements of the array to calculate the XOR value.

Space Complexity:

The space complexity of this algorithm is O(1) as we are not using any additional space to store the array elements or XOR values.

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