XOR Operation in an Array
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 1: 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. Where "^" corresponds to bitwise XOR operator. Example 2: Input: n = 4, start = 3 Output: 8 Explanation: Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8. Example 3: Input: n = 1, start = 7 Output: 7 Example 4: Input: n = 10, start = 5 Output: 2 Constraints: 1 <= n <= 1000 0 <= start <= 1000 n == nums.length class Solution { public int xorOperation(int n, int start) { int i = 1, result = 0; while(i<=n){ result ^= start; start+=2; i++; } return result; } } Here, we are going to solve using using XOR operator. In the problem,