Question

1486. XOR Operation in an Array

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Description

You are 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.

 

Constraints:

• 1 <= n <= 1000
• 0 <= start <= 1000
• n == nums.length

Solutions

Solution 1

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

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

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

func xorOperation(n int, start int) (ans int) {
  for i := 0; i < n; i++ {
    ans ^= start + 2*i
  }
  return
}

Solution 2

class Solution:
  def xorOperation(self, n: int, start: int) -> int:
    return reduce(xor, ((start + 2 * i) for i in range(n)))