LeetCode 1486. XOR Operation in an Array Solution in Java, C++, Python & More | Explanation + Code

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: Simulation

We can directly simulate to calculate the XOR result of all elements in the array.

The time complexity is O(n), where n is the length of the array. The space complexity is O(1).

PythonJavaC++GoTypeScript

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

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; } }(code-box)

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; } };(code-box)

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

function xorOperation(n: number, start: number): number { let ans = 0; for (let i = 0; i < n; ++i) { ans ^= start + 2 * i; } return ans; }(code-box)