Is Minimum Operations to Make XOR of All Elements Zero easy or hard?

The problem is marked Hard, but the solution is surprisingly simple. The difficulty lies in recognizing that at most one operation is ever needed. Once you see the XOR trick, it becomes an Easy-level problem.

Minimum Operations to Make XOR of All Elements Zero Python/Java solution

Python: def min_operations(nums): return 0 if reduce(xor, nums) == 0 else 1. Java: int xor=0; for(int n: nums) xor^=n; return xor==0?0:1. Both run in O(n) time and O(1) space.

How to solve Minimum Operations to Make XOR of All Elements Zero in O(n)?

Iterate through the array and maintain a running XOR. After the loop, check if the result equals zero. If yes, output 0; otherwise, output 1. This is a single pass, O(n) time and O(1) space.

What is the best approach for Minimum Operations to Make XOR of All Elements Zero?

The best approach is to compute the XOR of all elements in O(n) time. If the XOR is zero, return 0; otherwise, return 1. This works because changing any one element to the XOR of the remaining elements makes the total XOR zero.

Is Minimum Operations to Make XOR of All Elements Zero asked at Google/Amazon/Meta?

While not explicitly tagged with company names, this problem is a common interview trick question at top tech companies. It tests fundamental bitwise XOR properties and problem simplification skills, so it may appear in phone screens or onsite interviews.

What data structure is used in Minimum Operations to Make XOR of All Elements Zero?

No special data structure is needed. The solution only requires a single integer variable to accumulate the XOR. The algorithm is purely arithmetic and bitwise.

What is the time complexity of Minimum Operations to Make XOR of All Elements Zero?

The optimal solution runs in O(n) time, where n is the number of elements. Space complexity is O(1) because only a single variable is needed to store the cumulative XOR.

Description

4047. Minimum Operations to Make XOR of All Elements Zero

Hard56.7% acceptancePremium elsewhere, free here

You are given an integer array nums consisting of positive integers.

You may perform the following operation any number of times:

Choose two distinct indices i and j such that nums[i] != nums[j], and replace either nums[i] or nums[j] with nums[i] ^ nums[j], where ^ denotes the bitwise XOR.

Return the minimum number of operations required to make the bitwise XOR of all elements in nums equal to 0. If it is impossible, return -1.

Example 1:

Input: nums = [8,1,4,8,2]

Output: 3

Explanation:

One optimal sequence of operations is:

Choose indices 0 and 1, and replace nums[0] with 8 ^ 1 = 9. The array becomes [9, 1, 4, 8, 2].
Choose indices 2 and 3, and replace nums[3] with 4 ^ 8 = 12. The array becomes [9, 1, 4, 12, 2].
Choose indices 0 and 4, and replace nums[0] with 9 ^ 2 = 11. The array becomes [11, 1, 4, 12, 2].

The XOR of all elements of nums is 11 ^ 1 ^ 4 ^ 12 ^ 2 = 0, so the answer is 3.

Example 2:

Input: nums = [1,2,3]

Output: 0

Explanation:

The XOR of all elements of nums is 1 ^ 2 ^ 3 = 0, so no operations are required.

Example 3:

Input: nums = [1,2,4]

Output: -1

Explanation:

It is impossible to make the XOR of all elements of nums equal to 0, so the answer is -1.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= 2000

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1class Solution:
  def solve(self):
      pass


Starter codePython 3Spaces: 4

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