# 16.01.2025 [2425. Bitwise XOR of All Pairings]

Xor of all pairs xors #medium #xor

16.01.2025

2425. Bitwise XOR of All Pairings medium blog post substack youtube deep-dive

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Problem TLDR

Xor of all pairs xors #medium #xor

Intuition

Observe the all pairs xor:

    // 2 1 3
    // 10 2 5 0
    // 2^10 2^2 2^5 2^0
    // 1^10 1^2 1^5 1^0
    // 3^10 3^2 3^5 3^0

Even size will reduce other array to 0.

Approach

• we can use a single variable

Complexity

• Time complexity: O(n)

• Space complexity: O(1)

Code

    fun xorAllNums(nums1: IntArray, nums2: IntArray) =
        nums1.reduce(Int::xor) * (nums2.size % 2) xor
        nums2.reduce(Int::xor) * (nums1.size % 2)

    pub fn xor_all_nums(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let mut r = 0;
        if nums2.len() % 2 > 0 { for x in &nums1 { r ^= x }}
        if nums1.len() % 2 > 0 { for x in &nums2 { r ^= x }}; r
    }

    int xorAllNums(vector<int>& nums1, vector<int>& nums2) {
        int r = 0;
        if (nums2.size() % 2) for (int x: nums1) r ^= x;
        if (nums1.size() % 2) for (int x: nums2) r ^= x;
        return r;
    }