2533 Bitwise Xor Of All Pairings

Bitwise XOR of All Pairings

You are given two 0-indexed arrays, nums1 and nums2, consisting of non-negative integers. There exists another array, nums3, which contains the bitwise XOR of all pairings of integers between nums1 and nums2 (every integer in nums1 is paired with every integer in nums2 exactly once).

Return* the bitwise XOR of all integers in *nums3.

 

Example 1:

**Input:** nums1 = [2,1,3], nums2 = [10,2,5,0]
**Output:** 13
**Explanation:**
A possible nums3 array is [8,0,7,2,11,3,4,1,9,1,6,3].
The bitwise XOR of all these numbers is 13, so we return 13.

Example 2:

**Input:** nums1 = [1,2], nums2 = [3,4]
**Output:** 0
**Explanation:**
All possible pairs of bitwise XORs are nums1[0] ^ nums2[0], nums1[0] ^ nums2[1], nums1[1] ^ nums2[0],
and nums1[1] ^ nums2[1].
Thus, one possible nums3 array is [2,5,1,6].
2 ^ 5 ^ 1 ^ 6 = 0, so we return 0.

 

Constraints:

1 <= nums1.length, nums2.length <= 105
0 <= nums1[i], nums2[j] <= 109

func xorAllNums(nums1 []int, nums2 []int) int {
    
    // if num2 is even times and num 1 is odd times then num2 is answer
    // num2 is even and num1 is even then ans is 0
    // if both are odd then do all xor ^ firste elem ^ ...
    if len(nums1) % 2 == 0 && len(nums2) % 2 == 0 {
        return 0
    }

    

    temp2 := nums2[0]
    for i := 1; i < len(nums2) ; i ++ {
        temp2 ^= nums2[i]
    }

   

    temp1 := nums1[0]

    for i := 1 ; i < len(nums1) ; i ++ {
        temp1 ^= nums1[i]
    }
     if len(nums1) % 2 == 0 && len(nums2) % 2 != 0 {
        return temp1
    }

     if len(nums1) % 2 != 0 && len(nums2) % 2 == 0 {
        return temp2
    }
    return temp1 ^ temp2
}