2425. Bitwise XOR of All Pairings

class Solution {
public:
    int xorAllNums(vector<int>& nums1, vector<int>& nums2) {
        int n=nums1.size(), m=nums2.size();
         long long res=0;
         // because all have even pairs,  so all xor of all become zero
        if(n%2==0 && m%2==0){
            return res;
        }

        //all have odd pairs , so we need all of xor
        else if(n%2==1 && m%2==1)
        {
            for(int i=0; i<n; i++){
                res=res^nums1[i];
            }
            for(int i=0; i<m; i++){
                res=res^nums2[i];
            }
        }

        else if(n%2==0 && m%2==1)
        {
            for(int i=0; i<n; i++){
                res=res^nums1[i];
            }
        }

        else
        {
            for(int i=0; i<m; i++){
                res=res^nums2[i];
            }
        }

        return res;
    }
};