algorithm - sum of xor values of all pairs

Question

We have an array A (say [1,2,3]) . We need to find the XOR(^)SUM of all pairs of integers in the array. Though this can easily be done in O(n^2) but how can i improve the complexity of the solution ? E.g for the above array , A, the answer would be (1^2)+(1^3)+(2^3) = 6 Thanks.

Answer 1

You can separate the calculation to do one bit at a time.

For example, look at the rightmost bit of all the numbers in the array. Suppose that a numbers have a rightmost 0-bit, and b numbers have a 1-bit. Then out of the pairs, a*b of them will have 1 in the rightmost bit of the XOR. This is because there are a*b ways to choose one number that has a 0-bit and one that has a 1-bit. These bits will therefore contribute a*b towards the total of all the XORs.

In general, when looking at the nth bit (where the rightmost bit is the 0th), count how many numbers have 0 (call this a_n) and how many have 1 (call this b_n). The contribution towards the final sum will be a_n*b_n*2ⁿ. You need to do this for each bit and sum all these contributions together.

This can be done in O(kn) time, where k is the number of bits in the given values.