sum of xor values of all pairs

问题

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.

回答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.

回答2:

Here is a jsFiddle confirming interjay's answer which does the calculation using both methods O(N^2) vs O(Nk):

var list = [1,2,2,3,4,5]

function xorsum_on2( a )
{
    var sum = 0
    for( var i=0 ; i<a.length ; ++i )
        for( var j=i+1 ; j<a.length ; ++j )
            sum += a[i]^a[j]
    return sum
}

// This sets all elements of a to 0
function xorsum_onk( a )
{
    var allzeroes;
    var sum = 0;
    var power = 0;
    do {
        allzeroes = true;
        var bitcount = [0,0];
        for( var i=0 ; i<a.length ; ++i )
        {
            bitcount[a[i]&1]++;
            a[i] >>= 1;
            if( a[i] ) allzeroes = false;
        }
        sum += (bitcount[0]*bitcount[1]) << power++;
    } while( !allzeroes );
    return sum;
}


var onk = document.getElementById("onk")
var on2 = document.getElementById("on2")

on2.innerHTML = xorsum_on2(list)
onk.innerHTML = xorsum_onk(list)

回答3:

#include <iostream>
using namespace std;

int main() {
    long long int n,i,j,k;

    cin>>n; // number of elements in array 
    long long int arr[n],ans=0;

    for(i=0;i<n;i++)
        cin>>arr[i];

    // iterate through all the bits
    for(i=0;i<32;i++)
    {
        k=0; // k is the number of set bits in the array at ith psotion
        for(j=0;j<n;j++)
            if((arr[j] & (1<<i)))
                k++;
        /* there are k set bits and n-k unset bits.
            therefore number of pairs with one set bit and one unset bit is kC1 and n-kC1
            Every pair adds 2^i in the answer.
        */
        ans+= (1<<i)*(k*(n-k));     
    }
    cout<<ans<<endl;
    return 0;
}

来源：https://stackoverflow.com/questions/21388448/sum-of-xor-values-of-all-pairs

标签

algorithm

xor