Question

If we have an array of integers then how can we find the number of ways that they be XORed so that the result is 0 . Here at each step only one integer(i) can be decreased by any amount say d such that (i-d)>=0 . E.g for the integers 11,15,8 we can decrease 11 to 7 so that 7^15^8 =0 . Similary 15 can be reduced to 3 such that 11^3^8 = 0 and 8 can be reduced to 4 such that 11^15^4=0 . Hence total ways is 3

My approach: to this is for each integer , go on decreasing it and at each step XOR it with remaining integers in the array , if the result is 0 , break. Check this for all integers and get the total ways. But this is 0(n^2) . Is there any efficient way to do it?
Thanks.

Answer 1

You can XOR all integers in your array, then in loop XOR the result with each of your integers (it will “remove” that integer from all because x^x is always 0). As a result you’ll get XOR of the other members and that’s the number to be replaced (because x^x is always 0).

    int[] a = new int [size];
    //initialize 'a' array
    int[] b = new int [size];
    int all = 0;
    for(int i = 0; i<a.length; i++)
        all=all^a[i];

    for(int i =0; i< size; i++)
        b[i] = all^a[i];

In b[i] you’ll get the number to be replaced with a[i] to get zero.
I edited my answer and tried, it works. This is O(n).