PROBLEM LINK:

Author: Ankit Srivastava and Uttam Kanodia
Tester: Roman Rubanenko
Editorialist: Amit Pandey

DIFFICULTY:

Medium

PREREQUISITES:

Fast Fourier Transform.

PROBLEM:

Given an array of $N$ integers. You have to find count of all subarrays of size $m$, such that if a Nim Game is played on the subarray, second player will win. Do it for all possible values of $m$, $1 \le m \le N$.

EXPLANATION:

It can be claimed using Nim Game theory that second player will win the game if $XOR$ of the selected subarray is $0$. We can find XOR of any subarray in $O(1)$ time by creating a prefix XOR array.

$$ preXOR[0] = 0 \hspace{3 mm}\& \hspace{3 mm}preXOR[i] = A[0] \oplus A[1] \oplus \cdots \oplus A[i-1]$$

$$ A[i] \oplus A[i+1] \oplus \cdots A[j] = preXOR[i-1] \oplus preXOR[j]$$.

Now, let me present an $O(N^2)$ solution for the problem.

answer[n] = 0;
for i in range(0,n):
    for j in range(i+1,n):
        if preXOR[i] == preXOR[j]:      //subarray having size (j-i) has XOR 0.
            answer[j-i]++;

Consider a number $k$. If it appears $l$ times in the $preXOR$ array, then there will be $^lC_2$ subarrays in array $A$ whose XOR is zero. We can create an array corresponding to indices of $k$ in $preXOR$ array, and call it $S$. It means $preXOR[S[i]] = k$ for all $i < l$.

      for i = 0 to l-1:
          for j= i+1 to l-1:
             ans[j-i]++;       // there is a subarray of size (j-i) having XOR 0.

It can be explained further,Consider that preXOR array is $[1,2,3,1,1,4,1]$, then the array $S$ for $k=1$ will be $[0,3,4,6] \text{(0 based indexing)}$.

The above solution can take up to $O(N^2)$ time in the worst case, as size of $S$ can be $O(N)$. To improve it, we can do the following modifications:

1) If size of the array $S$ is lesser than $\sqrt{N}$, then we can use brute force as given above.

2) If size of the array $S$ is greater than $\sqrt{N}$, then the above solution will take more time. So, we can use the FFT to optimize the solution to time $Nlog(N)$.

Number of sub-arrays of size $K$ having XOR $0$, will be coefficient of $x^K$ in following polynomial:

$$(x^{S[0]} + x^{S[1]} + \cdots + x^{S[L-1]}) (x^{-S[0]} + x^{-S[1]} + \cdots + x^{-S[L-1]})$$

The powers of $x$ in the above polynomial are negative as well, therefore we do a small modification in above polynomial to make all the powers positive. It will be equivalent to the coefficient of $x^{N+K}$ in the following polynomial:

$$(x^{S[0]} + x^{S[1]} + \cdots + x^{S[L-1]}) (x^{N-S[0]} + x^{N-S[1]} + \cdots + x^{N-S[L-1]})$$

Which can be done in $O(Nlog(N))$ time using Fast Fourier transform. The FFT will give answer corresponding to all possible value of $m$ altogether.

We can repeat the above procedure for each distinct element $k$ of $preXOR$ array. The total count of subarrays corresponding to each value of $m$ will be the count of subarrays of size $m$ for each distinct value of $k$.

Worst case complexity of the above procedure will be $O(N \sqrt{N} \log{(N)})$.

Problems to solve: