Find sum of xor of all unordered triplets of the array

Given an array A, consisting of N non-negative integers, find the sum of xor of all unordered triplets of the array. For unordered triplets, the triplet (A[i], A[j], A[k]) is considered same as triplet (A[j], A[i], A[k]) and all the other permutations.
Since the answer can be large calculate its mod with 10037.

Examples:

Input : A = [3, 5, 2, 18, 7]
Output : 132

Input : A = [140, 1, 66]
Output : 207

Naive Approach
Iterate over all the unordered triplets and add xor of each to the sum.

Efficient Approach

  • An important point to observe is that xor is independent over all the bits. So we can do the required computation over each bit individually.
  • Let’s consider the k’th bit of all the array elements. If the number of unoredered triplets whose k’th bit xor to 1 be C, we can simply add C * 2k to the answer. Let the number of elements whose k’th bit is 1 be X and whose k’th bit is 0 be Y. Then to find the unordered triplets whose k’th bits xor to 1 can be formed using two cases:
    1. Only one of the three elements have k’th bit 1.
    2. All three of them have k’th bit 1.

    So we simply need to find the number of ways to select and this can be done using Permutation and Combination principles.

  • Number of ways to select 3 element having k’th bit 1 =  {X \choose 3}
    Number of ways to select 1 element with k’th bit 1 and rest with 0 =  X * {Y \choose 3}
  • We will use nCr mod p to compute the combinotorial function values.

Below is the implementation of the above approach.

CPP



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// C++ program to find sum of xor of 
// all unordered triplets of the array
#include <bits/stdc++.h>
  
using namespace std;
  
// Iterative Function to calculate 
// (x^y)%p in O(log y)
int power(int x, int y, int p)
{
    // Initialize result
    int res = 1; 
  
    // Update x if it is more than or
    // equal to p
    x = x % p; 
     
  
    while (y > 0)
    {
        // If y is odd, multiply x 
        // with result
        if (y & 1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Returns n^(-1) mod p
int modInverse(int n, int p)
{
    return power(n, p - 2, p);
}
  
// Returns nCr % p using Fermat's little
// theorem.
int nCrModPFermat(int n, int r, int p)
{
    // Base case
    if (r == 0)
        return 1;
    if (n < r)
        return 0;
      
    // Fill factorial array so that we
    // can find all factorial of r, n
    // and n-r
    int fac[n + 1];
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = fac[i - 1] * i % p;
  
    return (fac[n] * modInverse(fac[r], p) % p 
            * modInverse(fac[n - r], p) % p) % p;
}
  
// Function returns sum of xor of all
// unordered triplets of the array
int SumOfXor(int a[], int n)
{
  
    int mod = 10037;
  
    int answer = 0;
  
    // Iterating over the bits
    for (int k = 0; k < 32; k++) 
    {
        // Number of elements whith k'th bit 
        // 1 and 0 respectively
        int x = 0, y = 0;
  
        for (int i = 0; i < n; i++)
        {
            // Checking if k'th bit is 1
            if (a[i] & (1 << k))
                x++;
            else
                y++;
        }
        // Adding this bit's part to the answer
        answer += ((1 << k) % mod * 
                   (nCrModPFermat(x, 3, mod) 
                    + x * nCrModPFermat(y, 2, mod)) 
                   % mod) % mod;
    }
    return answer;
}
// Drivers code
int main()
{
    int n = 5;
    int A[n] = { 3, 5, 2, 18, 7 };
  
    cout << SumOfXor(A, n);
  
    return 0;
}

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Python3

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# Python3 program to find sum of xor of
# all unordered triplets of the array
  
# Iterative Function to calculate
# (x^y)%p in O(log y)
def power(x, y, p):
      
    # Initialize result
    res = 1
  
    # Update x if it is more than or
    # equal to p
    x = x % p
  
    while (y > 0):
        # If y is odd, multiply x
        # with result
        if (y & 1):
            res = (res * x) % p
  
        # y must be even now
        y = y >> 1#y = y/2
        x = (x * x) % p
    return res
  
# Returns n^(-1) mod p
def modInverse(n, p):
    return power(n, p - 2, p)
  
# Returns nCr % p using Fermat's little
# theorem.
def nCrModPFermat(n, r, p):
      
    # Base case
    if (r == 0):
        return 1
    if (n < r):
        return 0
  
    # Fill factorial array so that we
    # can find all factorial of r, n
    # and n-r
    fac = [0]*(n + 1)
    fac[0] = 1
    for i in range(1, n + 1):
        fac[i] = fac[i - 1] * i % p
  
    return (fac[n] * modInverse(fac[r], p) % p * 
            modInverse(fac[n - r], p) % p) % p
  
# Function returns sum of xor of all
# unordered triplets of the array
def SumOfXor(a, n):
  
    mod = 10037
  
    answer = 0
  
    # Iterating over the bits
    for k in range(32):
          
        # Number of elements whith k'th bit
        # 1 and 0 respectively
        x = 0
        y = 0
  
        for i in range(n):
              
            # Checking if k'th bit is 1
            if (a[i] & (1 << k)):
                x += 1
            else:
                y += 1
        # Adding this bit's part to the answer
        answer += ((1 << k) % mod * (nCrModPFermat(x, 3, mod)
                    + x * nCrModPFermat(y, 2, mod))
                % mod) % mod
  
    return answer
  
# Drivers code
if __name__ == '__main__':
    n = 5
    A=[3, 5, 2, 18, 7]
  
    print(SumOfXor(A, n))
  
# This code is contributed by mohit kumar 29

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Output:

132

Time Complexity : O(32 * N)

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