# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 = Number of ways to select 1 element with k’th bit 1 and rest with 0 = • We will use nCr mod p to compute the combinotorial function values.

Below is the implementation of the above approach.

## CPP

 `// C++ program to find sum of xor of  ` `// all unordered triplets of the array ` `#include ` ` `  `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 = 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; ` `} `

## Python3

 `# 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 `

Output:

```132
```

Time Complexity : O(32 * N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : mohit kumar 29