# Python3 Program for Number of unique triplets whose XOR is zero

Given N numbers with no duplicates, count the number of unique triplets **(a _{i}, a_{j}, a_{k})** such that their XOR is 0. A triplet is said to be unique if all of the three numbers in the triplet are unique.

**Examples:**

Input : a[] = {1, 3, 5, 10, 14, 15}; Output : 2 Explanation : {1, 14, 15} and {5, 10, 15} are the unique triplets whose XOR is 0. {1, 14, 15} and all other combinations of 1, 14, 15 are considered as 1 only. Input : a[] = {4, 7, 5, 8, 3, 9}; Output : 1 Explanation : {4, 7, 3} is the only triplet whose XOR is 0

**Naive Approach**: A naive approach is to run three nested loops, the first runs from 0 to n, the second from i+1 to n, and the last one from j+1 to n to get the unique triplets. Calculate the XOR of a_{i}, a_{j}, a_{k}, check if it equals 0. If so, then increase the count.

Time Complexity: O(n^{3})

**Efficient Approach**: An efficient approach is to use one of the properties of XOR: the XOR of two of the same numbers gives 0. So we need to calculate the XOR of unique pairs only, and if the calculated XOR is one of the array elements, then we get the triplet whose XOR is 0. Given below are the steps for counting the number of unique triplets:

Below is the complete algorithm for this approach:

- With the map, mark all the array elements.
- Run two nested loops, one from i-n-1, and the other from i+1-n to get all the pairs.
- Obtain the XOR of the pair.
- Check if the XOR is an array element and not one of a
_{i}or a_{j}. - Increase the count if the condition holds.
- Return count/3 as we only want unique triplets. Since i-n and j+1-n give us unique pairs but not triplets, we do a count/3 to remove the other two possible combinations.

Below is the implementation of the above idea:

## Python3

`# Python 3 program to count the number of ` `# unique triplets whose XOR is 0 ` ` ` `# function to count the number of ` `# unique triplets whose xor is 0 ` `def` `countTriplets(a, n): ` ` ` ` ` `# To store values that are present ` ` ` `s ` `=` `set` `() ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `s.add(a[i]) ` ` ` ` ` `# stores the count of unique triplets ` ` ` `count ` `=` `0` ` ` ` ` `# traverse for all i, j pairs such that j>i ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `for` `j ` `in` `range` `(i ` `+` `1` `, n, ` `1` `): ` ` ` ` ` `# xor of a[i] and a[j] ` ` ` `xr ` `=` `a[i] ^ a[j] ` ` ` ` ` `# if xr of two numbers is present, ` ` ` `# then increase the count ` ` ` `if` `(xr ` `in` `s ` `and` `xr !` `=` `a[i] ` `and` ` ` `xr !` `=` `a[j]): ` ` ` `count ` `+` `=` `1` `; ` ` ` ` ` `# returns answer ` ` ` `return` `int` `(count ` `/` `3` `) ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `a ` `=` `[` `1` `, ` `3` `, ` `5` `, ` `10` `, ` `14` `, ` `15` `] ` ` ` `n ` `=` `len` `(a) ` ` ` `print` `(countTriplets(a, n)) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar ` |

**Output:**

2

**Time Complexity:** O(n^{2})

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