# Total pairs in an array such that the bitwise AND, bitwise OR and bitwise XOR of LSB is 1

Given an array arr[] of size N. The task is to find the number of pairs (arr[i], arr[j]) as cntAND, cntOR and cntXOR such that:

1. cntAND: Count of pairs where bitwise AND of least significant bits is 1.
2. cntOR: Count of pairs where bitwise OR of least significant bits is 1.
3. cntXOR: Count of pairs where bitwise XOR of least significant bits is 1.

Examples:

Input: arr[] = {1, 2, 3}
Output:
cntXOR = 2
cntAND = 1
cntOR = 3
Array elements in binary are {01, 10, 11}
Total XOR pairs: 2 i.e., (1, 2) and (2, 3)
Total AND pairs: 1 i.e., (1, 3)
Total OR pairs: 3 i.e., (1, 2), (2, 3) and (1, 3)

Input: arr[] = {1, 3, 4, 2}
Output:
cntXOR = 4
cntAND = 1
cntOR = 5

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

Approach:

1. To get the LSB of the elements of the array, first we calculate total even and odd elements. Even elements have LSB as 0 and odd elements have LSB as 1.
2. In order for
• XOR to be 1, LSB of both the elements have to be different.
• AND to be 1, LSB of both the elements have to be 1.
• OR to be 1, at least one of the elements should have it’s LSB as 1.
3. Therefore, total number of required pairs
• For XOR: cntXOR = cntOdd * cntEven
• For AND: cntAND = cntOdd * (cntOdd – 1) / 2
• For OR: cntOR = (cntOdd * cntEven) + cntOdd * (cntOdd – 1) / 2

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the count of required pairs ` `void` `CalculatePairs(``int` `a[], ``int` `n) ` `{ ` ` `  `    ``// To store the count of elements which ` `    ``// give remainder 0 i.e. even values ` `    ``int` `cnt_zero = 0; ` ` `  `    ``// To store the count of elements which ` `    ``// give remainder 1 i.e. odd values ` `    ``int` `cnt_one = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``if` `(a[i] % 2 == 0) ` `            ``cnt_zero += 1; ` `        ``else` `            ``cnt_one += 1; ` `    ``} ` ` `  `    ``long` `int` `total_XOR_pairs = cnt_zero * cnt_one; ` `    ``long` `int` `total_AND_pairs = (cnt_one) * (cnt_one - 1) / 2; ` `    ``long` `int` `total_OR_pairs = cnt_zero * cnt_one ` `                              ``+ (cnt_one) * (cnt_one - 1) / 2; ` ` `  `    ``cout << ``"cntXOR = "` `<< total_XOR_pairs << endl; ` `    ``cout << ``"cntAND = "` `<< total_AND_pairs << endl; ` `    ``cout << ``"cntOR = "` `<< total_OR_pairs << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 3, 4, 2 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` `  `    ``CalculatePairs(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function to find the count of required pairs ` `    ``static` `void` `CalculatePairs(``int` `a[], ``int` `n) ` `    ``{ ` ` `  `        ``// To store the count of elements which ` `        ``// give remainder 0 i.e. even values ` `        ``int` `cnt_zero = ``0``; ` ` `  `        ``// To store the count of elements which ` `        ``// give remainder 1 i.e. odd values ` `        ``int` `cnt_one = ``0``; ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `            ``if` `(a[i] % ``2` `== ``0``) ` `                ``cnt_zero += ``1``; ` `            ``else` `                ``cnt_one += ``1``; ` `        ``} ` ` `  `        ``int` `total_XOR_pairs = cnt_zero * cnt_one; ` `        ``int` `total_AND_pairs = (cnt_one) * (cnt_one - ``1``) / ``2``; ` `        ``int` `total_OR_pairs = cnt_zero * cnt_one ` `                             ``+ (cnt_one) * (cnt_one - ``1``) / ``2``; ` ` `  `        ``System.out.println(``"cntXOR = "` `+ total_XOR_pairs); ` `        ``System.out.println(``"cntAND = "` `+ total_AND_pairs); ` `        ``System.out.println(``"cntOR = "` `+ total_OR_pairs); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `a[] = { ``1``, ``3``, ``4``, ``2` `}; ` `        ``int` `n = a.length; ` ` `  `        ``CalculatePairs(a, n); ` `    ``} ` `} `

## Python3

 `# Python3 program to find number of pairs ` ` `  `# Function to find the count of required pairs ` `def` `CalculatePairs(a, n): ` ` `  `    ``# To store the count of elements which ` `    ``# give remainder 0 i.e. even values ` `    ``cnt_zero ``=` `0` ` `  `    ``# To store the count of elements which ` `    ``# give remainder 1 i.e. odd values ` `    ``cnt_one ``=` `0` ` `  `    ``for` `i ``in` `range``(``0``, n): ` `        ``if` `(a[i] ``%` `2` `=``=` `0``): ` `            ``cnt_zero ``+``=` `1` `        ``else``: ` `            ``cnt_one ``+``=` `1` `     `  `    ``total_XOR_pairs ``=` `cnt_zero ``*` `cnt_one ` `    ``total_AND_pairs ``=` `(cnt_one) ``*` `(cnt_one ``-` `1``) ``/` `2` `    ``total_OR_pairs ``=` `cnt_zero ``*` `cnt_one ``+` `(cnt_one) ``*` `(cnt_one ``-` `1``) ``/` `2` ` `  `    ``print``(``"cntXOR = "``, ``int``(total_XOR_pairs)) ` `    ``print``(``"cntAND = "``, ``int``(total_AND_pairs)) ` `    ``print``(``"cntOR = "``, ``int``(total_OR_pairs)) ` `     `  ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``a ``=` `[``1``, ``3``, ``4``, ``2``] ` `    ``n ``=` `len``(a) ` `     `  `    ``# Print the count ` `    ``CalculatePairs(a, n) `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find the count of required pairs ` `    ``static` `void` `CalculatePairs(``int``[] a, ``int` `n) ` `    ``{ ` ` `  `        ``// To store the count of elements which ` `        ``// give remainder 0 i.e. even values ` `        ``int` `cnt_zero = 0; ` ` `  `        ``// To store the count of elements which ` `        ``// give remainder 1 i.e. odd values ` `        ``int` `cnt_one = 0; ` ` `  `        ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `            ``if` `(a[i] % 2 == 0) ` `                ``cnt_zero += 1; ` `            ``else` `                ``cnt_one += 1; ` `        ``} ` ` `  `        ``int` `total_XOR_pairs = cnt_zero * cnt_one; ` `        ``int` `total_AND_pairs = (cnt_one) * (cnt_one - 1) / 2; ` `        ``int` `total_OR_pairs = cnt_zero * cnt_one ` `                             ``+ (cnt_one) * (cnt_one - 1) / 2; ` ` `  `        ``Console.WriteLine(``"cntXOR = "` `+ total_XOR_pairs); ` `        ``Console.WriteLine(``"cntAND = "` `+ total_AND_pairs); ` `        ``Console.WriteLine(``"cntOR = "` `+ total_OR_pairs); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] a = { 1, 3, 4, 2 }; ` `        ``int` `n = a.Length; ` ` `  `        ``// Print the count ` `        ``CalculatePairs(a, n); ` `    ``} ` `} `

## PHP

 ` `

Output:

```cntXOR = 4
cntAND = 1
cntOR = 5
```

Time Complexity: O(N)

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