# Count of sub-arrays whose elements can be re-arranged to form palindromes

Given an array arr[] of size n. The task is to count the number of possible sub-arrays such that their elements can be re-arranged to form a palindrome.

Examples:

Input: arr[] = {1, 2, 1, 2}
Output: 7
{1}, {2}, {1}, {2}, {1, 2, 1}, {2, 1, 2} and {1, 2, 1, 2} are the valid sub-arrays.

Input: arr[] = {1, 2, 3, 1, 2, 3, 4}
Output: 11

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

Approach: There are a few observations:

• To create an even length palindrome all the distinct numbers need to have even occurrences.
• To create an odd length palindrome there has to be only one number of odd occurrence.

Now, the tricky part is to determine whether a particular section of the array can be made into a palindrome in O(1) complexity. We can use XOR to achieve this:

• For each number m, we can use it in the xor calculation as 2^n so that it contains a single set bit.
• If the xor of all the elements of a section is 0 then it means that occurrences of all the distinct numbers of this section is even.
• If the xor of all the elements of a section is greater than 0 then it means that:
• Either there are more than one distinct number with odd occurrences in which case the section cannot be re-arranged to form a palindrome
• Or exactly one number with odd occurrence (the binary representation of the number will have only 1 set bit).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `typedef` `signed` `long` `long` `ll; ` ` `  `// Function that returns true if n is a ` `// power of 2 i.e. n has only 1 set bit ` `bool` `is_power_of_two(ll n) ` `{ ` `    ``return` `!(n & (n - 1LL)); ` `} ` ` `  `// Function to return the count ` `// of all valid sub-arrays ` `int` `countSubArrays(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// To store the count of valid sub-arrays ` `    ``int` `cnt = 0; ` ` `  `    ``for` `(``int` `j = 0; j < n; j++) { ` `        ``ll xorval = 0LL; ` `        ``for` `(``int` `k = j; k < n; k++) { ` ` `  `            ``// num = 2 ^ arr[k] ` `            ``ll num = 1LL << arr[k]; ` `            ``xorval ^= num; ` ` `  `            ``// If frequency of all the elements of the ` `            ``// sub-array is even or there is only a ` `            ``// single element with odd frequency ` `            ``if` `(xorval == 0LL || is_power_of_two(xorval)) ` `                ``cnt++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the required count ` `    ``return` `cnt; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 1, 2, 3, 1, 2, 3, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << countSubArrays(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GfG  ` `{ ` `     `  `static` `long` `ll;  ` ` `  `// Function that returns true if n is a  ` `// power of 2 i.e. n has only 1 set bit  ` `static` `boolean` `is_power_of_two(``long` `n)  ` `{  ` `    ``//return !(n & (n - 1));  ` `    ``return` `false``; ` `}  ` ` `  `// Function to return the count  ` `// of all valid sub-arrays  ` `static` `int` `countSubArrays(``int` `arr[], ``int` `n)  ` `{  ` ` `  `    ``// To store the count of valid sub-arrays  ` `    ``int` `cnt = ``0``;  ` ` `  `    ``for` `(``int` `j = ``0``; j < n; j++)  ` `    ``{  ` `        ``long` `xorval = ``0``;  ` `        ``for` `(``int` `k = j; k < n; k++) ` `        ``{  ` ` `  `            ``// num = 2 ^ arr[k]  ` `            ``long` `num = ``1` `<< arr[k];  ` `            ``xorval ^= num;  ` ` `  `            ``// If frequency of all the elements of the  ` `            ``// sub-array is even or there is only a  ` `            ``// single element with odd frequency  ` `            ``if` `(xorval == ``0` `|| is_power_of_two(xorval))  ` `                ``cnt++;  ` `        ``}  ` `    ``}  ` ` `  `    ``// Return the required count  ` `    ``return` `cnt;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` ` `  `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``1``, ``2``, ``3``, ``4` `};  ` `    ``int` `n = arr.length;  ` `    ``System.out.println(countSubArrays(arr, n) + ``"1"``);  ` `} ` `}  ` ` `  `// This code is contributed by Prerna Saini `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function that returns true if n is a ` `# power of 2 i.e. n has only 1 set bit ` `def` `is_power_of_two(n): ` ` `  `    ``return` `0` `if``(n & (n ``-` `1``)) ``else` `1``; ` ` `  `# Function to return the count ` `# of all valid sub-arrays ` `def` `countSubArrays(arr, n): ` ` `  `    ``# To store the count of valid sub-arrays ` `    ``cnt ``=` `0``; ` ` `  `    ``for` `j ``in` `range``(n): ` `        ``xorval ``=` `0``; ` `        ``for` `k ``in` `range``(j, n): ` ` `  `            ``# num = 2 ^ arr[k] ` `            ``num ``=` `1` `<< arr[k]; ` `            ``xorval ^``=` `num; ` ` `  `            ``# If frequency of all the elements of the ` `            ``# sub-array is even or there is only a ` `            ``# single element with odd frequency ` `            ``if` `(xorval ``=``=` `0` `or` `is_power_of_two(xorval)): ` `                ``cnt ``+``=` `1``; ` ` `  `    ``# Return the required count ` `    ``return` `cnt; ` ` `  `# Driver code ` `arr ``=` `[ ``1``, ``2``, ``3``, ``1``, ``2``, ``3``, ``4` `]; ` `n ``=` `len``(arr); ` `print``(countSubArrays(arr, n)); ` ` `  `# This code is contributed by mits `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GfG  ` `{ ` `     `  `static` `long` `ll;  ` ` `  `// Function that returns true if n is a  ` `// power of 2 i.e. n has only 1 set bit  ` `static` `bool` `is_power_of_two(``long` `n)  ` `{  ` `    ``//return !(n & (n - 1));  ` `    ``return` `false``; ` `}  ` ` `  `// Function to return the count  ` `// of all valid sub-arrays  ` `static` `int` `countSubArrays(``int` `[]arr, ``int` `n)  ` `{  ` ` `  `    ``// To store the count of valid sub-arrays  ` `    ``int` `cnt = 0;  ` ` `  `    ``for` `(``int` `j = 0; j < n; j++)  ` `    ``{  ` `        ``long` `xorval = 0;  ` `        ``for` `(``int` `k = j; k < n; k++) ` `        ``{  ` ` `  `            ``// num = 2 ^ arr[k]  ` `            ``long` `num = 1 << arr[k];  ` `            ``xorval ^= num;  ` ` `  `            ``// If frequency of all the elements of the  ` `            ``// sub-array is even or there is only a  ` `            ``// single element with odd frequency  ` `            ``if` `(xorval == 0 || is_power_of_two(xorval))  ` `                ``cnt++;  ` `        ``}  ` `    ``}  ` ` `  `    ``// Return the required count  ` `    ``return` `cnt;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` ` `  `    ``int` `[]arr = { 1, 2, 3, 1, 2, 3, 4 };  ` `    ``int` `n = arr.Length;  ` `    ``Console.WriteLine(countSubArrays(arr, n) + ``"1"``);  ` `} ` `}  ` ` `  `// This code contributed by Rajput-Ji `

## PHP

 ` `

Output:

```11
```

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