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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