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



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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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[0]);
    cout << countSubArrays(arr, n);
  
    return 0;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

Output:

11


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