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

  • Difficulty Level : Expert
  • Last Updated : 27 Apr, 2021

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:
{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 
 

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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 occurrences.

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 are even.
  • If the xor of all the elements of a section is greater than 0 then it means that: 
    • Either there is 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 <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;
}

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




<?php
// PHP implementation of the approach
 
// Function that returns true if n is a
// power of 2 i.e. n has only 1 set bit
function is_power_of_two($n)
{
    return !($n & ($n - 1));
}
 
// Function to return the count
// of all valid sub-arrays
function countSubArrays($arr, $n)
{
 
    // To store the count of valid sub-arrays
    $cnt = 0;
 
    for ($j = 0; $j < $n; $j++) {
        $xorval = 0;
        for ($k = $j; $k < $n; $k++) {
 
            // 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 ($xorval == 0 || is_power_of_two($xorval))
                $cnt++;
        }
    }
 
    // Return the required count
    return $cnt;
}
 
// Driver code
 
    $arr = array( 1, 2, 3, 1, 2, 3, 4 );
    $n = count($arr);
    echo countSubArrays($arr, $n);
 
 
// This code is contributed by mits
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function that returns true if n is a
// power of 2 i.e. n has only 1 set bit
function is_power_of_two(n)
{
    return !(n & (n - 1));
}
 
// Function to return the count
// of all valid sub-arrays
function countSubArrays(arr, n)
{
 
    // To store the count of valid sub-arrays
    let cnt = 0;
 
    for (let j = 0; j < n; j++) {
        let xorval = 0;
        for (let k = j; k < n; k++) {
 
            // num = 2 ^ arr[k]
            let 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
 
    let arr = [ 1, 2, 3, 1, 2, 3, 4 ];
    let n = arr.length;
    document.write(countSubArrays(arr, n));
 
</script>
Output: 
11

 




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