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 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 bitbool is_power_of_two(ll n){ return !(n & (n - 1LL));}// Function to return the count// of all valid sub-arraysint 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 codeint 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!=0 && (n & (n - 1))==0; } // 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)); }} // 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 bitdef is_power_of_two(n): return 0 if(n & (n - 1)) else 1;# Function to return the count# of all valid sub-arraysdef 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 codearr = [ 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 approachusing 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 bitfunction is_power_of_two($n){ return !($n & ($n - 1));}// Function to return the count// of all valid sub-arraysfunction countSubArrays($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 bitfunction is_power_of_two(n){ return !(n & (n - 1));}// Function to return the count// of all valid sub-arraysfunction 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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