Quadruplet pair with XOR zero in the given Array

Given an array arr[] of N integers such that any two adjacent elements in the array differ at only one position in their binary representation. The task is to find whether there exists a quadruple (arr[i], arr[j], arr[k], arr[l]) such that arr[i] ^ arr[j] ^ arr[k] ^ arr[l] = 0. Here ^ denotes the bitwise xor operation and 1 ≤ i < j < k < l ≤ N.

Examples:

Input: arr[] = {1, 3, 7, 3}
Output: No
1 ^ 3 ^ 7 ^ 3 = 6

Input: arr[] = {1, 0, 2, 3, 7}
Output: Yes
1 ^ 0 ^ 2 ^ 3 = 0

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

Naive approach: Check for all possible quadruples whether their xor is zero or not. But the time complexity of such a solution would be N⁴, for all N.
Time Complexity:

Efficient Approach (O(N⁴), for N &leq; 130): We can Say that for array length more than or equal to 130 we can have at least 65 adjacent pairs each denoting xor of two elements. Here it is given that all adjacent elements differ at only one position in their binary form thus there would result in only one set bit. Since we have only 64 possible positions, we can say that at least two pairs will have the same xor. Thus xor of these 4 integers will be 0. For N < 130 we can use the naive approach.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
const int MAX = 130; 
  
// Function that returns true if the array 
// contains a valid quadruplet pair 
bool validQuadruple(int arr[], int n) 
{ 
  
    // We can always find a valid quadruplet pair 
    // for array size greater than MAX 
    if (n >= MAX) 
        return true; 
  
    // For smaller size arrays, perform brute force 
    for (int i = 0; i < n; i++) 
        for (int j = i + 1; j < n; j++) 
            for (int k = j + 1; k < n; k++) 
                for (int l = k + 1; l < n; l++) { 
                    if ((arr[i] ^ arr[j] ^ arr[k] ^ arr[l]) == 0) { 
                        return true; 
                    } 
                } 
    return false; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 1, 0, 2, 3, 7 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    if (validQuadruple(arr, n)) 
        cout << "Yes"; 
    else
        cout << "No"; 
  
    return 0; 
} 

Java

// Java implementation of the approach 
import java.util.*; 
import java.lang.*; 
import java.io.*; 
  
class GFG 
{ 
  
static int MAX = 130;  
  
// Function that returns true if the array  
// contains a valid quadruplet pair  
static boolean validQuadruple(int arr[], int n)  
{  
  
    // We can always find a valid quadruplet pair  
    // for array size greater than MAX  
    if (n >= MAX)  
        return true;  
  
    // For smaller size arrays, perform brute force  
    for (int i = 0; i < n; i++)  
        for (int j = i + 1; j < n; j++)  
            for (int k = j + 1; k < n; k++)  
                for (int l = k + 1; l < n; l++) 
                {  
                    if ((arr[i] ^ arr[j] ^  
                         arr[k] ^ arr[l]) == 0)  
                    {  
                        return true;  
                    }  
                }  
    return false;  
}  
  
// Driver code 
public static void main (String[] args)  
              throws java.lang.Exception 
{ 
    int arr[] = { 1, 0, 2, 3, 7 };  
    int n = arr.length; 
  
    if (validQuadruple(arr, n))  
        System.out.println("Yes");  
    else
        System.out.println("No");  
} 
} 
  
// This code is contributed by nidhiva 

Python3

# Python3 implementation of the approach 
MAX = 130
  
# Function that returns true if the array 
# contains a valid quadruplet pair 
def validQuadruple(arr, n): 
  
    # We can always find a valid quadruplet pair 
    # for array size greater than MAX 
    if (n >= MAX): 
        return True
  
    # For smaller size arrays,  
    # perform brute force 
    for i in range(n): 
        for j in range(i + 1, n): 
            for k in range(j + 1, n): 
                for l in range(k + 1, n): 
                    if ((arr[i] ^ arr[j] ^  
                         arr[k] ^ arr[l]) == 0): 
                        return True
  
    return False
  
# Driver code 
arr = [1, 0, 2, 3, 7] 
n = len(arr) 
  
if (validQuadruple(arr, n)): 
    print("Yes") 
else: 
    print("No") 
  
#  This code is contributed 
# by Mohit Kumar 

C#

// C# implementation of the approach 
using System; 
      
class GFG 
{ 
  
static int MAX = 130;  
  
// Function that returns true if the array  
// contains a valid quadruplet pair  
static Boolean validQuadruple(int []arr, int n)  
{  
  
    // We can always find a valid quadruplet pair  
    // for array size greater than MAX  
    if (n >= MAX)  
        return true;  
  
    // For smaller size arrays, perform brute force  
    for (int i = 0; i < n; i++)  
        for (int j = i + 1; j < n; j++)  
            for (int k = j + 1; k < n; k++)  
                for (int l = k + 1; l < n; l++) 
                {  
                    if ((arr[i] ^ arr[j] ^  
                         arr[k] ^ arr[l]) == 0)  
                    {  
                        return true;  
                    }  
                }  
    return false;  
}  
  
// Driver code 
public static void Main (String[] args) 
{ 
    int []arr = { 1, 0, 2, 3, 7 };  
    int n = arr.Length; 
  
    if (validQuadruple(arr, n))  
        Console.WriteLine("Yes");  
    else
        Console.WriteLine("No");  
} 
} 
  
// This code is contributed by 29AjayKumar 

PHP

<?php 
// PHP implementation of the approach  
const MAX = 130;  
  
// Function that returns true if the array  
// contains a valid quadruplet pair  
function validQuadruple($arr, $n)  
{  
  
    // We can always find a valid quadruplet pair  
    // for array size greater than MAX  
    if ($n >= MAX)  
        return true;  
  
    // For smaller size arrays, 
    // perform brute force  
    for ($i = 0; $i < $n; $i++)  
        for ($j = $i + 1; $j < $n; $j++)  
            for ($k = $j + 1; $k < $n; $k++)  
                for ($l = $k + 1; $l < $n; $l++)  
                {  
                    if (($arr[$i] ^ $arr[$j] ^  
                         $arr[$k] ^ $arr[$l]) == 0)  
                    {  
                        return true;  
                    }  
                }  
    return false;  
}  
  
// Driver code  
$arr = array(1, 0, 2, 3, 7);  
$n = count($arr);  
  
if (validQuadruple($arr, $n))  
    echo ("Yes");  
else
    echo ("No");  
  
// This code is contributed by Naman_Garg 
?> 

Output:

Yes

Time Complexity:

Another Efficient Approach (O(N²log N), for N &leq; 130):
Compute Xor of all pairs and hash it. ie, store indexes i and j in a list and Hash it in form <xor, list>. If the same xor is found again for different i and j, then we have a Quadruplet pair.

Below is the implementation of the above approach :

// Java implementation of the approach 
import java.util.HashMap; 
import java.util.LinkedList; 
import java.util.List; 
import java.util.Map; 
  
public class QuadrapuleXor { 
  
    static boolean check(int arr[]) 
    { 
        int n = arr.length; 
          
        if(n < 4) 
            return false; 
        if(n >=130) 
            return true; 
              
        Map<Integer,List<Integer>> map = new HashMap<>(); 
          
        for(int i=0;i<n-1;i++) 
        {     
            for(int j=i+1;j<n;j++) 
            { 
                int k = arr[i] ^ arr[j]; 
                if(!map.containsKey(k)) 
                    map.put(k,new LinkedList<>()); 
                  
                List<Integer> data = map.get(k); 
                if(!data.contains(i) && !data.contains(j)) 
                { 
                    data.add(i); 
                    data.add(j); 
                    if(data.size()>=4) 
                        return true; 
                    map.put(k, data); 
                } 
            }     
        } 
        return false; 
    } 
      
    // Driver code 
    public static void main (String[] args)  
                  throws java.lang.Exception 
    { 
        int arr[] = { 1, 0, 2, 3, 7 };  
       
        if (check(arr))  
            System.out.println("Yes");  
        else
            System.out.println("No");  
    } 
} 
  
//This code contributed by Pramod Hosahalli 

Output:

Yes

Time Complexity:

My Personal Notes

Suggest a Topic

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

C++

Java

Python3

C#

PHP

Recommended Posts: