Quadruplet pair with XOR zero in the given Array - GeeksforGeeks

18 Nov, 2021
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 = 6Input: arr[] = {1, 0, 2, 3, 7} Output: Yes 1 ^ 0 ^ 2 ^ 3 = 0  
 
 
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++
Java
Python3
C#
PHP
Javascript
// 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 pairbool 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 codeint 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 implementation of the approachimport 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 pairstatic 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 codepublic 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 implementation of the approachMAX = 130 # Function that returns true if the array# contains a valid quadruplet pairdef 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 codearr = [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# implementation of the approachusing System;     class GFG{ static int MAX = 130; // Function that returns true if the array// contains a valid quadruplet pairstatic 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 codepublic 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 implementation of the approachconst MAX = 130; // Function that returns true if the array// contains a valid quadruplet pairfunction 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?>
<script> // Javascript implementation// of the approach const MAX = 130; // Function that returns true if the array// contains a valid quadruplet pairfunction 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 (let i = 0; i < n; i++)        for (let j = i + 1; j < n; j++)            for (let k = j + 1; k < n; k++)                for (let l = k + 1; l < n; l++) {                    if ((arr[i] ^ arr[j] ^ arr[k] ^                         arr[l]) == 0) {                        return true;                    }                }    return false;} // Driver code    let arr = [ 1, 0, 2, 3, 7 ];    let n = arr.length;     if (validQuadruple(arr, n))        document.write("Yes");    else        document.write("No"); </script>
Yes
 
Time Complexity:  
Auxiliary Space: O(1)
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
Javascript
// Java implementation of the approachimport 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
<script>// Javascript implementation of the approach function check(arr){    let n = arr.length;                  if(n < 4)            return false;        if(n >=130)            return true;                      let map = new Map();                  for(let i=0;i<n-1;i++)        {              for(let j=i+1;j<n;j++)            {                let k = arr[i] ^ arr[j];                if(!map.has(k))                    map.set(k,[]);                                  let data = map.get(k);                if(!data.includes(i) && !data.includes(j))                {                    data.push(i);                    data.push(j);                    if(data.length>=4)                        return true;                    map.set(k, data);                }            }          }        return false;} // Driver codelet arr=[ 1, 0, 2, 3, 7 ];if (check(arr))    document.write("Yes<br>");else    document.write("No<br>");  // This code is contributed by rag2127</script>
Yes
 
Time Complexity:
Auxiliary Space: O(N2)
 
mohit kumar 29
nidhiva
29AjayKumar
PramodHosahalli
Naman_Garg
subham348
rag2127
subhammahato348
Bitwise-XOR
Arrays
Bit Magic
Mathematical
Arrays
Mathematical
Bit Magic
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