Count all pairs with given XOR

Given an array of distinct positive integers and a number x, find the number of pairs of integers in the array whose XOR is equal to x. 

Examples:

Input : arr[] = {5, 4, 10, 15, 7, 6}, x = 5
Output : 1
Explanation :  (10 ^ 15) = 5

Input : arr[] = {3, 6, 8, 10, 15, 50}, x = 5
Output : 2
Explanation : (3 ^ 6) = 5 and (10 ^ 15) = 5 

A Simple solution is to traverse each element and check if there’s another number whose XOR with it is equal to x. This solution takes O(n²) time.  An efficient solution to this problem takes O(n) time. The idea is based on the fact that arr[i] ^ arr[j] is equal to x if and only if arr[i] ^ x is equal to arr[j].  

1) Initialize result as 0.
2) Create an empty hash set "s".
3) Do following for each element arr[i] in arr[]
   (a)    If x ^ arr[i] is in "s", then increment result by 1.
   (b)    Insert arr[i] into the hash set "s".
3) return result.

Implementation:

C++

// C++ program to Count all pair with given XOR
// value x
#include<bits/stdc++.h>

using namespace std;

// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
int xorPairCount(int arr[], int n, int x)
{
    int result = 0; // Initialize result

    // create empty set that stores the visiting
    // element of array.
    // Refer below post for details of unordered_set
    // https://www.geeksforgeeks.org/unorderd_set-stl-uses/
    unordered_set<int> s;

    for (int i=0; i<n ; i++)
    {
        // If there exist an element in set s
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (s.find(x^arr[i]) != s.end())
            result++;

        // Make element visited
        s.insert(arr[i]);
    }

    // return total count of pairs with XOR equal to x
    return result;
}

// driver program
int main()
{
    int arr[] = {5 , 4 ,10, 15, 7, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    int x = 5;
    cout << "Count of pairs with given XOR = "
         << xorPairCount(arr, n, x);
    return 0;
}

Java

// Java program to Count all pair with
// given XOR value x
import java.util.*;

class GFG
{

    // Returns count of pairs in arr[0..n-1] with XOR
    // value equals to x.
    static int xorPairCount(int arr[], int n, int x)
    {
        int result = 0; // Initialize result

        // create empty set that stores the visiting
        // element of array.
        // Refer below post for details of unordered_set
        // https://www.geeksforgeeks.org/unorderd_set-stl-uses/
        HashSet<Integer> s = new HashSet<Integer>();

        for (int i = 0; i < n; i++)
        {
            // If there exist an element in set s
            // with XOR equals to x^arr[i], that means
            // there exist an element such that the
            // XOR of element with arr[i] is equal to
            // x, then increment count.
            if (s.contains(x ^ arr[i]) &&
                    (x ^ arr[i]) == (int) s.toArray()[s.size() - 1])
            {
                result++;
            }

            // Make element visited
            s.add(arr[i]);
        }

        // return total count of
        // pairs with XOR equal to x
        return result;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {5, 4, 10, 15, 7, 6};
        int n = arr.length;
        int x = 5;
        System.out.print("Count of pairs with given XOR = "
                + xorPairCount(arr, n, x));
    }
}

// This code contributed by Rajput-Ji

Python3

# Python3 program to count all the pair
# with given xor

# Returns count of pairs in arr[0..n-1]
# with XOR value equals to x.
def xorPairCount(arr, n, x):
    result = 0 # Initialize result

    # create empty set that stores the
    # visiting element of array.
    s = set()
    for i in range(0, n):

        # If there exist an element in set s
        # with XOR equals to x^arr[i], that
        # means there exist an element such
        # that the XOR of element with arr[i]
        # is equal to x, then increment count.
        if(x ^ arr[i] in s):
            result = result + 1

        # Make element visited
        s.add(arr[i])
    return result

# Driver Code
if __name__ == "__main__":
    arr = [5, 4, 10, 15, 7, 6]
    n = len(arr)
    x = 5
    print("Count of pair with given XOR = " +
                str(xorPairCount(arr, n, x)))

# This code is contributed by Anubhav Natani

C#

// C# program to Count all pair with
// given XOR value x
using System;
using System.Collections.Generic;

class GFG
{

    // Returns count of pairs in arr[0..n-1] with XOR
    // value equals to x.
    static int xorPairCount(int []arr, int n, int x)
    {
        int result = 0; // Initialize result

        // create empty set that stores the visiting
        // element of array.
        // Refer below post for details of unordered_set
        // https://www.geeksforgeeks.org/unorderd_set-stl-uses/
        HashSet<int> s = new HashSet<int>();

        for (int i = 0; i < n; i++)
        {
            // If there exist an element in set s
            // with XOR equals to x^arr[i], that means
            // there exist an element such that the
            // XOR of element with arr[i] is equal to
            // x, then increment count.
            if (s.Contains(x ^ arr[i]))
            {
                result++;
            }

            // Make element visited
            s.Add(arr[i]);
        }

        // return total count of
        // pairs with XOR equal to x
        return result;
    }

    // Driver code
    public static void Main()
    {
        int []arr = {5, 4, 10, 15, 7, 6};
        int n = arr.Length;
        int x = 5;
        Console.WriteLine("Count of pairs with given XOR = "
                + xorPairCount(arr, n, x));
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>
// Javascript program to Count all pair with
// given XOR value x

     // Returns count of pairs in arr[0..n-1] with XOR
    // value equals to x.
    function xorPairCount(arr,n,x)
    {
        let result = 0; // Initialize result

        // create empty set that stores the visiting
        // element of array.
        // Refer below post for details of unordered_set
        // https://www.geeksforgeeks.org/unorderd_set-stl-uses/
        let s = new Set();

        for (let i = 0; i < n; i++)
        {
            // If there exist an element in set s
            // with XOR equals to x^arr[i], that means
            // there exist an element such that the
            // XOR of element with arr[i] is equal to
            // x, then increment count.
            if (s.has(x ^ arr[i]) )
            {
                result++;
            }

            // Make element visited
            s.add(arr[i]);
        }

        // return total count of
        // pairs with XOR equal to x
        return result;
    }

    // Driver code
    let arr=[5, 4, 10, 15, 7, 6];
    let n = arr.length;
    let x = 5;
    document.write("Count of pairs with given XOR = "
                + xorPairCount(arr, n, x));



    // This code is contributed by unknown2108
</script>

Output

Count of pairs with given XOR = 1

Time complexity : O(n)
Auxiliary Space: O(n)

How to handle duplicates?
The above efficient solution doesn’t work if there are duplicates in the input array. For example, the above solution produces different results for {2, 2, 5} and {5, 2, 2}. To handle duplicates, we store counts of occurrences of all elements. We use unordered_map instead of unordered_set.  

Implementation:

C++

// C++ program to Count all pair with given XOR
// value x
#include<bits/stdc++.h>

using namespace std;

// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
int xorPairCount(int arr[], int n, int x)
{
    int result = 0; // Initialize result

    // create empty map that stores counts of
    // individual elements of array.
    unordered_map<int, int> m;

    for (int i=0; i<n ; i++)
    {
        int curr_xor =  x^arr[i];

        // If there exist an element in map m
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (m.find(curr_xor) != m.end())
            result += m[curr_xor];

        // Increment count of current element
        m[arr[i]]++;
    }


    // return total count of pairs with XOR equal to x
    return result;
}

// driver program
int main()
{
    int arr[] = {2, 5, 2};
    int n = sizeof(arr)/sizeof(arr[0]);
    int x = 0;
    cout << "Count of pairs with given XOR = "
         << xorPairCount(arr, n, x);
    return 0;
}

Java

// Java program to Count all pair with given XOR
// value x
import java.util.*;

class GFG
{

// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
static int xorPairCount(int arr[], int n, int x)
{
    int result = 0; // Initialize result

    // create empty map that stores counts of
    // individual elements of array.
    Map<Integer,Integer> m = new HashMap<>();

    for (int i = 0;  i < n ; i++)
    {
        int curr_xor = x^arr[i];

        // If there exist an element in map m
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (m.containsKey(curr_xor))
            result += m.get(curr_xor);

        // Increment count of current element
        if(m.containsKey(arr[i]))
        {
            m.put(arr[i], m.get(arr[i]) + 1);
        }
        else{
            m.put(arr[i], 1);
        }
    }
    // return total count of pairs with XOR equal to x
    return result;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = {2, 5, 2};
    int n = arr.length;
    int x = 0;
    System.out.println("Count of pairs with given XOR = "
        + xorPairCount(arr, n, x));
}
}

// This code has been contributed by 29AjayKumar

Python3

# Python3 program to Count all pair with
# given XOR value x

# Returns count of pairs in arr[0..n-1]
# with XOR value equals to x.
def xorPairCount(arr, n, x):

    result = 0 # Initialize result

    # create empty map that stores counts
    # of individual elements of array.
    m = dict()

    for i in range(n):

        curr_xor = x ^ arr[i]

        # If there exist an element in map m
        # with XOR equals to x^arr[i], that
        # means there exist an element such that
        # the XOR of element with arr[i] is equal
        # to x, then increment count.
        if (curr_xor in m.keys()):
            result += m[curr_xor]

        # Increment count of current element
        if arr[i] in m.keys():
            m[arr[i]] += 1
        else:
            m[arr[i]] = 1

    # return total count of pairs
    # with XOR equal to x
    return result

# Driver Code
arr = [2, 5, 2]
n = len(arr)
x = 0
print("Count of pairs with given XOR = ",
                 xorPairCount(arr, n, x))

# This code is contributed by Mohit Kumar

C#

// C# program to Count all pair with given XOR
// value x
using System;
using System.Collections.Generic;

class GFG
{

// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
static int xorPairCount(int []arr, int n, int x)
{
    int result = 0; // Initialize result

    // create empty map that stores counts of
    // individual elements of array.
    Dictionary<int,int> m = new Dictionary<int,int>();

    for (int i = 0; i < n ; i++)
    {
        int curr_xor = x^arr[i];

        // If there exist an element in map m
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (m.ContainsKey(curr_xor))
            result += m[curr_xor];

        // Increment count of current element
        if(m.ContainsKey(arr[i]))
        {
            var val = m[arr[i]];
            m.Remove(arr[i]);
            m.Add(arr[i], val + 1);
        }
        else
        {
            m.Add(arr[i], 1);
        }
    }

    // return total count of pairs with XOR equal to x
    return result;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = {2, 5, 2};
    int n = arr.Length;
    int x = 0;
    Console.WriteLine("Count of pairs with given XOR = "
        + xorPairCount(arr, n, x));
}
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

// Javascript program to Count all pair with given XOR
// value x

// Returns count of pairs in arr[0..n-1] with XOR
// value equals to x.
function xorPairCount(arr, n, x)
{
    let result = 0; // Initialize result

    // create empty map that stores counts of
    // individual elements of array.
    let m = new Map();

    for (let i = 0;  i < n ; i++)
    {
        let curr_xor = x^arr[i];

        // If there exist an element in map m
        // with XOR equals to x^arr[i], that means
        // there exist an element such that the
        // XOR of element with arr[i] is equal to
        // x, then increment count.
        if (m.has(curr_xor))
            result += m.get(curr_xor);

        // Increment count of current element
        if(m.has(arr[i]))
        {
            m.set(arr[i], m.get(arr[i]) + 1);
        }
        else{
            m.set(arr[i], 1);
        }
    }
    // return total count of pairs with XOR equal to x
    return result;
}

// Driver program

    let arr = [2, 5, 2];
    let n = arr.length;
    let x = 0;
    document.write("Count of pairs with given XOR = "
        + xorPairCount(arr, n, x));

</script>

Output

Count of pairs with given XOR = 1

Time complexity : O(n)
Auxiliary Space: O(n)

This article is contributed by Nishant_singh(pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.