Count all pairs with given XOR - GeeksforGeeks

05 Nov, 2021
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.
C++
Java
Python3
C#
Javascript
// 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 programint 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 program to Count all pair with// given XOR value ximport 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 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 Codeif __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# program to Count all pair with// given XOR value xusing 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 */
<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.  
C++
Java
Python3
C#
Javascript
// 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 programint 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 program to Count all pair with given XOR// value ximport 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 codepublic 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 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 Codearr = [2, 5, 2]n = len(arr)x = 0print("Count of pairs with given XOR = ",                 xorPairCount(arr, n, x)) # This code is contributed by Mohit Kumar
// C# program to Count all pair with given XOR// value xusing 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 codepublic 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
<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)
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