Sum of XOR of all pairs in an array
Given an array of n integers, find the sum of xor of all pairs of numbers in the array.
Examples :
Input : arr[] = {7, 3, 5}
Output : 12
7 ^ 3 = 4
3 ^ 5 = 6
7 ^ 5 = 2
Sum = 4 + 6 + 2
= 12
Input : arr[] = {5, 9, 7, 6}
Output : 47
5 ^ 9 = 12
9 ^ 7 = 14
7 ^ 6 = 1
5 ^ 7 = 2
5 ^ 6 = 3
9 ^ 6 = 15
Sum = 12 + 14 + 1 + 2 + 3 + 15
= 47
Naive Solution: A Brute Force approach is to run two loops and time complexity is O(n2).
C++
// A Simple C++ program to compute // sum of bitwise OR of all pairs #include <bits/stdc++.h> using namespace std; // Returns sum of bitwise OR // of all pairs int pairORSum( int arr[], int n) { int ans = 0; // Initialize result // Consider all pairs (arr[i], arr[j) such that // i < j for ( int i = 0; i < n; i++) for ( int j = i + 1; j < n; j++) ans += arr[i] ^ arr[j]; return ans; } // Driver program to test above function int main() { int arr[] = { 5, 9, 7, 6 }; int n = sizeof (arr) / sizeof (arr[0]); cout << pairORSum(arr, n) << endl; return 0; } |
Java
// A Simple Java program to compute // sum of bitwise OR of all pairs import java.io.*; class GFG { // Returns sum of bitwise OR // of all pairs static int pairORSum( int arr[], int n) { // Initialize result int ans = 0 ; // Consider all pairs (arr[i], arr[j) // such that i < j for ( int i = 0 ; i < n; i++) for ( int j = i + 1 ; j < n; j++) ans += arr[i] ^ arr[j]; return ans; } // Driver program to test above function public static void main (String[] args) { int arr[] = { 5 , 9 , 7 , 6 }; int n = arr.length; System.out.println(pairORSum(arr, arr.length)); } } //This code is contributed by vt_m |
Python3
# A Simple Python 3 program to compute # sum of bitwise OR of all pairs # Returns sum of bitwise OR # of all pairs def pairORSum(arr, n) : ans = 0 # Initialize result # Consider all pairs (arr[i], arr[j) # such that i < j for i in range ( 0 , n) : for j in range (i + 1 , n) : ans = ans + (arr[i] ^ arr[j]) return ans # Driver Code arr = [ 5 , 9 , 7 , 6 ] n = len (arr) print (pairORSum(arr, n)) # This code is contributed by Nikita Tiwari. |
C#
// A Simple C# program to compute // sum of bitwise OR of all pairs using System; class GFG { // Returns sum of bitwise OR // of all pairs static int pairORSum( int []arr, int n) { // Initialize result int ans = 0; // Consider all pairs (arr[i], arr[j) // such that i < j for ( int i = 0; i < n; i++) for ( int j = i + 1; j < n; j++) ans += arr[i] ^ arr[j]; return ans; } // Driver program to test above function public static void Main () { int []arr = { 5, 9, 7, 6 }; int n = arr.Length; Console.WriteLine(pairORSum(arr, arr.Length)); } } // This code is contributed by vt_m |
Javascript
// A Simple Javascript program to compute // sum of bitwise OR of all pairs // Returns sum of bitwise OR // of all pairs const pairORSum = (arr, n) => { let ans = 0; // Initialize result // Consider all pairs (arr[i], arr[j) such that // i < j for (let i = 0; i < n; i++) for (let j = i + 1; j < n; j++) ans += arr[i] ^ arr[j]; return ans; } // Driver program to test above function let arr = [5, 9, 7, 6]; let n = arr.length; document.write(pairORSum(arr, n)); // This code is contributed by rakeshsahni |
PHP
<?php // A Simple PHP program to compute // sum of bitwise OR of all pairs // Returns sum of bitwise OR // of all pairs function pairORSum( $arr , $n ) { // Initialize result $ans = 0; // Consider all pairs // (arr[i], arr[j) such that // i < j for ( $i = 0; $i < $n ; $i ++) for ( $j = $i + 1; $j < $n ; $j ++) $ans += $arr [ $i ] ^ $arr [ $j ]; return $ans ; } // Driver Code $arr = array ( 5, 9, 7, 6 ); $n = count ( $arr ); echo pairORSum( $arr , $n ) ; // This code is contributed by anuj_67. ?> |
47
Efficient Solution: An Efficient Solution can solve this problem in O(n) time. The assumption here is that integers are represented using 32 bits. Optimized solution will be to try bit manipulation. To implement the solution, we consider all bits which are 1 and which are 0 and store their count in two different variables. Next multiple those counts along with the power of 2 raised to that bit position. Do this for all the bit positions of the numbers. Their sum would be our answer.
How this actually works?
For example, look at the rightmost bit of all the numbers in the array. Suppose that numbers have a rightmost 0-bit, and b numbers have a 1-bit. Then out of the pairs, a*b of them will have 1 in the rightmost bit of the XOR. This is because there are a*b ways to choose one number that has a 0-bit and one that has a 1-bit. These bits will therefore contribute a*b towards the total of all the XORs.In general, when looking at the nth bit (where the rightmost bit is the 0th), count how many numbers have 0 (call this an) and how many have 1 (call this bn). The contribution towards the final sum will be an*bn*pow(2,n). You need to do this for each bit and sum all these contributions together.This can be done in O(kn) time, where k is the number of bits in the given values.
Explanation : arr[] = { 7, 3, 5 }
7 = 1 1 1
3 = 0 1 1
5 = 1 0 1
For bit position 0 :
Bits with zero = 0
Bits with one = 3
Answer = 0 * 3 * 2 ^ 0 = 0
Similarly, for bit position 1 :
Bits with zero = 1
Bits with one = 2
Answer = 1 * 2 * 2 ^ 1 = 4
Similarly, for bit position 2 :
Bits with zero = 1
Bits with one = 2
Answer = 1 * 2 * 2 ^ 2 = 8
Final answer = 0 + 4 + 8 = 12
CPP
// An efficient C++ program to compute // sum of bitwise OR of all pairs #include <bits/stdc++.h> using namespace std; // Returns sum of bitwise OR // of all pairs long long int sumXOR( int arr[], int n) { long long int sum = 0; for ( int i = 0; i < 32; i++) { // Count of zeros and ones int zc = 0, oc = 0; // Individual sum at each bit position long long int idsum = 0; for ( int j = 0; j < n; j++) { if (arr[j] % 2 == 0) zc++; else oc++; arr[j] /= 2; } // calculating individual bit sum idsum = oc * zc * (1 << i); // final sum sum += idsum; } return sum; } int main() { long long int sum = 0; int arr[] = { 5, 9, 7, 6 }; int n = sizeof (arr) / sizeof (arr[0]); sum = sumXOR(arr, n); cout << sum; return 0; } |
Java
// An efficient Java program to compute // sum of bitwise OR of all pairs import java.io.*; class GFG { // Returns sum of bitwise OR // of all pairs static long sumXOR( int arr[], int n) { long sum = 0 ; for ( int i = 0 ; i < 32 ; i++) { // Count of zeros and ones int zc = 0 , oc = 0 ; // Individual sum at each bit position long idsum = 0 ; for ( int j = 0 ; j < n; j++) { if (arr[j] % 2 == 0 ) zc++; else oc++; arr[j] /= 2 ; } // calculating individual bit sum idsum = oc * zc * ( 1 << i); // final sum sum += idsum; } return sum; } // Driver Code public static void main(String args[]) { long sum = 0 ; int arr[] = { 5 , 9 , 7 , 6 }; int n = arr.length; sum = sumXOR(arr, n); System.out.println(sum); } } // This code is contributed by Nikita Tiwari. |
Python3
# An efficient Python3 program to compute # sum of bitwise OR of all pair # Returns sum of bitwise OR # of all pairs def sumXOR( arr, n): sum = 0 for i in range ( 0 , 32 ): # Count of zeros and ones zc = 0 oc = 0 # Individual sum at each bit position idsum = 0 for j in range ( 0 , n): if (arr[j] % 2 = = 0 ): zc = zc + 1 else : oc = oc + 1 arr[j] = int (arr[j] / 2 ) # calculating individual bit sum idsum = oc * zc * ( 1 << i) # final sum sum = sum + idsum; return sum # driver function sum = 0 arr = [ 5 , 9 , 7 , 6 ] n = len (arr) sum = sumXOR(arr, n); print ( sum ) # This code is contributed by saloni1297 |
C#
// An efficient C# program to compute // sum of bitwise OR of all pairs using System; class GFG { // Returns sum of bitwise OR // of all pairs static long sumXOR( int []arr, int n) { long sum = 0; for ( int i = 0; i < 32; i++) { // Count of zeros and ones int zc = 0, oc = 0; // Individual sum at each bit position long idsum = 0; for ( int j = 0; j < n; j++) { if (arr[j] % 2 == 0) zc++; else oc++; arr[j] /= 2; } // calculating individual bit sum idsum = oc * zc * (1 << i); // final sum sum += idsum; } return sum; } // Driver Code public static void Main() { long sum = 0; int []arr = { 5, 9, 7, 6 }; int n = arr.Length; sum = sumXOR(arr, n); Console.WriteLine(sum); } } // This code is contributed by vt_m. |
JavaScript
<script> // An efficient JavaScript program to compute // sum of bitwise OR of all pairs // Returns sum of bitwise OR // of all pairs const sumXOR = (arr, n) => { let sum = 0; for (let i = 0; i < 32; i++) { // Count of zeros and ones let zc = 0, oc = 0; // Individual sum at each bit position let idsum = 0; for (let j = 0; j < n; j++) { if (arr[j] % 2 == 0) zc++; else oc++; arr[j] = parseInt(arr[j] / 2); } // calculating individual bit sum idsum = oc * zc * (1 << i); // final sum sum += idsum; } return sum; } let sum = 0; let arr = [5, 9, 7, 6]; let n = arr.length; sum = sumXOR(arr, n); document.write(sum); // This code is contributed by rakeshsahni </script> ?> |
PHP
<?php // An efficient PHP program to compute // sum of bitwise OR of all pairs // Returns sum of bitwise OR // of all pairs function sumXOR( $arr , $n ) { $sum = 0; for ( $i = 0; $i < 32; $i ++) { // Count of zeros and ones $zc = 0; $oc = 0; // Individual sum at each // bit position $idsum = 0; for ( $j = 0; $j < $n ; $j ++) { if ( $arr [ $j ] % 2 == 0) $zc ++; else $oc ++; $arr [ $j ] /= 2; } // calculating individual bit sum $idsum = $oc * $zc * (1 << $i ); // final sum $sum += $idsum ; } return $sum ; } // Driver code $sum = 0; $arr = array ( 5, 9, 7, 6 ); $n = count ( $arr ); $sum = sumXOR( $arr , $n ); echo $sum ; // This code is contributed by anuj_67 |
Output:
47