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Sum of XOR of all pairs in an array in C++
In this problem, we are given an array arr[] of n integers. Our task is to create a program to find the sum of XOR of all pairs in an array.
Let’s take an example to understand the problem,
Input: arr[] = {5, 1, 4}
Output: 10
Explanation: the sum of all pairs:
5 ^ 1 = 4
1 ^ 4 = 5
5 ^ 4 = 1
sum = 4 + 5 + 1 = 10One simple approach to solve this problem is to run nested loops and find all pairs of numbers. Find XOR of each pair and add them to the sum.
Algorithm
Initialise sum = 0 Step 1: for(i -> 0 to n). Do: Step 1.1: for( j -> i to n ). Do: Step 1.1.1: update sum. i.e. sum += arr[i] ^ arr[j]. Step 2: return sum.
Example
Program to illustrate the working of our solution,
#include <iostream>
using namespace std;
int findXORSum(int arr[], int n) {
int sum = 0;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
sum += (arr[i]^arr[j]);
return sum;
}
int main() {
int arr[] = { 5, 1, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"Sum of XOR of all pairs in an array is "<<findXORSum(arr, n);
return 0;
}Output
Sum of XOR of all pairs in an array is 10
The time complexity of the algorithm is O(n2) and is not the most efficient solution to the problem.
An efficient solution to the problem is using the bit manipulation technique.
Here, we will consider bits of the number and at each position. And apply the below formula find the intermediate sum,
(number of set bits) * (number of unset bits) * (2^(bit_position))
To find the final sum, we will add the intermediate sum of all bits.
Our solution is for the 64-bit integer value. For this approach, we need the number of bits.
Algorithm
Initialize sum = 0, setBits = 0, unsetBits = 0. Step 1: Loop for i -> 0 to 64. repeat steps 2, 3. Step 2: Reset setBits and unsetBits to 0. Step 3: For each element of the array find the value of setBits and unsetBits. At ith position. Step 4: update sum += (setBits * unsetBits * (2i))
Example
Program to illustrate the working of our solution,
#include <iostream>
#include <math.h>
using namespace std;
long findXORSum(int arr[], int n) {
long sum = 0;
int unsetBits = 0, setBits = 0;
for (int i = 0; i < 32; i++) {
unsetBits = 0; setBits = 0;
for (int j = 0; j < n; j++) {
if (arr[j] % 2 == 0)
unsetBits++;
else
setBits++;
arr[j] /= 2;
}
sum += ( unsetBits*setBits* (pow(2,i)) );
}
return sum;
}
int main() {
int arr[] = { 5, 1, 4, 7, 9};
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"Sum of XOR of all pairs in an array is "<<findXORSum(arr, n);
return 0;
}Output
Sum of XOR of all pairs in an array is 68
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