Given an array arr[] of non-negative integers, the task is to find an integer X such that (arr[0] XOR X) + (arr[1] XOR X) + … + arr[n – 1] XOR X is minimum possible.
Examples:
Input: arr[] = {3, 9, 6, 2, 4}
Output: X = 2, Sum = 22
Input: arr[] = {6, 56, 78, 34}
Output: X = 2, Sum = 170
Approach: We will check ‘i’th bit of every number of the array in binary representation and count those numbers containing that ‘i’th bit set to ‘1’ because these set bits will contribute to maximize the sum rather than minimize. So we have to make this set ‘i’th bit to ‘0’ if the count is greater than N/2 and if the count is less than N/2 then the numbers having ‘i’th bit set are less and so it will not affect the answer. According to XOR operation on two bits, we know that when A XOR B and both A and B are the same then it gives the result as ‘0’ so we will make that ‘i’th bit in our number (num) to ‘1’, so that (1 XOR 1) will give ‘0’ and minimize the sum.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>#include <cmath>using namespace std;
// Function to find an integer X such that// the sum of all the array elements after// getting XORed with X is minimumvoid findX(int arr[], int n)
{ // Finding Maximum element of array
int* itr = max_element(arr, arr + n);
// Find Maximum number of bits required
// in the binary representation
// of maximum number
// so log2 is calculated
int p = log2(*itr) + 1;
// Running loop from p times which is
// the number of bits required to represent
// all the elements of the array
int X = 0;
for (int i = 0; i < p; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
// If the bits in same position are set
// then count
if (arr[j] & (1 << i)) {
count++;
}
}
// If count becomes greater than half of
// size of array then we need to make
// that bit '0' by setting X bit to '1'
if (count > (n / 2)) {
// Again using shift operation to calculate
// the required number
X += 1 << i;
}
}
// Calculate minimized sum
long long int sum = 0;
for (int i = 0; i < n; i++)
sum += (X ^ arr[i]);
// Print solution
cout << "X = " << X << ", Sum = " << sum;
}// Driver codeint main()
{ int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
findX(arr, n);
return 0;
} |
Java
// Java implementation of above approachimport java.lang.Math;
import java.util.*;
class GFG
{ // Function to find an integer X such that
// the sum of all the array elements after
// getting XORed with X is minimum
public static void findX(int[] a, int n)
{
// Finding Maximum element of array
Collections.sort(Arrays.asList(a), null);
int itr = a[n-1];
// Find Maximum number of bits required
// in the binary representation
// of maximum number
// so log2 is calculated
int p = (int)(Math.log(itr)/Math.log(2)) + 1;
// Running loop from p times which is
// the number of bits required to represent
// all the elements of the array
int x = 0;
for (int i = 0; i < p; i++)
{
int count = 0;
for (int j = 0; j < n; j++)
{
// If the bits in same position are set
// then count
if ((a[j] & (1 << i)) != 0)
count++;
}
// If count becomes greater than half of
// size of array then we need to make
// that bit '0' by setting X bit to '1'
if (count > (n / 2))
{
// Again using shift operation to calculate
// the required number
x += 1 << i;
}
}
// Calculate minimized sum
long sum = 0;
for (int i = 0; i < n; i++)
sum += (x ^ a[i]);
// Print solution
System.out.println("X = " + x + ", Sum = " + sum);
}
// Driver Code
public static void main(String[] args)
{
int[] a = {2, 3, 4, 5, 6};
int n = a.length;
findX(a, n);
}
}// This code is contributed by// sanjeev2552 |
Python3
# Python 3 implementation of the approachfrom math import log2
# Function to find an integer X such that# the sum of all the array elements after# getting XORed with X is minimumdef findX(arr, n):
# Finding Maximum element of array
itr = arr[0]
for i in range(len(arr)):
# Find Maximum number of bits required
# in the binary representation
# of maximum number
# so log2 is calculated
if(arr[i] > itr):
itr = arr[i]
p = int(log2(itr)) + 1
# Running loop from p times which is
# the number of bits required to represent
# all the elements of the array
X = 0
for i in range(p):
count = 0
for j in range(n):
# If the bits in same position are set
# then increase count
if (arr[j] & (1 << i)):
count += 1
# If count becomes greater than half of
# size of array then we need to make
# that bit '0' by setting X bit to '1'
if (count > int(n / 2)):
# Again using shift operation to calculate
# the required number
X += 1 << i
# Calculate minimized sum
sum = 0
for i in range(n):
sum += (X ^ arr[i])
# Print solution
print("X =", X, ", Sum =", sum)
# Driver codeif __name__=='__main__':
arr = [2, 3, 4, 5, 6]
n = len(arr)
findX(arr, n)
# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of above approach using System;
using System.Linq;
class GFG
{ // Function to find an integer X such that
// the sum of all the array elements after
// getting XORed with X is minimum
public static void findX(int[] a, int n)
{
// Finding Maximum element of array
int itr = a.Max();
// Find Maximum number of bits required
// in the binary representation
// of maximum number
// so log2 is calculated
int p = (int) Math.Log(itr, 2) + 1;
// Running loop from p times which is
// the number of bits required to represent
// all the elements of the array
int x = 0;
for (int i = 0; i < p; i++)
{
int count = 0;
for (int j = 0; j < n; j++)
{
// If the bits in same position are set
// then count
if ((a[j] & (1 << i)) != 0)
count++;
}
// If count becomes greater than half of
// size of array then we need to make
// that bit '0' by setting X bit to '1'
if (count > (n / 2))
{
// Again using shift operation to calculate
// the required number
x += 1 << i;
}
}
// Calculate minimized sum
long sum = 0;
for (int i = 0; i < n; i++)
sum += (x ^ a[i]);
// Print solution
Console.Write("X = " + x + ", Sum = " + sum);
}
// Driver Code
public static void Main(String[] args)
{
int[] a = {2, 3, 4, 5, 6};
int n = a.Length;
findX(a, n);
}
} // This code is contributed by ravikishor |
Javascript
<script>// Javascript implementation of the approach// Function to find an integer X such that// the sum of all the array elements after// getting XORed with X is minimumfunction findX(arr, n)
{ // Finding Maximum element of array
let itr = Math.max(...arr);
// Find Maximum number of bits required
// in the binary representation
// of maximum number
// so log2 is calculated
let p = parseInt(Math.log(itr) / Math.log(2)) + 1;
// Running loop from p times which is
// the number of bits required to represent
// all the elements of the array
let X = 0;
for (let i = 0; i < p; i++) {
let count = 0;
for (let j = 0; j < n; j++) {
// If the bits in same position are set
// then count
if (arr[j] & (1 << i)) {
count++;
}
}
// If count becomes greater than half of
// size of array then we need to make
// that bit '0' by setting X bit to '1'
if (count > parseInt(n / 2)) {
// Again using shift
// operation to calculate
// the required number
X += 1 << i;
}
}
// Calculate minimized sum
let sum = 0;
for (let i = 0; i < n; i++)
sum += (X ^ arr[i]);
// Print solution
document.write("X = " + X + ", Sum = " + sum);
}// Driver code let arr = [ 2, 3, 4, 5, 6 ];
let n = arr.length;
findX(arr, n);
</script> |
Output:
X = 6, Sum = 14
Time Complexity: O(N * log(A))
Auxiliary Space: O(1)