Find a number which give minimum sum when XOR with every number of array of integers
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 minimum void 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 code int 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 approach import 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 approach from math import log2 # Function to find an integer X such that # the sum of all the array elements after # getting XORed with X is minimum def 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 code if __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 minimum function 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> |
X = 6, Sum = 14
Time Complexity: O(N * log(A))
Auxiliary Space: O(1)
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