Maximum XOR of Two Binary Numbers after shuffling their bits.
Last Updated :
27 Mar, 2024
Given two n-bit integers a and b, find the maximum possible value of (a’ xor b’) where a’ is a shuffle-a, b’ is a shuffle-b and xor is the bitwise xor operator.
Note: Consider a n-bit integer a. We call an integer a’ as shuffle-a if a’ can be obtained by shuffling the bits of a in its binary representation. For eg. if n = 5 and a = 6 = (00110)2, a’ can be any 5-bit integer having exactly two 1s in it i.e., any of (00011)2, (00101)2, (00110)2, (01010)2, …., (11000)2.
Examples:
Input: n = 3, a = 5, b = 4
Output: 7
Explanation: a = (101)2, b = (100)2, a’ = (110)2 , b’ = (001)2 and a’ xor b’ = (111)2
Input: n = 5, a = 0, b = 1
Output: 16
Explanation: a = (00000)2 , b = (00001)2 a’ = (00000)2 , b = (10000)2 and a’ xor b’ = (10000)2
Approach: To solve the problem follow the below steps:
- Count the number of set bits in both a and b.
- Then, compute minimum (set bit count in a, zeroes in b) and minimum (set bit count in b, zeroes in a).
- The summation(sum) of these two values gives the total number of set bits possible in a’ xor b’.
The above formula works for every possible a and b because:
- We want to maximize our answer, hence we find out the maximum number of 1’s it can contain. In XOR operations, if a particular bit position has a 1 in one number and a 0 in the other number, the result will have a 1 in that position (1^0=1 and 0^1=1). So, we find the maximum number of 1-0 and 0-1 pairs. The function min(1’s in a , 0’s in b) gives the maximum 1-0 pairs possible and the function min(0’s in a , 1’s in b) gives the maximum 0-1 pairs possible. Their sum will give the maximum number of 1’s possible in our answer.
- Example: n = 4, a = 10, b = 13 a’ = 1010, b’ = 1101
- Minimum(set bit count in a, zeroes in b) = 1, Minimum(set bit count in b, zeroes in a) = 2
- Total set bits in a’ xor b'(sum) -> 1 + 2 = 3
- One of the possible a’ and b’ combinations is a’=1001, b’= 0111 a’ xor b’ = 1110 = 14, which is the maximum possible value.
- It is clear that 3 is the total number of bits to maximize xor.
- Inside the for loop, it computes the maximum possible value of a’ xor b’ by placing all the set bits(= to sum) in the beginning followed by zeroes
Below is the implementation of the above approach:
C++
// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
int MaximumXor(int a, int b, int n)
{
int count1 = 0, count2 = 0;
// Count the number of set bits in a
while (a) {
if (a & 1)
count1++;
a = a >> 1;
}
// Count the number of set bits in b
while (b) {
if (b & 1)
count2++;
b = b >> 1;
}
// Minimum of set bit count in a
// and zeroes in b
int m1 = min(count1, n - count2);
// Minimum of set bit count in b and
// zeroes in a
int m2 = min(n - count1, count2);
// The total number of set bits in
// a' xor b'
int sum = (m1 + m2);
double XOR = 0;
// Compute the maximum possible
// value of a' xor b'
for (int i = 1; i <= sum; i++) {
XOR = XOR + pow(2, n - 1);
n--;
}
return (int)XOR;
}
// Drivers code
int main()
{
int n, a, b;
n = 3, a = 5, b = 4;
// Functional call
cout << MaximumXor(a, b, n) << "\n";
return 0;
}
import java.util.*;
public class Main {
public static void main(String[] args) {
int n, a, b;
n = 3;
a = 5;
b = 4;
// Functional call
System.out.println(MaximumXor(a, b, n));
}
public static int MaximumXor(int a, int b, int n) {
int count1 = 0, count2 = 0;
// Count the number of set bits in a
while (a != 0) {
if ((a & 1) == 1)
count1++;
a = a >> 1;
}
// Count the number of set bits in b
while (b != 0) {
if ((b & 1) == 1)
count2++;
b = b >> 1;
}
// Minimum of set bit count in a
// and zeroes in b
int m1 = Math.min(count1, n - count2);
// Minimum of set bit count in b and
// zeroes in a
int m2 = Math.min(n - count1, count2);
// The total number of set bits in
// a' xor b'
int sum = (m1 + m2);
double XOR = 0;
// Compute the maximum possible
// value of a' xor b'
for (int i = 1; i <= sum; i++) {
XOR = XOR + Math.pow(2, n - 1);
n--;
}
return (int) XOR;
}
}
using System;
namespace GFG
{
class Geek
{
static int MaximumXor(int a, int b, int n)
{
int count1 = 0, count2 = 0;
// Count the number of
// set bits in a
while (a > 0)
{
if ((a & 1) == 1)
count1++;
a >>= 1;
}
// Count the number of
// set bits in b
while (b > 0)
{
if ((b & 1) == 1)
count2++;
b >>= 1;
}
// Minimum of set bit count in a
// and zeroes in b
int m1 = Math.Min(count1, n - count2);
// Minimum of set bit count in b and
// zeroes in a
int m2 = Math.Min(n - count1, count2);
int sum = (m1 + m2);
double XOR = 0;
// Compute the maximum possible
// value of a' xor b'
for (int i = 1; i <= sum; i++)
{
XOR += Math.Pow(2, n - 1);
n--;
}
return (int)XOR;
}
static void Main(string[] args)
{
int n = 3, a = 5, b = 4;
// Functional call
Console.WriteLine(MaximumXor(a, b, n));
}
}
}
// Javascript code for the above approach
function MaximumXor(a, b, n) {
let count1 = 0,
count2 = 0;
// Count the number of set bits in a
while (a !== 0) {
if ((a & 1) === 1)
count1++;
a = a >> 1;
}
// Count the number of set bits in b
while (b !== 0) {
if ((b & 1) === 1)
count2++;
b = b >> 1;
}
// Minimum of set bit count in a
// and zeroes in b
let m1 = Math.min(count1, n - count2);
// Minimum of set bit count in b and
// zeroes in a
let m2 = Math.min(n - count1, count2);
// The total number of set bits in
// a' xor b'
let sum = (m1 + m2);
let XOR = 0;
// Compute the maximum possible
// value of a' xor b'
for (let i = 1; i <= sum; i++) {
XOR = XOR + Math.pow(2, n - 1);
n--;
}
return XOR;
}
// Drivers code
let n, a, b;
n = 3;
a = 5;
b = 4;
// Functional call
console.log(MaximumXor(a, b, n));
// This code is contributed by ragul21
def MaximumXor(a, b, n):
count1 = 0
count2 = 0
# Count the number of set bits in a
while a:
if a & 1:
count1 += 1
a = a >> 1
# Count the number of set bits in b
while b:
if b & 1:
count2 += 1
b = b >> 1
# Minimum of set bit count in a
# and zeroes in b
m1 = min(count1, n - count2)
# Minimum of set bit count in b and
# zeroes in a
m2 = min(n - count1, count2)
# The total number of set bits in
# a' xor b'
sum = m1 + m2
XOR = 0
# Compute the maximum possible
# value of a' xor b'
for i in range(1, sum + 1):
XOR = XOR + pow(2, n - 1)
n -= 1
return int(XOR)
# Drivers code
n = 3
a = 5
b = 4
# Functional call
print(MaximumXor(a, b, n))
# This code is contributed by Sakshi
Time complexity: O(n), where n is the length of the binary number
Auxiliary Space: O(1) as constant space is used.
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