Maximize the expression (A AND X) * (B AND X) | Bit Manipulation
Last Updated :
24 Mar, 2022
Given two positive integers A and B such that A != B, the task is to find a positive integer X which maximizes the expression (A AND X) * (B AND X).
Example:
Input: A = 9 B = 8
Output: 8
(9 AND 8) * (8 AND 8) = 8 * 8 = 64 (maximum possible)
Input: A = 11 and B = 13
Output: 9
Naive approach: One can run a loop from 1 to max(A, B) and can easily find X which maximizes the given expression.
Efficient approach: It is known that,
(a – b)2 ? 0
which implies (a + b)2 – 4*a*b ? 0
which implies a * b ? (a + b)2 / 4
Hence, it concludes that a * b will be maximum when a * b = (a + b)2 / 4
which implies a = b
From the above result, (A AND X) * (B AND X) will be maximum when (A AND X) = (B AND X)
Now X can be found as:
A = 11 = 1011
B = 13 = 1101
X = ? = abcd
At 0th place: (1 AND d) = (1 AND d) implies d = 0, 1 but to maximize (A AND X) * (B AND X) d = 1
At 1st place: (1 AND d) = (0 AND d) implies c = 0
At 2nd place: (0 AND d) = (1 AND d) implies b = 0
At 3rd place: (1 AND d) = (1 AND d) implies a = 0, 1 but to maximize (A AND X) * (B AND X) a = 1
Hence, X = 1001 = 9
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
#define MAX 32
int findX( int A, int B)
{
int X = 0;
for ( int bit = 0; bit < MAX; bit++) {
int tempBit = 1 << bit;
int bitOfX = A & B & tempBit;
X += bitOfX;
}
return X;
}
int main()
{
int A = 11, B = 13;
cout << findX(A, B);
return 0;
}
|
C
#include <stdio.h>
#define MAX 32
int findX( int A, int B)
{
int X = 0;
for ( int bit = 0; bit < MAX; bit++) {
int tempBit = 1 << bit;
int bitOfX = A & B & tempBit;
X += bitOfX;
}
return X;
}
int main()
{
int A = 11, B = 13;
printf ( "%d" , findX(A, B));
return 0;
}
|
Java
class GFG
{
static int MAX = 32 ;
static int findX( int A, int B)
{
int X = 0 ;
for ( int bit = 0 ; bit < MAX; bit++)
{
int tempBit = 1 << bit;
int bitOfX = A & B & tempBit;
X += bitOfX;
}
return X;
}
public static void main(String []args)
{
int A = 11 , B = 13 ;
System.out.println(findX(A, B));
}
}
|
Python3
MAX = 32
def findX(A, B) :
X = 0 ;
for bit in range ( MAX ) :
tempBit = 1 << bit;
bitOfX = A & B & tempBit;
X + = bitOfX;
return X;
if __name__ = = "__main__" :
A = 11 ; B = 13 ;
print (findX(A, B));
|
C#
using System;
class GFG
{
static int MAX = 32;
static int findX( int A, int B)
{
int X = 0;
for ( int bit = 0; bit < MAX; bit++)
{
int tempBit = 1 << bit;
int bitOfX = A & B & tempBit;
X += bitOfX;
}
return X;
}
public static void Main(String []args)
{
int A = 11, B = 13;
Console.WriteLine(findX(A, B));
}
}
|
Javascript
<script>
function findX( A, B)
{
var X = 0;
var MAX = 32;
for ( var bit = 0; bit < MAX; bit++)
{
var tempBit = 1 << bit;
var bitOfX = A & B & tempBit;
X += bitOfX;
}
return X;
}
var A = 11, B = 13;
document.write(findX(A, B));
</script>
|
Time Complexity: O(MAX)
Auxiliary Space: O(1)
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