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Maximize the expression (A AND X) * (B AND X) | Bit Manipulation
  • Difficulty Level : Easy
  • Last Updated : 12 May, 2021

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:
(9 AND 8) * (8 AND 8) = 8 * 8 = 64 (maximum possible)
Input: A = 11 and B = 13 
Output:
 

 

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++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 32
 
// Function to find X according
// to the given conditions
int findX(int A, int B)
{
    int X = 0;
 
    // int can have 32 bits
    for (int bit = 0; bit < MAX; bit++) {
 
        // Temporary ith bit
        int tempBit = 1 << bit;
 
        // Compute ith bit of X according to
        // given conditions
        // Expression below is the direct
        // conclusion from the illustration
        // we had taken earlier
        int bitOfX = A & B & tempBit;
 
        // Add the ith bit of X to X
        X += bitOfX;
    }
 
    return X;
}
 
// Driver code
int main()
{
    int A = 11, B = 13;
 
    cout << findX(A, B);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
static int MAX = 32;
 
// Function to find X according
// to the given conditions
static int findX(int A, int B)
{
    int X = 0;
 
    // int can have 32 bits
    for (int bit = 0; bit < MAX; bit++)
    {
 
        // Temporary ith bit
        int tempBit = 1 << bit;
 
        // Compute ith bit of X according to
        // given conditions
        // Expression below is the direct
        // conclusion from the illustration
        // we had taken earlier
        int bitOfX = A & B & tempBit;
 
        // Add the ith bit of X to X
        X += bitOfX;
    }
    return X;
}
 
// Driver code
public static void main(String []args)
{
    int A = 11, B = 13;
 
    System.out.println(findX(A, B));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
MAX = 32
 
# Function to find X according
# to the given conditions
def findX(A, B) :
 
    X = 0;
 
    # int can have 32 bits
    for bit in range(MAX) :
 
        # Temporary ith bit
        tempBit = 1 << bit;
 
        # Compute ith bit of X according to
        # given conditions
        # Expression below is the direct
        # conclusion from the illustration
        # we had taken earlier
        bitOfX = A & B & tempBit;
 
        # Add the ith bit of X to X
        X += bitOfX;
 
    return X;
 
# Driver code
if __name__ == "__main__" :
     
    A = 11; B = 13;
    print(findX(A, B));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
     
class GFG
{
static int MAX = 32;
 
// Function to find X according
// to the given conditions
static int findX(int A, int B)
{
    int X = 0;
 
    // int can have 32 bits
    for (int bit = 0; bit < MAX; bit++)
    {
 
        // Temporary ith bit
        int tempBit = 1 << bit;
 
        // Compute ith bit of X according to
        // given conditions
        // Expression below is the direct
        // conclusion from the illustration
        // we had taken earlier
        int bitOfX = A & B & tempBit;
 
        // Add the ith bit of X to X
        X += bitOfX;
    }
    return X;
}
 
// Driver code
public static void Main(String []args)
{
    int A = 11, B = 13;
 
    Console.WriteLine(findX(A, B));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to find X according
// to the given conditions
function findX( A,  B)
{
    var X = 0;
    var MAX = 32;
 
   
    // int can have 32 bits
    for (var bit = 0; bit < MAX; bit++)
    {
   
        // Temporary ith bit
         
        var tempBit = 1 << bit;
   
        // Compute ith bit of X according to
        // given conditions
        // Expression below is the direct
        // conclusion from the illustration
        // we had taken earlier
         
        var bitOfX = A & B & tempBit;
   
        // Add the ith bit of X to X
        X += bitOfX;
    }
    return X;
}
   
// Driver code
    var A = 11, B = 13;
    document.write(findX(A, B));
   
  // This code is contributed by bunnyram19.
</script>
Output: 
9

 

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