Given a natural number N, the task is to find a number M smaller than N such that the difference between their bitwise XOR (N ^ M) and bitwise AND (N & M) is maximum.
Examples:
Input: N = 4
Output: 3
Explanation:
(4 ^ 0) – (4 & 0) = 4
(4 ^ 1) – (4 & 1) = 5
(4 ^ 2) – (4 & 2) = 6
(4 ^ 3) – (4 & 3) = 7
Hence, the value of M is 3.Input: N = 6
Output: 1
Explanation:
The difference between N ^ M and N & M is maximum when M = 1.
Naive Approach: The idea is to iterate for every element less than N and find M for which N^M – N&M is maximum. Below are the steps:
- Initialize a variable let’s say maxDiff with 0 and M with -1.
- Iterate from 0 to N-1 and calculate diff = N^i -N&i.
- If diff is greater or equal to maxDiff assign M = i and maxDiff = diff.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to return M<N such that // N^M - N&M is maximum int getMaxDifference( int N)
{ // Initialize variables
int M = -1;
int maxDiff = 0;
// Iterate for all values < N
for ( int i = 0; i < N; i++) {
// Find the difference between
// Bitwise XOR and AND
int diff = (N ^ i) - (N & i);
// Check if new difference is
// greater than previous maximum
if (diff >= maxDiff) {
// Update variables
maxDiff = diff;
M = i;
}
}
// Return the answer
return M;
} // Driver Code int main()
{ // Given Number N
int N = 6;
// Function Call
cout << getMaxDifference(N);
return 0;
} |
// Java program for the above approach class GFG{
// Function to return M<N such that // N^M - N&M is maximum static int getMaxDifference( int N)
{ // Initialize variables
int M = - 1 ;
int maxDiff = 0 ;
// Iterate for all values < N
for ( int i = 0 ; i < N; i++)
{
// Find the difference between
// Bitwise XOR and AND
int diff = (N ^ i) - (N & i);
// Check if new difference is
// greater than previous maximum
if (diff >= maxDiff)
{
// Update variables
maxDiff = diff;
M = i;
}
}
// Return the answer
return M;
} // Driver Code public static void main(String[] args)
{ // Given Number N
int N = 6 ;
// Function Call
System.out.print(getMaxDifference(N));
} } // This code is contributed by Rohit_ranjan |
# Python3 program for the above approach # Function to return M<N such that # N^M - N&M is maximum def getMaxDifference(N):
# Initialize variables
M = - 1 ;
maxDiff = 0 ;
# Iterate for all values < N
for i in range (N):
# Find the difference between
# Bitwise XOR and AND
diff = (N ^ i) - (N & i);
# Check if new difference is
# greater than previous maximum
if (diff > = maxDiff):
# Update variables
maxDiff = diff;
M = i;
# Return the answer
return M;
# Driver Code if __name__ = = '__main__' :
# Given number N
N = 6 ;
# Function call
print (getMaxDifference(N));
# This code is contributed by amal kumar choubey |
// C# program for the above approach using System;
class GFG{
// Function to return M<N such that // N^M - N&M is maximum static int getMaxDifference( int N)
{ // Initialize variables
int M = -1;
int maxDiff = 0;
// Iterate for all values < N
for ( int i = 0; i < N; i++)
{
// Find the difference between
// Bitwise XOR and AND
int diff = (N ^ i) - (N & i);
// Check if new difference is
// greater than previous maximum
if (diff >= maxDiff)
{
// Update variables
maxDiff = diff;
M = i;
}
}
// Return the answer
return M;
} // Driver Code public static void Main(String[] args)
{ // Given number N
int N = 6;
// Function call
Console.Write(getMaxDifference(N));
} } // This code is contributed by Rajput-Ji |
<script> // javascript program for the above approach // Function to return M<N such that // N^M - N&M is maximum
function getMaxDifference(N) {
// Initialize variables
var M = -1;
var maxDiff = 0;
// Iterate for all values < N
for (i = 0; i < N; i++) {
// Find the difference between
// Bitwise XOR and AND
var diff = (N ^ i) - (N & i);
// Check if new difference is
// greater than previous maximum
if (diff >= maxDiff) {
// Update variables
maxDiff = diff;
M = i;
}
}
// Return the answer
return M;
}
// Driver Code
// Given Number N
var N = 6;
// Function Call
document.write(getMaxDifference(N));
// This code contributed by aashish1995 </script> |
1
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: The idea is to observe that the difference between Bitwise XOR and Bitwise AND is maximum if Bitwise AND of the two numbers is the minimum possible number and the minimum possible number is 0.
The Bitwise AND between the two numbers is zero if and only if they complement each other. Therefore, the possible value of M must be the complement of the given number N.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to flip all bits of N int findM( int N)
{ int M = 0;
// Finding most significant bit of N
int MSB = ( int )log2(N);
// Calculating required number
for ( int i = 0; i < MSB; i++) {
if (!(N & (1 << i)))
M += (1 << i);
}
// Return the answer
return M;
} // Driver Code int main()
{ // Given Number
int N = 6;
// Function Call
cout << findM(N);
return 0;
} |
// Java program for the above approach class GFG{
// Function to flip all bits of N static int findM( int N)
{ int M = 0 ;
// Finding most significant bit of N
int MSB = ( int )Math.log(N);
// Calculating required number
for ( int i = 0 ; i < MSB; i++)
{
if ((N & ( 1 << i)) == 0 )
M += ( 1 << i);
}
// Return the answer
return M;
} // Driver Code public static void main(String[] args)
{ // Given number
int N = 6 ;
// Function call
System.out.print(findM(N));
} } // This code is contributed by Rajput-Ji |
# Python3 program for the above approach import math
# Function to flip all bits of N def findM(N):
M = 0 ;
# Finding most significant bit of N
MSB = int (math.log(N));
# Calculating required number
for i in range (MSB):
if ((N & ( 1 << i)) = = 0 ):
M + = ( 1 << i);
# Return the answer
return M;
# Driver Code if __name__ = = '__main__' :
# Given number
N = 6 ;
# Function call
print (findM(N));
# This code is contributed by Amit Katiyar |
// C# program for the above approach using System;
class GFG{
// Function to flip all bits of N static int findM( int N)
{ int M = 0;
// Finding most significant bit of N
int MSB = ( int )Math.Log(N);
// Calculating required number
for ( int i = 0; i < MSB; i++)
{
if ((N & (1 << i)) == 0)
M += (1 << i);
}
// Return the answer
return M;
} // Driver Code public static void Main(String[] args)
{ // Given number
int N = 6;
// Function call
Console.Write(findM(N));
} } // This code is contributed by Amit Katiyar |
<script> // javascript program for the above approach // Function to flip all bits of N function findM(N) {
var M = 0;
// Finding most significant bit of N
var MSB = parseInt( Math.log(N));
// Calculating required number
for (i = 0; i < MSB; i++) {
if ((N & (1 << i)) == 0)
M += (1 << i);
}
// Return the answer
return M;
}
// Driver Code
// Given number
var N = 6;
// Function call
document.write(findM(N));
// This code is contributed by Rajput-Ji </script> |
1
Time Complexity: O(log2N)
Auxiliary Space: O(1)