Skip to content
Related Articles

Related Articles

Improve Article

Largest possible value of M not exceeding N having equal Bitwise OR and XOR between them

  • Difficulty Level : Easy
  • Last Updated : 30 Apr, 2021
Geek Week

Given an integer N, the task is to find the largest number M, where (M < N), such that N(XOR)M is equal to N(OR)M i.e. (N ^ M) = (N | M).

Examples: 
 

Input: N = 5 
Output:
5 ^ 4 = 1 and 5 | 4 = 5. Therefore, XOR and OR between them are not equal. 
5 ^ 3 = 6 and 5 | 3 = 7. Therefore, XOR and OR between them are not equal. 
5 ^ 2 = 7 and 5 | 2 = 7. Therefore, XOR and OR between them are equal.
Input: N = 14 
Output:
 

Approach: 
To get the required number M, traverse all the bits of N from its Least Significant Bit (LSB) to Most Significant Bit (MSB). Two cases arise here: 
 

  1. If the ith bit of N is 1 then: 
    • If the ith bit of M is set to 1, then N^M will not be equal to N|M as (1^1 = 0) and (1|1 = 1).
    • If the ith bit is set of M to 0, then N^M will be equal to N|M as (1^0 = 1) and (1|0 = 1).
    • So if the ith bit of N is 1, set the ith bit of M to 0.
  2. If the ith bit of N is 0 then: 
    • If the ith bit of M is set to 1, then N^M will be equal to N|M as (0^1 = 1) and (0|1 = 1).
    • If we set the ith bit of M to 0, then N^M will be equal to N|M as (0^0 = 0) and (0|0 = 0).
    • So, if the ith bit of M is set to either 0 or 1, N^M will always be equal to N|M.
    • As the largest value of M which is less than N has to be found out, always set the ith bit of M to 1.

Illustration: 
 



  • N = 5
  • 32-bit representation of 5 = 00000000000000000000000000000101
  • LSB index of 5 = 31
  • MSB index of 5 = 29
  • Traversing from LSB to MSB i.e. from 31 to 29:
    • For index 31, N[31] = 1. So M[31] should be set to 0.
    • For index 30, N[30] = 0. So M[30] should be set to 1.
    • For index 29, N[29] = 1. So M[29] should be set to 0.
  • Thus the 32-bit representation of M is 00000000000000000000000000000010, which is equal to 2 in decimal representation.

Below is the implementation of the above approach:
 

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find required
// number M
int equalXORandOR(int n)
{
    // Initialising m
    int m = 0;
 
    // Finding the index of the
    // 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 m;
}
 
// Driver Code
int main()
{
    int n = 14;
    cout << equalXORandOR(n);
    return 0;
}

Java




// Java program to implement
// the above approach
class GFG{
 
// Function to find required
// number M
static int equalXORandOR(int n)
{
     
    // Initialising m
    int m = 0;
 
    // Finding the index of the
    // 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 m;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 14;
     
    System.out.print(equalXORandOR(n));
}
}
 
// This code is contributed by amal kumar choubey

Python3




# Python3 program to implement
# the above approach
from math import log2
 
# Function to find required
# number M
def equalXORandOR(n):
 
    # Initialising m
    m = 0
 
    # Finding the index of the
    # most significant bit of N
    MSB = int(log2(n))
 
    # Calculating required number
    for i in range(MSB + 1):
        if(not(n & (1 << i))):
            m += (1 << i)
 
    return m
 
# Driver Code
n = 14
 
# Function call
print(equalXORandOR(n))
 
# This code is contributed by Shivam Singh

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to find required
// number M
static int equalXORandOR(int n)
{
     
    // Initialising m
    int m = 0;
 
    // Finding the index of the
    // 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 m;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 14;
     
    Console.Write(equalXORandOR(n));
}
}
 
// This code is contributed by amal kumar choubey

Javascript




<script>
 
// javascript program to implement
// the above approach   
// Function to find required
    // number M
    function equalXORandOR(n) {
 
        // Initialising m
        var m = 0;
 
        // Finding the index of the
        // 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 m;
    }
 
    // Driver Code
     
        var n = 14;
 
        document.write(equalXORandOR(n));
 
// This code contributed by Rajput-Ji
 
</script>
Output: 
1

 

Time Complexity: O(log2 N) 
Auxiliary Space: O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :