Maximum possible Bitwise OR of the two numbers from the range [L, R]

Given a range [L, R], the task is to find the maximum bitwise OR of some pair (a, b) from the given range.

Examples:

Input: L = 10, R = 20
Output: 31



Input: L = 56, R = 77
Output: 127

Approach: First, convert both the given integers L and R to their binary representations. Now, If L has less number of bits than R then push zero to the MSB side of L to make the number of bits of both L and R to be equal.
Now from the MSB side compare the individual bits of both L and R. As R is greater than L, we will find the case when ith bit of R is 1 and ith bit of L is 0. So after ith bit, make all the bits of L to be 1. This ensures that the modifications done in the bits of L will not exceed R, so it will be between L and R only. And doing this also ensures maximum bitwise or.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximum bitwise
// OR of any pair from the given range
long long int max_bitwise_or(long long int L, long long int R)
{
    vector<long long int> v1, v2, v3;
    long long int z = 0, i, ans = 0, cnt = 1;
  
    // Converting L to its binary representation
    while (L > 0) {
        v1.push_back(L % 2);
        L = L / 2;
    }
  
    // Converting R to its binary representation
    while (R > 0) {
        v2.push_back(R % 2);
        R = R / 2;
    }
  
    // In order to make the number
    // of bits of L and R same
    while (v1.size() != v2.size()) {
  
        // Push 0 to the MSB
        v1.push_back(0);
    }
  
    for (i = v2.size() - 1; i >= 0; i--) {
  
        // When ith bit of R is 1
        // and ith bit of L is 0
        if (v2[i] == 1 && v1[i] == 0 && z == 0) {
  
            z = 1;
            continue;
        }
  
        // From MSB side set all bits of L to be 1
        if (z == 1) {
  
            // From (i+1)th bit, all bits
            // of L changed to be 1
            v1[i] = 1;
        }
    }
  
    for (i = 0; i < v2.size(); i++) {
        v3.push_back(v2[i] | v1[i]);
    }
  
    for (i = 0; i < v2.size(); i++) {
        if (v3[i] == 1) {
            ans += cnt;
        }
        cnt *= 2;
    }
    return ans;
}
  
// Driver code
int main()
{
    long long int L = 10, R = 20;
  
    cout << max_bitwise_or(L, R);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
class GFG 
{
  
// Function to return the maximum bitwise
// OR of any pair from the given range
static int max_bitwise_or(int L, int R)
{
    Vector<Integer> v1 = new Vector<Integer>(),
                    v2 = new Vector<Integer>(),
                    v3 = new Vector<Integer>();
  
    int z = 0, i, ans = 0, cnt = 1;
  
    // Converting L to its binary representation
    while (L > 0)
    {
        v1.add(L % 2);
        L = L / 2;
    }
  
    // Converting R to its binary representation
    while (R > 0)
    {
        v2.add(R % 2);
        R = R / 2;
    }
  
    // In order to make the number
    // of bits of L and R same
    while (v1.size() != v2.size())
    {
  
        // Push 0 to the MSB
        v1.add(0);
    }
  
    for (i = v2.size() - 1; i >= 0; i--) 
    {
  
        // When ith bit of R is 1
        // and ith bit of L is 0
        if (v2.get(i) == 1 && v1.get(i) == 0 && z == 0)
        {
            z = 1;
            continue;
        }
  
        // From MSB side set all bits of L to be 1
        if (z == 1
        {
  
            // From (i+1)th bit, all bits
            // of L changed to be 1
            v1.remove(i);
            v1.add(i,1);
        }
    }
  
    for (i = 0; i < v2.size(); i++)
    {
        v3.add(v2.get(i) | v1.get(i));
    }
  
    for (i = 0; i < v2.size(); i++)
    {
        if (v3.get(i) == 1
        {
            ans += cnt;
        }
        cnt *= 2;
    }
    return ans;
}
  
// Driver code
public static void main(String []args)
{
    int L = 10, R = 20;
  
    System.out.println(max_bitwise_or(L, R));
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 implementation of the approach
  
# Function to return the maximum bitwise
# OR of any pair from the given range
def max_bitwise_or(L, R):
    v1 = []
    v2 = []
    v3 = []
    z = 0
    i = 0
    ans = 0
    cnt = 1
  
    # Converting L to its binary representation
    while (L > 0):
        v1.append(L % 2)
        L = L // 2
  
    # Converting R to its binary representation
    while (R > 0):
        v2.append(R % 2)
        R = R // 2
  
    # In order to make the number
    # of bits of L and R same
    while (len(v1) != len(v2)):
  
        # Push 0 to the MSB
        v1.append(0)
  
    for i in range(len(v2) - 1, -1, -1):
  
        # When ith bit of R is 1
        # and ith bit of L is 0
        if (v2[i] == 1 and 
            v1[i] == 0 and z == 0):
            z = 1
            continue
  
        # From MSB side set all bits of L to be 1
        if (z == 1):
  
            # From (i+1)th bit, all bits
            # of L changed to be 1
            v1[i] = 1
  
    for i in range(len(v2)):
        v3.append(v2[i] | v1[i])
  
    for i in range(len(v2)):
        if (v3[i] == 1):
            ans += cnt
        cnt *= 2
  
    return ans
  
# Driver code
L = 10
R = 20
  
print(max_bitwise_or(L, R))
  
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
      
class GFG 
{
  
// Function to return the maximum bitwise
// OR of any pair from the given range
static int max_bitwise_or(int L, int R)
{
    List<int> v1 = new List<int>(),
              v2 = new List<int>(),
              v3 = new List<int>();
  
    int z = 0, i, ans = 0, cnt = 1;
  
    // Converting L to its binary representation
    while (L > 0)
    {
        v1.Add(L % 2);
        L = L / 2;
    }
  
    // Converting R to its binary representation
    while (R > 0)
    {
        v2.Add(R % 2);
        R = R / 2;
    }
  
    // In order to make the number
    // of bits of L and R same
    while (v1.Count != v2.Count)
    {
  
        // Push 0 to the MSB
        v1.Add(0);
    }
  
    for (i = v2.Count - 1; i >= 0; i--) 
    {
  
        // When ith bit of R is 1
        // and ith bit of L is 0
        if (v2[i] == 1 && v1[i] == 0 && z == 0)
        {
            z = 1;
            continue;
        }
  
        // From MSB side set all bits of L to be 1
        if (z == 1) 
        {
  
            // From (i+1)th bit, all bits
            // of L changed to be 1
            v1.RemoveAt(i);
            v1.Insert(i, 1);
        }
    }
  
    for (i = 0; i < v2.Count; i++)
    {
        v3.Add(v2[i] | v1[i]);
    }
  
    for (i = 0; i < v2.Count; i++)
    {
        if (v3[i] == 1) 
        {
            ans += cnt;
        }
        cnt *= 2;
    }
    return ans;
}
  
// Driver code
public static void Main(String []args)
{
    int L = 10, R = 20;
  
    Console.WriteLine(max_bitwise_or(L, R));
}
}
  
// This code is contributed by Rajput-Ji

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Output:

31


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