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Smallest number greater or equals to N such that it has no odd positioned bit set

  • Last Updated : 10 May, 2021
Geek Week

Given an integer N, the task is to find the smallest integer X such that it has no odd position set and X ≥ N
Note: The positioning of bits is assumed from the right side and the first bit is assumed to be the 0th bit. 

Examples:  

Input: N = 9 
Output: 16 
16’s binary representation is 10000, which has its 4th bit 
set which is the smallest number possible satisfying the given condition. 
Input: N = 5 
Output: 5
Input: N = 19 
Output: 20 

Approach: The problem can be solved using a greedy approach and some bit properties. The property that if smaller powers of two are taken exactly once and added up they can never exceed a higher power of two (e.g., (1 + 2 + 4) < 8). The following greedy approach is used to solve the above problem: 
 

  • Initially count the number of bits.
  • Get the leftmost index of the set bit.
  • If the leftmost set bit is at an odd index, then answer will always be (1 << (leftmost_bit_index + 1)).
  • Else, greedily form a number by setting all the even bits from 0 to leftmost_bit_index. Now greedily remove a power of two from right to check if we get a number that satisfies the given conditions.
  • If the above condition does not give us our number, then we simply set the next leftmost even bit and return the answer.

Below is the implementation of the above approach: 
 



C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the total bits
int countBits(int n)
{
    int count = 0;
 
    // Iterate and find the
    // number of set bits
    while (n) {
        count++;
 
        // Right shift the number by 1
        n >>= 1;
    }
    return count;
}
 
// Function to find the nearest number
int findNearestNumber(int n)
{
 
    // Count the total number of bits
    int cnt = countBits(n);
 
    // To get the position
    cnt -= 1;
 
    // If the last set bit is
    // at odd position then
    // answer will always be a number
    // with the left bit set
    if (cnt % 2) {
        return 1 << (cnt + 1);
    }
 
    else {
 
        int tempnum = 0;
 
        // Set all the even bits which
        // are possible
        for (int i = 0; i <= cnt; i += 2)
            tempnum += 1 << i;
 
        // If the number still is less than N
        if (tempnum < n) {
 
            // Return the number by setting the
            // next even set bit
            return (1 << (cnt + 2));
        }
 
        else if (tempnum == n)
            return n;
 
        // If we have reached this position
        // it means tempsum > n
        // hence turn off even bits to get the
        // first possible number
        for (int i = 0; i <= cnt; i += 2) {
 
            // Turn off the bit
            tempnum -= (1 << i);
 
            // If it gets lower than N
            // then set it and return that number
            if (tempnum < n)
                return tempnum += (1 << i);
        }
    }
}
 
// Driver code
int main()
{
    int n = 19;
    cout << findNearestNumber(n);
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    // Function to count the total bits
    static int countBits(int n)
    {
        int count = 0;
 
        // Iterate and find the
        // number of set bits
        while (n > 0)
        {
            count++;
 
            // Right shift the number by 1
            n >>= 1;
        }
        return count;
    }
 
    // Function to find the nearest number
    static int findNearestNumber(int n)
    {
 
        // Count the total number of bits
        int cnt = countBits(n);
 
        // To get the position
        cnt -= 1;
 
        // If the last set bit is
        // at odd position then
        // answer will always be a number
        // with the left bit set
        if (cnt % 2 == 1)
        {
            return 1 << (cnt + 1);
        }
        else
        {
 
            int tempnum = 0;
 
            // Set all the even bits which
            // are possible
            for (int i = 0; i <= cnt; i += 2)
            {
                tempnum += 1 << i;
            }
 
            // If the number still is less than N
            if (tempnum < n)
            {
 
                // Return the number by setting the
                // next even set bit
                return (1 << (cnt + 2));
            }
            else
            if (tempnum == n)
            {
                return n;
            }
 
            // If we have reached this position
            // it means tempsum > n
            // hence turn off even bits to get the
            // first possible number
            for (int i = 0; i <= cnt; i += 2)
            {
 
                // Turn off the bit
                tempnum -= (1 << i);
 
                // If it gets lower than N
                // then set it and return that number
                if (tempnum < n)
                {
                    return tempnum += (1 << i);
                }
            }
        }
        return Integer.MIN_VALUE;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 19;
 
        System.out.println(findNearestNumber(n));
    }
}
 
// This code is contributed by 29AjayKumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to count the total bits
    static int countBits(int n)
    {
        int count = 0;
 
        // Iterate and find the
        // number of set bits
        while (n > 0)
        {
            count++;
 
            // Right shift the number by 1
            n >>= 1;
        }
        return count;
    }
 
    // Function to find the nearest number
    static int findNearestNumber(int n)
    {
 
        // Count the total number of bits
        int cnt = countBits(n);
 
        // To get the position
        cnt -= 1;
 
        // If the last set bit is
        // at odd position then
        // answer will always be a number
        // with the left bit set
        if (cnt % 2 == 1)
        {
            return 1 << (cnt + 1);
        }
        else
        {
 
            int tempnum = 0;
 
            // Set all the even bits which
            // are possible
            for (int i = 0; i <= cnt; i += 2)
            {
                tempnum += 1 << i;
            }
 
            // If the number still is less than N
            if (tempnum < n)
            {
 
                // Return the number by setting the
                // next even set bit
                return (1 << (cnt + 2));
            }
            else
            if (tempnum == n)
            {
                return n;
            }
 
            // If we have reached this position
            // it means tempsum > n
            // hence turn off even bits to get the
            // first possible number
            for (int i = 0; i <= cnt; i += 2)
            {
 
                // Turn off the bit
                tempnum -= (1 << i);
 
                // If it gets lower than N
                // then set it and return that number
                if (tempnum < n)
                {
                    return tempnum += (1 << i);
                }
            }
        }
        return int.MinValue;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 19;
 
        Console.WriteLine(findNearestNumber(n));
    }
}
 
// This code is contributed by anuj_67..

Python3




# Python implementation of the above approach
 
  
# Function to count the total bits
def countBits(n):
    count = 0;
  
    # Iterate and find the
    # number of set bits
    while (n>0):
        count+=1;
  
        # Right shift the number by 1
        n >>= 1;
    return count;
  
# Function to find the nearest number
def findNearestNumber(n):
  
    # Count the total number of bits
    cnt = countBits(n);
  
    # To get the position
    cnt -= 1;
  
    # If the last set bit is
    # at odd position then
    # answer will always be a number
    # with the left bit set
    if (cnt % 2):
        return 1 << (cnt + 1);
  
    else:
  
        tempnum = 0;
  
        # Set all the even bits which
        # are possible
        for i in range(0,cnt+1,2):
            tempnum += 1 << i;
  
        # If the number still is less than N
        if (tempnum < n):
  
            # Return the number by setting the
            # next even set bit
            return (1 << (cnt + 2));
  
        elif (tempnum == n):
            return n;
  
        # If we have reached this position
        # it means tempsum > n
        # hence turn off even bits to get the
        # first possible number
        for i in range(0,cnt+1,2):
  
            # Turn off the bit
            tempnum -= (1 << i);
  
            # If it gets lower than N
            # then set it and return that number
            if (tempnum < n):
                tempnum += (1 << i);
                return tempnum;
# Driver code
n = 19;
print(findNearestNumber(n));
 
# This code contributed by PrinciRaj1992

Javascript




<script>
// Javascript implementation of the above approach
 
// Function to count the total bits
function countBits(n)
{
    let count = 0;
 
    // Iterate and find the
    // number of set bits
    while (n) {
        count++;
 
        // Right shift the number by 1
        n >>= 1;
    }
    return count;
}
 
// Function to find the nearest number
function findNearestNumber(n)
{
 
    // Count the total number of bits
    let cnt = countBits(n);
 
    // To get the position
    cnt -= 1;
 
    // If the last set bit is
    // at odd position then
    // answer will always be a number
    // with the left bit set
    if (cnt % 2) {
        return 1 << (cnt + 1);
    }
 
    else {
 
        let tempnum = 0;
 
        // Set all the even bits which
        // are possible
        for (let i = 0; i <= cnt; i += 2)
            tempnum += 1 << i;
 
        // If the number still is less than N
        if (tempnum < n) {
 
            // Return the number by setting the
            // next even set bit
            return (1 << (cnt + 2));
        }
 
        else if (tempnum == n)
            return n;
 
        // If we have reached this position
        // it means tempsum > n
        // hence turn off even bits to get the
        // first possible number
        for (let i = 0; i <= cnt; i += 2) {
 
            // Turn off the bit
            tempnum -= (1 << i);
 
            // If it gets lower than N
            // then set it and return that number
            if (tempnum < n)
                return tempnum += (1 << i);
        }
    }
}
 
// Driver code
    let n = 19;
    document.write(findNearestNumber(n));
 
</script>
Output: 
20

 

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