Smallest number greater or equals to N such that it has no odd positioned bit set
Last Updated :
05 Apr, 2023
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++
#include <bits/stdc++.h>
using namespace std;
int countBits( int n)
{
int count = 0;
while (n) {
count++;
n >>= 1;
}
return count;
}
int findNearestNumber( int n)
{
int cnt = countBits(n);
cnt -= 1;
if (cnt % 2) {
return 1 << (cnt + 1);
}
else {
int tempnum = 0;
for ( int i = 0; i <= cnt; i += 2)
tempnum += 1 << i;
if (tempnum < n) {
return (1 << (cnt + 2));
}
else if (tempnum == n)
return n;
for ( int i = 0; i <= cnt; i += 2) {
tempnum -= (1 << i);
if (tempnum < n)
return tempnum += (1 << i);
}
}
}
int main()
{
int n = 19;
cout << findNearestNumber(n);
}
|
Java
import java.util.*;
class GFG
{
static int countBits( int n)
{
int count = 0 ;
while (n > 0 )
{
count++;
n >>= 1 ;
}
return count;
}
static int findNearestNumber( int n)
{
int cnt = countBits(n);
cnt -= 1 ;
if (cnt % 2 == 1 )
{
return 1 << (cnt + 1 );
}
else
{
int tempnum = 0 ;
for ( int i = 0 ; i <= cnt; i += 2 )
{
tempnum += 1 << i;
}
if (tempnum < n)
{
return ( 1 << (cnt + 2 ));
}
else
if (tempnum == n)
{
return n;
}
for ( int i = 0 ; i <= cnt; i += 2 )
{
tempnum -= ( 1 << i);
if (tempnum < n)
{
return tempnum += ( 1 << i);
}
}
}
return Integer.MIN_VALUE;
}
public static void main(String[] args)
{
int n = 19 ;
System.out.println(findNearestNumber(n));
}
}
|
C#
using System;
class GFG
{
static int countBits( int n)
{
int count = 0;
while (n > 0)
{
count++;
n >>= 1;
}
return count;
}
static int findNearestNumber( int n)
{
int cnt = countBits(n);
cnt -= 1;
if (cnt % 2 == 1)
{
return 1 << (cnt + 1);
}
else
{
int tempnum = 0;
for ( int i = 0; i <= cnt; i += 2)
{
tempnum += 1 << i;
}
if (tempnum < n)
{
return (1 << (cnt + 2));
}
else
if (tempnum == n)
{
return n;
}
for ( int i = 0; i <= cnt; i += 2)
{
tempnum -= (1 << i);
if (tempnum < n)
{
return tempnum += (1 << i);
}
}
}
return int .MinValue;
}
public static void Main()
{
int n = 19;
Console.WriteLine(findNearestNumber(n));
}
}
|
Python3
def countBits(n):
count = 0 ;
while (n> 0 ):
count + = 1 ;
n >> = 1 ;
return count;
def findNearestNumber(n):
cnt = countBits(n);
cnt - = 1 ;
if (cnt % 2 ):
return 1 << (cnt + 1 );
else :
tempnum = 0 ;
for i in range ( 0 ,cnt + 1 , 2 ):
tempnum + = 1 << i;
if (tempnum < n):
return ( 1 << (cnt + 2 ));
elif (tempnum = = n):
return n;
for i in range ( 0 ,cnt + 1 , 2 ):
tempnum - = ( 1 << i);
if (tempnum < n):
tempnum + = ( 1 << i);
return tempnum;
n = 19 ;
print (findNearestNumber(n));
|
Javascript
<script>
function countBits(n)
{
let count = 0;
while (n) {
count++;
n >>= 1;
}
return count;
}
function findNearestNumber(n)
{
let cnt = countBits(n);
cnt -= 1;
if (cnt % 2) {
return 1 << (cnt + 1);
}
else {
let tempnum = 0;
for (let i = 0; i <= cnt; i += 2)
tempnum += 1 << i;
if (tempnum < n) {
return (1 << (cnt + 2));
}
else if (tempnum == n)
return n;
for (let i = 0; i <= cnt; i += 2) {
tempnum -= (1 << i);
if (tempnum < n)
return tempnum += (1 << i);
}
}
}
let n = 19;
document.write(findNearestNumber(n));
</script>
|
Time Complexity: O(log N)
Auxiliary Space: O(1)
Another Approach: Another approach to solving this problem is to check each number starting from N until we find a number that satisfies the given conditions. Bitwise operators can be used to check if the number has odd-positioned bits set or not.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findNearestNumber( int n)
{
while ( true ) {
bool hasOddBitSet = false ;
int temp = n;
while (temp) {
if (temp & 0x2)
hasOddBitSet = true ;
temp >>= 2;
}
if (!hasOddBitSet)
return n;
n++;
}
}
int main()
{
int n = 19;
cout << findNearestNumber(n);
}
|
C#
using System;
public class Program
{
public static int FindNearestNumber( int n)
{
while ( true )
{
bool hasOddBitSet = false ;
int temp = n;
while (temp > 0)
{
if ((temp & 0x2) != 0)
{
hasOddBitSet = true ;
}
temp >>= 2;
}
if (!hasOddBitSet)
{
return n;
}
n++;
}
}
public static void Main()
{
int n = 19;
Console.WriteLine(FindNearestNumber(n));
}
}
|
Java
import java.util.*;
public class Main {
public static int findNearestNumber( int n) {
while ( true ) {
boolean hasOddBitSet = false ;
int temp = n;
while (temp != 0 ) {
if ((temp & 0x2 ) != 0 )
hasOddBitSet = true ;
temp >>= 2 ;
}
if (!hasOddBitSet)
return n;
n++;
}
}
public static void main(String[] args) {
int n = 19 ;
System.out.println(findNearestNumber(n));
}
}
|
Javascript
function findNearestNumber(n) {
while ( true ) {
let hasOddBitSet = false ;
let temp = n;
while (temp) {
if (temp & 0x2)
hasOddBitSet = true ;
temp >>= 2;
}
if (!hasOddBitSet)
return n;
n++;
}
}
let n = 19;
console.log(findNearestNumber(n));
|
Python3
def findNearestNumber(n):
while True :
hasOddBitSet = False
temp = n
while temp ! = 0 :
if temp & 0x2 :
hasOddBitSet = True
temp >> = 2
if not hasOddBitSet:
return n
n + = 1
n = 19
print (findNearestNumber(n))
|
Time Complexity: The time complexity of the given approach is O(log N) in the worst-case scenario. This is because we only need to iterate through the bits of the number once to count the total number of bits, and then perform constant time operations on the bits. The worst-case scenario would occur if we need to iterate through all the bits of the number to find the leftmost set bit.
Auxiliary Space: The auxiliary space used by the given approach is O(1) as we are not using any additional data structures, and all the operations are performed on the given number itself. Therefore, the space complexity is constant.
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