Minimum absolute difference between N and any power of 2

Given a positive integer N, the task is to find the minimum absolute difference between N and any power of 2.

Examples:

Input: N = 3
Output: 1
Smaller power of 2 nearest to 3 is 2, abs(3 – 2) = 1
Higher power of 2 nearest to 3 is 4, abs(4 – 3) = 1



Input: N = 6
Output: 2

Approach:

  1. Find the highest power of 2 less than or equal to N and store it in a variable low.
  2. Find the smallest power of 2 greater than or equal to N and store it in a variable high.
  3. Now, the answer will be max(N – low, high – N).

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 highest power
// of 2 less than or equal to n
int prevPowerof2(int n)
{
    int p = (int)log2(n);
    return (int)pow(2, p);
}
  
// Function to return the smallest power
// of 2 greater than or equal to n
int nextPowerOf2(int n)
{
    int p = 1;
    if (n && !(n & (n - 1)))
        return n;
  
    while (p < n)
        p <<= 1;
  
    return p;
}
  
// Function that returns the minimum
// absolute difference between n
// and any power of 2
int minDiff(int n)
{
    int low = prevPowerof2(n);
    int high = nextPowerOf2(n);
  
    return min(n - low, high - n);
}
  
// Driver code
int main()
{
    int n = 6;
  
    cout << minDiff(n);
  
    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 highest power 
    // of 2 less than or equal to n 
    static int prevPowerof2(int n) 
    
        int p = (int)(Math.log(n) / Math.log(2)); 
          
        return (int)Math.pow(2, p); 
    
      
    // Function to return the smallest power 
    // of 2 greater than or equal to n 
    static int nextPowerOf2(int n) 
    
        int p = 1
        if ((n == 0) && !((n & (n - 1)) == 0))
            return n; 
      
        while (p < n) 
            p <<= 1
      
        return p; 
    
      
    // Function that returns the minimum 
    // absolute difference between n 
    // and any power of 2 
    static int minDiff(int n) 
    
        int low = prevPowerof2(n); 
        int high = nextPowerOf2(n); 
      
        return Math.min(n - low, high - n); 
    
      
    // Driver code 
    public static void main (String[] args) 
    
        int n = 6
      
        System.out.println(minDiff(n)); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach
from math import log
  
# Function to return the highest power
# of 2 less than or equal to n
def prevPowerof2(n):
    p = int(log(n))
    return pow(2, p)
  
# Function to return the smallest power
# of 2 greater than or equal to n
def nextPowerOf2(n):
    p = 1
    if (n and (n & (n - 1)) == 0):
        return n
  
    while (p < n):
        p <<= 1
  
    return p
  
# Function that returns the minimum
# absolute difference between n
# and any power of 2
def minDiff(n):
    low = prevPowerof2(n)
    high = nextPowerOf2(n)
  
    return min(n - low, high - n)
  
# Driver code
n = 6
  
print(minDiff(n))
  
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of the approach
using System; 
  
class GFG 
{
      
    // Function to return the highest power 
    // of 2 less than or equal to n 
    static int prevPowerof2(int n) 
    
        int p = (int)(Math.Log(n) / Math.Log(2)); 
          
        return (int)Math.Pow(2, p); 
    
      
    // Function to return the smallest power 
    // of 2 greater than or equal to n 
    static int nextPowerOf2(int n) 
    
        int p = 1; 
        if ((n == 0) && !((n & (n - 1)) == 0))
            return n; 
      
        while (p < n) 
            p <<= 1; 
      
        return p; 
    
      
    // Function that returns the minimum 
    // absolute difference between n 
    // and any power of 2 
    static int minDiff(int n) 
    
        int low = prevPowerof2(n); 
        int high = nextPowerOf2(n); 
      
        return Math.Min(n - low, high - n); 
    
      
    // Driver code 
    public static void Main (String []args) 
    
        int n = 6; 
      
        Console.WriteLine(minDiff(n)); 
    
}
  
// This code is contributed by Arnab Kundu

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Output:

2


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