# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

Output:

```2
```

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