Minimum absolute difference between N and a power of 2

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

Examples:

Input: N = 4
Output: 0
Power of 2 closest to 4 is 4. Therefore the minimum difference possible is 0.

Input: N = 9
Output: 1
Power of 2 closest to 9 is 8 and 9 – 8 = 1

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

Approach: Find the power of 2 closest to N on its left, left = 2floor(log2(N)) then the closest power of 2 on N’s right will be left * 2. Now the minimum absolute difference will be the minimum of N – left and right – N.

Below is the implementation of the above approach:

C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum difference  ` `// between N and a power of 2 ` `int` `minAbsDiff(``int` `n) ` `{ ` `    ``// Power of 2 closest to n on its left ` `    ``int` `left = ``pow``(2, ``floor``(log2(n))); ` ` `  `    ``// Power of 2 closest to n on its right ` `    ``int` `right = left * 2; ` ` `  `    ``// Return the minimum abs difference ` `    ``return` `min((n - left), (right - n)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 15; ` `    ``cout << minAbsDiff(n); ` `    ``return` `0; ` `} `

Java

 `// Java implementation of the above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the minimum difference  ` `// between N and a power of 2 ` `static` `int` `minAbsDiff(``int` `n) ` `{ ` `    ``// Power of 2 closest to n on its left ` `    ``int` `left = (``int``)Math.pow(``2``, (``int``)(Math.log(n) / ` `                                ``Math.log(``2``))); ` ` `  `    ``// Power of 2 closest to n on its right ` `    ``int` `right = left * ``2``; ` ` `  `    ``// Return the minimum abs difference ` `    ``return` `Math.min((n - left), (right - n)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `n = ``15``; ` `    ``System.out.println(minAbsDiff(n)); ` `} ` `} ` ` `  `// This code is contributed by  ` `// Surendra_Gangwar `

Python3

 `# Python3 implementation of the  ` `# above approach  ` ` `  `# from math lib import floor ` `# and log2 function ` `from` `math ``import` `floor, log2 ` ` `  `# Function to return the minimum  ` `# difference between N and a power of 2  ` `def` `minAbsDiff(n) : ` `     `  `    ``# Power of 2 closest to n on its left  ` `    ``left ``=` `pow``(``2``, floor(log2(n))) ` ` `  `    ``# Power of 2 closest to n on its right  ` `    ``right ``=` `left ``*` `2` ` `  `    ``# Return the minimum abs difference  ` `    ``return` `min``((n ``-` `left), (right ``-` `n))  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``n ``=` `15` `    ``print``(minAbsDiff(n)) ` ` `  `# This code is contributed by Ryuga `

C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Function to return the minimum difference  ` `// between N and a power of 2 ` `static` `double` `minAbsDiff(``double` `n) ` `{ ` `    ``// Power of 2 closest to n on its left ` `    ``double` `left = Math.Pow(2,  ` `                ``Math.Floor(Math.Log(n, 2))); ` ` `  `    ``// Power of 2 closest to n on its right ` `    ``double` `right = left * 2; ` ` `  `    ``// Return the minimum abs difference ` `    ``return` `Math.Min((n - left), (right - n)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``double` `n = 15; ` `    ``Console.Write(minAbsDiff(n)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Akanksha Rai `

PHP

 `

Output:

```1
```

We can use left shift operator to optimize the implementation.

C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum difference  ` `// between N and a power of 2 ` `int` `minAbsDiff(``int` `n) ` `{ ` `    ``// Power of 2 closest to n on its left ` `    ``int` `left = 1 << ((``int``)``floor``(log2(n))); ` ` `  `    ``// Power of 2 closest to n on its right ` `    ``int` `right = left * 2; ` ` `  `    ``// Return the minimum abs difference ` `    ``return` `min((n - left), (right - n)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 15; ` `    ``cout << minAbsDiff(n); ` `    ``return` `0; ` `} `

Java

 `// Java implementation of the above approach ` `class` `GFG ` `{ ` `     `  `// Function to return the minimum difference  ` `// between N and a power of 2 ` `static` `int` `minAbsDiff(``int` `n) ` `{ ` `    ``// Power of 2 closest to n on its left ` `    ``int` `left = ``1` `<< ((``int``)Math.floor(Math.log(n) / Math.log(``2``))); ` ` `  `    ``// Power of 2 closest to n on its right ` `    ``int` `right = left * ``2``; ` ` `  `    ``// Return the minimum abs difference ` `    ``return` `Math.min((n - left), (right - n)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `n = ``15``; ` `    ``System.out.println(minAbsDiff(n)); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

Python3

 `# Python3 implementation of the ` `# above approach ` `import` `math ` ` `  `# Function to return the minimum  ` `# difference between N and a power of 2 ` `def` `minAbsDiff(n): ` `     `  `    ``# Power of 2 closest to n on its left ` `    ``left ``=` `1` `<< (``int``)(math.floor(math.log2(n))) ` ` `  `    ``# Power of 2 closest to n on its right ` `    ``right ``=` `left ``*` `2` ` `  `    ``# Return the minimum abs difference ` `    ``return` `min``((n ``-` `left), (right ``-` `n)) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``n ``=` `15` `    ``print``(minAbsDiff(n)) ` ` `  `# This code is contributed  ` `# by 29AjayKumar `

C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `public` `class` `GFG ` `{ ` ` `  `// Function to return the minimum difference  ` `// between N and a power of 2 ` `static` `int` `minAbsDiff(``int` `n) ` `{ ` `    ``// Power of 2 closest to n on its left ` `    ``int` `left = 1 << ((``int``)Math.Floor(Math.Log(n) / Math.Log(2))); ` ` `  `    ``// Power of 2 closest to n on its right ` `    ``int` `right = left * 2; ` ` `  `    ``// Return the minimum abs difference ` `    ``return` `Math.Min((n - left), (right - n)); ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` `    ``int` `n = 15; ` `    ``Console.WriteLine(minAbsDiff(n)); ` `} ` `} ` ` `  `// This code is contributed by jit_t. `

PHP

 ` `

Output:

```1
```

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