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) = 1Input: N = 6
Output: 2
Approach:
- Find the highest power of 2 less than or equal to N and store it in a variable low.
- Find the smallest power of 2 greater than or equal to N and store it in a variable high.
- 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 <bits/stdc++.h>using namespace std;// Function to return the highest power// of 2 less than or equal to nint 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 nint 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 2int minDiff(int n){ int low = prevPowerof2(n); int high = nextPowerOf2(n); return min(n - low, high - n);}// Driver codeint 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 approachfrom math import log# Function to return the highest power# of 2 less than or equal to ndef prevPowerof2(n): p = int(log(n)) return pow(2, p)# Function to return the smallest power# of 2 greater than or equal to ndef 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 2def minDiff(n): low = prevPowerof2(n) high = nextPowerOf2(n) return min(n - low, high - n)# Driver coden = 6print(minDiff(n))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing 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 |
Javascript
<script>// Javascript implementation of the approach// Function to return the highest power// of 2 less than or equal to nfunction prevPowerof2(n){ var p = parseInt(Math.log(n)/Math.log(2)); return parseInt(Math.pow(2, p));}// Function to return the smallest power// of 2 greater than or equal to nfunction nextPowerOf2(n){ var 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 2function minDiff(n){ var low = prevPowerof2(n); var high = nextPowerOf2(n); return Math.min(n - low, high - n);}// Driver codevar n = 6;document.write(minDiff(n));// This code is contributed by noob2000.</script> |
Output
2
Time Complexity: O(log N)
Auxiliary Space: O(1)
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