Given an integer N, the task is to find the highest power of 2 that is smaller than or equal to N.
Examples:
Input: N = 9
Output: 8
Explanation:
Highest power of 2 less than 9 is 8.Input: N = -20
Output: -32
Explanation:
Highest power of 2 less than -20 is -32.Input: N = -84
Output: -128
Approach: The idea is to use logarithm to solve the above problem. For any given number N, it can be either positive or negative. The following task can be performed for each case:
- If the input is positive: floor(log2(N)) can be calculated.
- If the input is negative: ceil(log2(N)) can be calculated and a -ve sign can be added to the value.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to return the lowest power of 2 close to given// positive numberint powOfPositive(int n){ // Floor function is used to determine the value close // to the number int pos = floor(log2(n)); return pow(2, pos);}// Function to return the lowest power of 2 close to given// negative numberint powOfNegative(int n){ // Ceil function is used for negative numbers as -1 > // -4. It would be opposite to positive numbers where 1 < 4 int pos = ceil(log2(n)); return (-1 * pow(2, pos));}// Function to find the highest power of 2void highestPowerOf2(int n){ // To check if the given number is positive or negative if (n > 0) cout << powOfPositive(n); else { // If the number is negative, then the ceil of the // positive number is calculated and negative sign // is added n = -n; cout << powOfNegative(n); }}// Driver codeint main(){ int n = -24; highestPowerOf2(n); return 0;}// This code is contributed by Sania Kumari Gupta |
C
// C implementation of the above approach#include <math.h>#include <stdio.h>// Function to return the lowest power of 2 close to given// positive numberint powOfPositive(int n){ // Floor function is used to determine the value close // to the number int pos = floor(log2(n)); return pow(2, pos);}// Function to return the lowest power of 2 close to given// negative numberint powOfNegative(int n){ // Ceil function is used for negative numbers as -1 > // -4. It would be opposite to positive numbers where 1 < 4 int pos = ceil(log2(n)); return (-1 * pow(2, pos));}// Function to find the highest power of 2void highestPowerOf2(int n){ // To check if the given number is positive or negative if (n > 0) printf("%d", powOfPositive(n)); else { // If the number is negative, then the ceil of the // positive number is calculated and negative sign // is added n = -n; printf("%d", powOfNegative(n)); }}// Driver codeint main(){ int n = -24; highestPowerOf2(n); return 0;}// This code is contributed by Sania Kumari Gupta |
Java
// Java implementation of the above approach class GFG { // Function to return the lowest power // of 2 close to given positive number static int powOfPositive(int n) { // Floor function is used to determine // the value close to the number int pos = (int)Math.floor((Math.log(n)/Math.log(2))); return (int)Math.pow(2, pos); } // Function to return the lowest power // of 2 close to given negative number static int powOfNegative(int n) { // Ceil function is used for negative numbers // as -1 > -4. It would be opposite // to positive numbers where 1 < 4 int pos = (int)Math.ceil((Math.log(n)/Math.log(2))); return (int)(-1 * Math.pow(2, pos)); } // Function to find the highest power of 2 static void highestPowerOf2(int n) { // To check if the given number // is positive or negative if (n > 0) { System.out.println(powOfPositive(n)); } else { // If the number is negative, // then the ceil of the positive number // is calculated and // negative sign is added n = -n; System.out.println(powOfNegative(n)); } } // Driver code public static void main (String[] args) { int n = -24; highestPowerOf2(n); } }// This code is contributed by AnkitRai01 |
C#
// C# implementation of the above approach using System;class GFG { // Function to return the lowest power // of 2 close to given positive number static int powOfPositive(int n) { // Floor function is used to determine // the value close to the number int pos = (int)Math.Floor((Math.Log(n)/Math.Log(2))); return (int)Math.Pow(2, pos); } // Function to return the lowest power // of 2 close to given negative number static int powOfNegative(int n) { // Ceil function is used for negative numbers // as -1 > -4. It would be opposite // to positive numbers where 1 < 4 int pos = (int)Math.Ceiling((Math.Log(n)/Math.Log(2))); return (int)(-1 * Math.Pow(2, pos)); } // Function to find the highest power of 2 static void highestPowerOf2(int n) { // To check if the given number // is positive or negative if (n > 0) { Console.WriteLine(powOfPositive(n)); } else { // If the number is negative, // then the ceil of the positive number // is calculated and // negative sign is added n = -n; Console.WriteLine(powOfNegative(n)); } } // Driver code public static void Main() { int n = -24; highestPowerOf2(n); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the above approach from math import floor,ceil,log2# Function to return the lowest power # of 2 close to given positive number def powOfPositive(n) : # Floor function is used to determine # the value close to the number pos = floor(log2(n)); return 2**pos; # Function to return the lowest power # of 2 close to given negative number def powOfNegative(n) : # Ceil function is used for negative numbers # as -1 > -4. It would be opposite # to positive numbers where 1 < 4 pos = ceil(log2(n)); return (-1 * pow(2, pos)); # Function to find the highest power of 2 def highestPowerOf2(n) : # To check if the given number # is positive or negative if (n > 0) : print(powOfPositive(n)); else : # If the number is negative, # then the ceil of the positive number # is calculated and # negative sign is added n = -n; print(powOfNegative(n)); # Driver code if __name__ == "__main__" : n = -24; highestPowerOf2(n); # This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the above approach // Function to return the lowest power// of 2 close to given positive numberfunction powOfPositive(n){ // Floor function is used to determine // the value close to the number let pos = Math.floor(Math.log2(n)); return Math.pow(2, pos);}// Function to return the lowest power// of 2 close to given negative numberfunction powOfNegative(n){ // Ceil function is used for negative numbers // as -1 > -4. It would be opposite // to positive numbers where 1 < 4 let pos = Math.ceil(Math.log2(n)); return (-1 * Math.pow(2, pos));}// Function to find the highest power of 2function highestPowerOf2(n){ // To check if the given number // is positive or negative if (n > 0) { document.write(powOfPositive(n)); } else { // If the number is negative, // then the ceil of the positive number // is calculated and // negative sign is added n = -n; document.write(powOfNegative(n)); }}// Driver codelet n = -24;highestPowerOf2(n);// This code is contributed by mohit kumar 29</script> |
Output
-32
Time Complexity: O(log2(log2n))
Auxiliary Space: O(1)
Another Approach: we can use Bits manipulation in this way :
1.If number is positive- Then our answer will be pow(2,i) where i is leftmost set bit . Because if number is equal to pow(2,i), Then number is already power of 2, and if number > pow(2,i) , any power of 2 can not be greater than pow(2,i) as in bit concepts.
2.If number is negative- Let i is leftmost set bit .Then our answer will be pow(2,i) if(pow(2,i)==num). else our answer will be pow(2,i+1) because we are taking in negative.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function find highestpowerof2 <= nvoid highestPowerOf2(int n){ // leftmost- keep track of leftmost set bit int set_bits = 0, leftmost, highest; // iterate all sits of integer for (int i = 0; i <= 32; i++) { int val = pow(2, i); int Bitwise_AND = val & abs(n); // Bitwise AND if (Bitwise_AND > 0) // If Bitwise_AND > 0 means sit is set { leftmost = i; // update leftmost set bit highest = pow( 2, leftmost); // Update // highest(highestpowerof2) set_bits++; // count set bits } if (val >= abs(n)) { break; } // if we have found leftmost set bit ,break } if (n < 0) // make +highest to -highest { if (set_bits > 1) // means -n=-pow(2,i)-pow(2,j)..... { highest = pow(2, leftmost + 1); } highest = -1 * highest; } cout << highest; // Print highestPowerOf2 <= n}// Driver codeint main(){ int n = -24; // Function call highestPowerOf2(n); return 0;}// This Approach is contributed by nikhilsainiofficial546 |
Java
import java.lang.Math;class Main { // Function to find highest power of 2 <= n static void highestPowerOf2(int n) { int set_bits = 0, leftmost = 0, highest = 0; // iterate over all bits of integer for (int i = 0; i <= 31; i++) { int val = (int) Math.pow(2, i); int Bitwise_AND = val & Math.abs(n); // Bitwise AND if (Bitwise_AND > 0) { // If Bitwise_AND > 0 means bit is set leftmost = i; // update leftmost set bit highest = (int) Math.pow(2, leftmost); // Update highest(highest power of 2) set_bits++; // count set bits } if (val >= Math.abs(n)) { break; } // if we have found leftmost set bit, break } if (n < 0) { // make +highest to -highest if (set_bits > 1) { // means -n=-pow(2,i)-pow(2,j)..... highest = (int) Math.pow(2, leftmost + 1); } highest = -1 * highest; } System.out.println(highest); // Print highest power of 2 <= n } // Driver code public static void main(String[] args) { int n = -24; highestPowerOf2(n); // Function call }} |
Python3
import mathdef highestPowerOf2(n): set_bits = 0 leftmost = 0 highest = 0 for i in range(32): # iterate over all bits of integer val = int(math.pow(2, i)) Bitwise_AND = val & abs(n) # Bitwise AND if (Bitwise_AND > 0): # If Bitwise_AND > 0 means bit is set leftmost = i # update leftmost set bit highest = int(math.pow(2, leftmost)) # Update highest(highest power of 2) set_bits += 1 # count set bits if (val >= abs(n)): break # if we have found leftmost set bit, break if (n < 0): # make +highest to -highest if (set_bits > 1): # means -n=-pow(2,i)-pow(2,j)..... highest = int(math.pow(2, leftmost + 1)) highest = -1 * highest print(highest) # Print highest power of 2 <= n# Driver coden = -24highestPowerOf2(n) # Function call |
C#
// C# implementation of the above approachusing System;class GFG { // Function find highestpowerof2 <= n static void highestPowerOf2(int n) { // leftmost- keep track of leftmost set bit int set_bits = 0, leftmost = 0, highest = 0; // iterate all sits of integer for (int i = 0; i <= 32; i++) { int val = (int)Math.Pow(2, i); int bitwiseAnd = val & Math.Abs(n); // Bitwise AND // If Bitwise_AND > 0 means sit is set if (bitwiseAnd > 0) { leftmost = i; // update leftmost set bit // Update highest(highestpowerof2) highest = (int)Math.Pow(2, leftmost); // count set bits set_bits++; } // if we have found leftmost set bit ,break if (val >= Math.Abs(n)) { break; } } // make +highest to -highest if (n < 0) { if (set_bits > 1) { highest = (int)Math.Pow( 2, leftmost + 1); // means // -n=-pow(2,i)-pow(2,j)..... } highest = -1 * highest; } Console.WriteLine(highest); } // Driver code public static void Main() { int n = -24; // Function call highestPowerOf2(n); }} |
Javascript
// JavaScript implementation of the above approach// Function to find highest power of 2 <= nfunction highestPowerOf2(n){ let set_bits = 0, leftmost, highest; // Iterate all bits of the integer for (let i = 0; i <= 32; i++) { let val = Math.pow(2, i); let Bitwise_AND = val & Math.abs(n); // Bitwise AND with // absolute value of n // If Bitwise_AND > 0 means bit is set if (Bitwise_AND > 0) { leftmost = i; // Update leftmost set bit highest = Math.pow( 2, leftmost); // Update highest (highest // power of 2) set_bits++; // Count set bits } // If we have found leftmost set bit, break if (val >= Math.abs(n)) { break; } } // If n is negative, make +highest to -highest if (n < 0) { if (set_bits > 1) { // Means -n = -pow(2,i) - pow(2,j) ..... highest = Math.pow(2, leftmost + 1); } highest = -1 * highest; } console.log(highest); // Print highest power of 2 <= n}// Driver codelet n = -24;highestPowerOf2(n); // Output: -32 |
Output
-32
Time Complexity: O(log2n) , Because we are breaking loop , if leftmost set bit is found
Auxiliary Space: O(1)
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