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# Highest power of 2 less than or equal to given Integer

• Difficulty Level : Basic
• Last Updated : 28 Feb, 2023

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:
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:

1. If the input is positive: floor(log2(N)) can be calculated.
2. 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 ` `using` `namespace` `std;`   `// Function to return the lowest power of 2 close to given` `// positive number` `int` `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 number` `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 = ``ceil``(log2(n));` `    ``return` `(-1 * ``pow``(2, pos));` `}`   `// Function to find the highest power of 2` `void` `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 code` `int` `main()` `{` `    ``int` `n = -24;` `    ``highestPowerOf2(n);` `    ``return` `0;` `}`   `// This code is contributed by Sania Kumari Gupta`

## C

 `// C implementation of the above approach` `#include ` `#include `   `// Function to return the lowest power of 2 close to given` `// positive number` `int` `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 number` `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 = ``ceil``(log2(n));` `    ``return` `(-1 * ``pow``(2, pos));` `}`   `// Function to find the highest power of 2` `void` `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 code` `int` `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

 ``

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 ` `using` `namespace` `std;`   `// Function find highestpowerof2 <= n` `void` `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 code` `int` `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` `math`   `def` `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 code` `n ``=` `-``24` `highestPowerOf2(n) ``# Function call`

## C#

 `// C# implementation of the above approach` `using` `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 <= n` `function` `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 code` `let 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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