Highest power of 2 that divides a number represented in binary

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Given binary string str, the task is to find the largest power of 2 that divides the decimal equivalent of the given binary number.

Examples:

Input: str = “100100”
Output: 2
2² = 4 is the highest power of 2 that divides 36 (100100).
Input: str = “10010”
Output: 1

Approach: Starting from the right, count the number of 0s in the binary representation which is the highest power of 2 which will divide the number.

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
// which divides the given binary number
int highestPower(string str, int len)
{
 
    // To store the highest required power of 2
    int ans = 0;
 
    // Counting number of consecutive zeros
    // from the end in the given binary string
    for (int i = len - 1; i >= 0; i--) {
        if (str[i] == '0')
            ans++;
        else
            break;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    string str = "100100";
    int len = str.length();
    cout << highestPower(str, len);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function to return the highest power of 2
// which divides the given binary number
static int highestPower(String str, int len)
{
 
    // To store the highest required power of 2
    int ans = 0;
 
    // Counting number of consecutive zeros
    // from the end in the given binary string
    for (int i = len - 1; i >= 0; i--) 
    {
        if (str.charAt(i) == '0')
            ans++;
        else
            break;
    }
 
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String str = "100100";
    int len = str.length();
    System.out.println(highestPower(str, len));
}
}
 
// This code is contributed by Code_Mech.

Python3

# Python3 implementation of the approach
 
# Function to return the highest power of 2
# which divides the given binary number
def highestPower(str, length):
 
    # To store the highest required power of 2
    ans = 0;
 
    # Counting number of consecutive zeros
    # from the end in the given binary string
    for i in range(length-1,-1,-1):
        if (str[i] == '0'):
            ans+=1;
        else:
            break;
    return ans;
 
# Driver code
def main():
    str = "100100";
    length = len(str);
    print(highestPower(str, length));
 
if __name__ == '__main__':
    main()
 
# This code contributed by PrinciRaj1992

C#

// C# implementation of the approach
using System;
     
class GFG
{
     
// Function to return the highest power of 2
// which divides the given binary number
static int highestPower(String str, int len)
{
 
    // To store the highest required power of 2
    int ans = 0;
 
    // Counting number of consecutive zeros
    // from the end in the given binary string
    for (int i = len - 1; i >= 0; i--) 
    {
        if (str[i] == '0')
            ans++;
        else
            break;
    }
 
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "100100";
    int len = str.Length;
    Console.WriteLine(highestPower(str, len));
}
}
 
/* This code contributed by PrinciRaj1992 */

PHP

<?php
// PHP implementation of the approach 
 
// Function to return the highest power of 2 
// which divides the given binary number 
function highestPower($str, $len) 
{ 
 
    // To store the highest required power of 2 
    $ans = 0; 
 
    // Counting number of consecutive zeros 
    // from the end in the given binary string 
    for ($i = $len - 1; $i >= 0; $i--)
    { 
        if ($str[$i] == '0') 
            $ans++; 
        else
            break; 
    } 
 
    return $ans; 
} 
 
// Driver code 
$str = "100100"; 
$len = strlen($str); 
echo highestPower($str, $len); 
 
// This code is contributed by Ryuga
?>

Javascript

<script>
// Javascript implementation of the approach
 
// Function to return the highest power of 2
// which divides the given binary number
function highestPower(str, len)
{
 
    // To store the highest required power of 2
    let ans = 0;
 
    // Counting number of consecutive zeros
    // from the end in the given binary string
    for (let i = len - 1; i >= 0; i--) {
        if (str[i] == '0')
            ans++;
        else
            break;
    }
 
    return ans;
}
 
// Driver code
    let str = "100100";
    let len = str.length;
    document.write(highestPower(str, len));
     
</script>

Output

Time Complexity: O(n) where n is number of elements in given array. As, we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(1), as we are not using any extra space.

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