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 numberint 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 codeint main(){ string str = "100100"; int len = str.length(); cout << highestPower(str, len); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to return the highest power of 2// which divides the given binary numberstatic 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 codepublic 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 numberdef 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 codedef main(): str = "100100"; length = len(str); print(highestPower(str, length));if __name__ == '__main__': main()# This code contributed by PrinciRaj1992 |
C#
// C# implementation of the approachusing System; class GFG{ // Function to return the highest power of 2// which divides the given binary numberstatic 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 codepublic 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 numberfunction 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
2
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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
