Given a binary string and a number k, the task is to check whether the binary string is evenly divisible by 2k or not.
Examples :
Input : 11000 k = 2
Output : Yes
Explanation :
(11000)2 = (24)10
24 is evenly divisible by 2k i.e. 4.
Input : 10101 k = 3
Output : No
Explanation :
(10101)2 = (21)10
21 is not evenly divisible by 2k i.e. 8.
Naive Approach: Compute the binary string into decimal and then simply check whether it is divisible by 2k. But, this approach is not feasible for the long binary strings as time complexity will be high for computing decimal number from binary string and then dividing it by 2k.
Step-by-step approach:
- Convert the binary string to decimal by iterating over each digit from left to right.
- Calculate the value of 2^k using bitwise left shift operation (1 << k).
- Check if the decimal value obtained in step 1 is divisible by the value obtained in step 2 using the modulus operator.
- If the modulus is zero, return true, otherwise return false.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// function to check whether// given binary number is// evenly divisible by 2^k or notbool isDivisible(char str[], int k){ // convert binary string to decimal int decimal_num = 0; int base = 1; int n = strlen(str); for (int i = n - 1; i >= 0; i--) { if (str[i] == '1') { decimal_num += base; } base *= 2; } // check if decimal_num is divisible by 2^k return (decimal_num % (1 << k)) == 0;}// Driver program to test aboveint main(){ // first example char str1[] = "10101100"; int k = 2; if (isDivisible(str1, k)) cout << "Yes" << endl; else cout << "No" << "\n"; // Second example char str2[] = "111010100"; k = 2; if (isDivisible(str2, k)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
import java.util.Arrays;public class Main { // function to check whether // given binary number is // evenly divisible by 2^k or not public static boolean isDivisible(String str, int k) { // convert binary string to decimal int decimal_num = 0; int base = 1; int n = str.length(); for (int i = n - 1; i >= 0; i--) { if (str.charAt(i) == '1') { decimal_num += base; } base *= 2; } // check if decimal_num is divisible by 2^k return (decimal_num % (1 << k)) == 0; } // Driver program to test above public static void main(String[] args) { // first example String str1 = "10101100"; int k = 2; if (isDivisible(str1, k)) System.out.println("Yes"); else System.out.println("No"); // Second example String str2 = "111010100"; k = 2; if (isDivisible(str2, k)) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# function to check whether# given binary number is# evenly divisible by 2^k or notdef is_divisible(binary_str, k): # convert binary string to decimal decimal_num = 0 base = 1 n = len(binary_str) for i in range(n - 1, -1, -1): if binary_str[i] == '1': decimal_num += base base *= 2 # check if decimal_num is divisible by 2^k return decimal_num % (1 << k) == 0# Driver program to test aboveif __name__ == "__main__": # first example str1 = "10101100" k = 2 if is_divisible(str1, k): print("Yes") else: print("No") # Second example str2 = "111010100" k = 2 if is_divisible(str2, k): print("Yes") else: print("No")# This code is contributed by shivamgupta0987654321 |
C#
using System;class GFG{ // function to check whether // given binary number is // evenly divisible by 2^k or not static bool IsDivisible(string binaryStr, int k) { // convert binary string to decimal int decimalNum = 0; int baseValue = 1; int n = binaryStr.Length; for (int i = n - 1; i >= 0; i--) { if (binaryStr[i] == '1') { decimalNum += baseValue; } baseValue *= 2; } // check if decimalNum is divisible by 2^k return (decimalNum % (1 << k)) == 0; } // Driver program to test above static void Main() { // first example string str1 = "10101100"; int k = 2; if (IsDivisible(str1, k)) Console.WriteLine("Yes"); else Console.WriteLine("No"); // Second example string str2 = "111010100"; k = 2; if (IsDivisible(str2, k)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }} |
Javascript
// Javascript code// Function to check whether the given binary number is evenly divisible by 2^k or notfunction isDivisible(binaryStr, k) { // Convert binary string to decimal let decimalNum = 0; let base = 1; for (let i = binaryStr.length - 1; i >= 0; i--) { if (binaryStr[i] === '1') { decimalNum += base; } base *= 2; } // Check if decimalNum is divisible by 2^k return decimalNum % (1 << k) === 0;}// Driver program to test aboveconst str1 = "10101100";let k = 2;if (isDivisible(str1, k)) { console.log("Yes");} else { console.log("No");}const str2 = "111010100";k = 2;if (isDivisible(str2, k)) { console.log("Yes");} else { console.log("No");}// This code is contributed by guptapratik |
Output
Yes Yes
Time Complexity: O(n), where n is the length of the binary string.
Auxiliary Space: O(1), as we are not using any extra space.
Efficient Approach: In the binary string, check for the last k bits. If all the last k bits are 0, then the binary number is evenly divisible by 2k else it is not evenly divisible. Time complexity using this approach is O(k).
Below is the implementation of the approach.
C++
// C++ implementation to check whether// given binary number is evenly// divisible by 2^k or not#include <bits/stdc++.h>using namespace std;// function to check whether// given binary number is // evenly divisible by 2^k or notbool isDivisible(char str[], int k){ int n = strlen(str); int c = 0; // count of number of 0 from last for (int i = 0; i < k; i++) if (str[n - i - 1] == '0') c++; // if count = k, number is evenly // divisible, so returns true else // false return (c == k);}// Driver program to test aboveint main(){ // first example char str1[] = "10101100"; int k = 2; if (isDivisible(str1, k)) cout << "Yes" << endl; else cout << "No" << "\n"; // Second example char str2[] = "111010100"; k = 2; if (isDivisible(str2, k)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java implementation to check whether// given binary number is evenly// divisible by 2^k or notclass GFG { // function to check whether // given binary number is // evenly divisible by 2^k or not static boolean isDivisible(String str, int k) { int n = str.length(); int c = 0; // count of number of 0 from last for (int i = 0; i < k; i++) if (str.charAt(n - i - 1) == '0') c++; // if count = k, number is evenly // divisible, so returns true else // false return (c == k); } // Driver program to test above public static void main(String args[]) { // first example String str1 = "10101100"; int k = 2; if (isDivisible(str1, k) == true) System.out.println("Yes"); else System.out.println("No"); // Second example String str2 = "111010100"; k = 2; if (isDivisible(str2, k) == true) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by JaideepPyne. |
Python3
# Python 3 implementation to check# whether given binary number is# evenly divisible by 2^k or not# function to check whether# given binary number is # evenly divisible by 2^k or notdef isDivisible(str, k): n = len(str) c = 0 # count of number of 0 from last for i in range(0, k): if (str[n - i - 1] == '0'): c += 1 # if count = k, number is evenly # divisible, so returns true else # false return (c == k)# Driver program to test above# first examplestr1 = "10101100"k = 2if (isDivisible(str1, k)): print("Yes")else: print("No")# Second examplestr2 = "111010100"k = 2if (isDivisible(str2, k)): print("Yes")else: print("No")# This code is contributed by Smitha |
C#
// C# implementation to check whether// given binary number is evenly// divisible by 2^k or notusing System;class GFG { // function to check whether // given binary number is // evenly divisible by 2^k or not static bool isDivisible(String str, int k) { int n = str.Length; int c = 0; // count of number of 0 from last for (int i = 0; i < k; i++) if (str[n - i - 1] == '0') c++; // if count = k, number is evenly // divisible, so returns true else // false return (c == k); } // Driver program to test above public static void Main() { // first example String str1 = "10101100"; int k = 2; if (isDivisible(str1, k) == true) Console.Write("Yes\n"); else Console.Write("No"); // Second example String str2 = "111010100"; k = 2; if (isDivisible(str2, k) == true) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by Smitha. |
Javascript
<script>// Javascript implementation to check whether// given binary number is evenly// divisible by 2^k or not// Function to check whether// given binary number is // evenly divisible by 2^k or notfunction isDivisible(str, k){ let n = str.length; let c = 0; // Count of number of 0 from last for(let i = 0; i < k; i++) if (str[n - i - 1] == '0') c++; // If count = k, number is evenly // divisible, so returns true else // false return (c == k);}// Driver code// First examplelet str1 = "10101100";let k = 2;if (isDivisible(str1, k) == true) document.write("Yes" + "</br>");else document.write("No");// Second examplelet str2 = "111010100";k = 2;if (isDivisible(str2, k) == true) document.write("Yes");else document.write("No"); // This code is contributed by rameshtravel07</script> |
PHP
<?php// PHP implementation to check whether// given binary number is evenly// function to check whether// given binary number is // evenly divisible by 2^k or notfunction isDivisible($str, $k){ $n = strlen($str); $c = 0; // count of number // of 0 from last for ($i = 0; $i < $k; $i++) if ($str[$n - $i - 1] == '0') $c++; // if count = k, // number is evenly // divisible, so // returns true else // false return ($c == $k);} // Driver Code // first example $str1 = "10101100"; $k = 2; if (isDivisible($str1, $k)) echo "Yes", "\n"; else echo "No", "\n"; // Second example $str2= "111010100"; $k = 2; if (isDivisible($str2, $k)) echo "Yes", "\n"; else echo "No", "\n";// This code is contributed by Ajit?> |
Output
Yes Yes
Complexity Analysis:
- Time Complexity: O(k)
- Auxiliary Space: O(1)
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