Given a binary string, we need to check whether that number is divisible by 64 or not after removing of some bits. If yes then print “possible” else “not possible”. We cannot make number 0 to make it divisible.
Example :
Input: 100010001 Output: Possible Explanation: We can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system. Input: 100 Output: Not possible Explanation : The number is 4 which is not divisible by 64 or cannot be made possible my removing some digits.
If we have 6 zeros after any one, then we can remove other bits represent it as a multiple of 64. So we just need to check if there is a 1 before six zeros.
C++
// CPP program to find if given binary string can// become divisible by 64 after removing some bits.#include <iostream>using namespace std;// function to check if it is possible// to make it a multiple of 64.bool checking(string s){ int c = 0; // counter to count 0's int n = s.length(); // length of the string // loop which traverses right to left and // calculates the number of zeros before 1. for (int i = n - 1; i >= 0; i--) { if (s[i] == '0') c++; if (c >= 6 and s[i] == '1') return true; } return false;}// driver codeint main(){ string s = "100010001"; if (checking(s)) cout << "Possible"; else cout << "Not possible"; return 0;} |
Java
// Java program to find if // given binary string can// become divisible by // 64 after removing some bitsimport java.io.*;class GFG { // function to check if it is possible // to make it a multiple of 64. static boolean checking(String s) { // counter to count 0's int c = 0; // length of the string int n = s.length(); // loop which traverses right to left and // calculates the number of zeros before 1. for (int i = n - 1; i >= 0; i--) { if (s.charAt(i) == '0') c++; if (c >= 6 && s.charAt(i) == '1') return true; } return false; } // Driver code public static void main (String[] args) { String s = "100010001"; if (checking(s)) System.out.println ( "Possible"); else System.out.println ( "Not possible"); }}// This code is contributed by vt_m |
Python3
# Python 3 program to find if given binary # string can become divisible by 64 after# removing some bits.# function to check if it is possible# to make it a multiple of 64.def checking(s): c = 0 # counter to count 0's n = len(s) # length of the string # loop which traverses right to left and # calculates the number of zeros before 1. i = n - 1 while(i >= 0): if (s[i] == '0'): c += 1 if (c >= 6 and s[i] == '1'): return True i -= 1 return False# Driver codeif __name__ == '__main__': s = "100010001" if (checking(s)): print("Possible") else: print("Not possible") # This code is contributed by# Surendra_Gangwar |
C#
// C# program to find if given binary// string can become divisible by 64// after removing some bitsusing System;class GFG { // function to check if it is possible // to make it a multiple of 64. static bool checking(string s) { // counter to count 0's int c = 0; // length of the string int n = s.Length; // loop which traverses right to // left and calculates the number // of zeros before 1. for (int i = n - 1; i >= 0; i--) { if (s[i] == '0') c++; if (c >= 6 && s[i] == '1') return true; } return false; } // Driver code public static void Main () { String s = "100010001"; if (checking(s)) Console.WriteLine ( "Possible"); else Console.WriteLine( "Not possible"); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to find if // given binary string can// become divisible by 64 // after removing some bits.// function to check if // it is possible// to make it a multiple of 64.function checking($s){ // counter to count 0's $c = 0; // length of the string $n = strlen($s); // loop which traverses right // to left and calculates the // number of zeros before 1. for ($i = $n - 1; $i >= 0; $i--) { if ($s[$i] == '0') $c++; if ($c >= 6 and $s[$i] == '1') return true; } return false;}// Driver Code$s = "100010001";if (checking($s)) echo "Possible";else echo "Not possible";// This code is contributed by ajit?> |
Javascript
<script>// Javascript program to find if given binary// string can become divisible by 64// after removing some bits // function to check if it is possible // to make it a multiple of 64. function checking(s) { // counter to count 0's let c = 0; // length of the string let n = s.length; // loop which traverses right to // left and calculates the number // of zeros before 1. for (let i = n - 1; i >= 0; i--) { if (s[i] == '0') c++; if (c >= 6 && s[i] == '1') return true; } return false; } // Driver code let s = "100010001"; if (checking(s)) document.write( "Possible"); else document.write( "Not possible"); // This code is contributed by code_hunt.</script> |
Output :
Possible
Time Complexity: O(length of string)
Space Complexity: O(1)
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