Given a number N and base b if N in base b representation starts with 1 print Yes else print No
Examples :
Input : n = 6, b = 4 Output : Yes 6 can be written asin base 4so answer is Yes as it starts with 1Input : n = 24, b = 2Output : Yes24 can be written as
in base 2so answer is Yes as it starts with 1Input : n = 24, b = 7Output : No24 can be written as
in base 7so answer is No as it starts with 3
in base 4so answer is Yes as it starts with 1Input : n = 24, b = 2Output : Yes24 can be written as 
in base 2so answer is Yes as it starts with 1Input : n = 24, b = 7Output : No24 can be written as 
in base 7so answer is No as it starts with 3 When a number N is represented in base ‘b’ it gets converted to m+1 length sequence 
[Tex]b_{m-1} [/Tex]…..
which implies *
+ 
*
…..+*
= N
The smallest number in base b and starting with ‘1’ i.e. 100..00 and m+1 digits in base is
and largest number is 2*-1.So N should lie in this range. <= N <= 2*-1
Now another thing to notice is that m cannot exceed floor(
(N)) because when we represent any number in base-2 it gets converted into a sequence of only 1s and 0s so the length of this sequence will always be greater than any other base representation and its length will be equal to floor((N))+1.
So to check for a given base ‘b’ if N starts with 1 or not we will traverse from m = 1 to m = floor((N)) and check if for any m N lies in the range <= N <= 2*-1 or not and accordingly print “Yes” or “No”.
C++
// C++ program to check if number starts with // one in base b representation #include <bits/stdc++.h> using namespace std; // Returns true if n starts with 1 in // base b representation bool CheckIfstartsWithOne(int n, int b) { // highest m can be log2(n) int m = log2(n); for (int i = 1; i <= m; i++) { // if b^m <= N <= 2*b^m - 1, // return true if (n >= pow(b, i) && n <= 2 * pow(b, i) - 1) return true; } return false; } // printing yes or no void printYesORno(int n, int b) { if (CheckIfstartsWithOne(n, b) == true) cout << "Yes" << endl; else if (CheckIfstartsWithOne(n, b) == false) cout << "No" << endl; } // driver function int main() { printYesORno(6, 4); printYesORno(24, 2); printYesORno(24, 7); printYesORno(24, 15); } |
Java
// Java program to check if number starts with // one in base b representation class GFG { // Returns true if n starts with 1 in // base b representation static boolean CheckIfstartsWithOne(int n, int b) { // highest m can be log2(n) int m = (int)(Math.log10(n) / Math.log10(2)); for (int i = 1; i <= m; i++) { // if b^m <= N <= 2*b^m - 1, // return true if (n >= (int)Math.pow(b, i) && n <= 2 * (int)Math.pow(b, i) - 1) return true; } return false; } // Driver method public static void main(String args[]) { System.out.println( CheckIfstartsWithOne(6, 4) ? "Yes" : "No"); System.out.println( CheckIfstartsWithOne(24, 2) ? "Yes" : "No"); System.out.println( CheckIfstartsWithOne(24, 7) ? "Yes" : "No"); System.out.println( CheckIfstartsWithOne(24, 15) ? "Yes" : "No"); } } |
Python3
# Python3 program to check # if number starts with one # in base b representation import math # Returns true if n # starts with 1 in # base b representation def CheckIfstartsWithOne(n, b): # highest m can be log2(n) m = (int)(math.log2(n)); for i in range (1, m + 1): # if b^m <= N <= 2*b^m - 1, #return true x = (int)(math.pow(b, i)); if n >= x and n <= 2 * x - 1: return 1; return 0; # printing yes or no def printYesORno(n, b): if CheckIfstartsWithOne(n, b) == 1: print("Yes"); if CheckIfstartsWithOne(n, b) == 0: print("No"); # Driver Code printYesORno(6, 4); printYesORno(24, 2); printYesORno(24, 7); printYesORno(24, 15); # This code is contributed by mits. |
C#
// C# program to check if number starts with // one in base b representation using System; class GFG{ // Returns true if n starts with 1 in // base b representation static bool CheckIfstartsWithOne(int n, int b) { // highest m can be log2(n) int m = (int)(Math.Log10(n) / Math.Log10(2)); for(int i = 1; i <= m; i++) { // if b^m <= N <= 2*b^m - 1, // return true if (n >= (int)Math.Pow(b, i) && n <= 2 * (int)Math.Pow(b, i) - 1) return true; } return false; } // Driver code public static void Main(String []args) { Console.WriteLine( CheckIfstartsWithOne(6, 4) ? "Yes" : "No"); Console.WriteLine( CheckIfstartsWithOne(24, 2) ? "Yes" : "No"); Console.WriteLine( CheckIfstartsWithOne(24, 7) ? "Yes" : "No"); Console.WriteLine( CheckIfstartsWithOne(24, 15) ? "Yes" : "No"); } } // This code is contributed by Princi Singh |
PHP
<?php // PHP program to check if // number starts with one // in base b representation // Returns true if n starts // with 1 in base b representation function CheckIfstartsWithOne($n, $b) { // highest m can be log2(n) $m = log($n, 2); for ($i = 1; $i <= $m; $i++) { // if b^m <= N <= 2*b^m - 1, // return true if ($n >= pow($b, $i) && $n <= 2 * pow($b, $i) - 1) return true; } return false; } // printing yes or no function printYesORno($n, $b) { if (CheckIfstartsWithOne($n, $b) == true) echo "Yes" ,"\n"; else if (CheckIfstartsWithOne($n, $b) == false) echo "No" ,"\n"; } // Driver Code printYesORno(6, 4); printYesORno(24, 2); printYesORno(24, 7); printYesORno(24, 15); // This code is contributed by ajit ?> |
Javascript
<script> // Javascript program to check if number starts // with one in base b representation // Returns true if n starts with // 1 in base b representation function CheckIfstartsWithOne(n, b) { // highest m can be log2(n) let m = parseInt(Math.log10(n) / Math.log10(2), 10); for (let i = 1; i <= m; i++) { // if b^m <= N <= 2*b^m - 1, // return true if (n >= Math.pow(b, i) && n <= 2 * Math.pow(b, i) - 1) return true; } return false; } document.write(CheckIfstartsWithOne(6, 4) ? "Yes" + "</br>" : "No" + "</br>"); document.write(CheckIfstartsWithOne(24, 2) ? "Yes" + "</br>" : "No" + "</br>"); document.write(CheckIfstartsWithOne(24, 7) ? "Yes" + "</br>" : "No" + "</br>"); document.write(CheckIfstartsWithOne(24, 15) ? "Yes" : "No"); </script> |
Output :
Yes Yes No Yes
Time Complexity: O(m), where m is calculated as log2(n)
Auxiliary Space: O(1), as no extra space is required
This article is contributed by Ayush Jha. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
