Given n, find the greatest number which is strictly not more then n and whose binary representation consists of m consecutive ones, then m-1 consecutive zeros and nothing else
Examples:
Input : n = 7 Output : 6 Explanation: 6's binary representation is 110, and 7's is 111, so 6 consists of 2 consecutive 1's and then 1 consecutive 0. Input : 130 Output : 120 Explanation: 28 and 120 are the only numbers <=120, 28 is 11100 consists of 3 consecutive 1's and then 2 consecutive 0's. 120 is 1111000 consists of 4 consecutive 1's and then 3 consecutive 0's. So 120 is the greatest of number<=120 which meets the given condition.
A naive approach will be to traverse from 1 to N and check for every binary representation which consists of m consecutive 1’s and m-1 consecutive 0’s and store the largest of them which meets the given condition.
An efficient approach is to observe a pattern of numbers,
[1(1), 6(110), 28(11100), 120(1111000), 496(111110000), ….]
To get the formula for the numbers which satisfies the conditions we take 120 as an example-
120 is represented as 1111000 which has m = 4 1’s and m = 3 0’s. Converting 1111000 to decimal we get:
2^3+2^4+2^5+2^6 which can be represented as (2^m-1 + 2^m+ 2^m+1 + … 2^m+2, 2^2*m)
2^3*(1+2+2^2+2^3) which can be represented as (2^(m-1)*(1+2+2^2+2^3+..2^(m-1))
2^3*(2^4-1) which can be represented as [2^(m-1) * (2^m -1)].
So all the numbers that meet the given condition can be represented as
[2^(m-1) * (2^m -1)]
We can iterate till the number does not exceeds N and print the largest of all possible elements. A closer observation will shows that at m = 33 it will exceed the 10^18 mark , so we are calculating the number in unit’s time as log(32) is near to constant which is required in calculating the pow .
So, the overall complexity will be O(1).
C++
// CPP program to find largest number // smaller than equal to n with m set // bits then m-1 0 bits. #include <bits/stdc++.h> using namespace std; // Returns largest number with m set // bits then m-1 0 bits. long long answer(long long n) { // Start with 2 bits. long m = 2; // initial answer is 1 // which meets the given condition long long ans = 1; long long r = 1; // check for all numbers while (r < n) { // compute the number r = (int)(pow(2, m) - 1) * (pow(2, m - 1)); // if less then N if (r < n) ans = r; // increment m to get the next number m++; } return ans; } // driver code to check the above condition int main() { long long n = 7; cout << answer(n); return 0; } |
Java
// java program to find largest number // smaller than equal to n with m set // bits then m-1 0 bits. public class GFG { // Returns largest number with // m set bits then m-1 0 bits. static long answer(long n) { // Start with 2 bits. long m = 2; // initial answer is 1 which // meets the given condition long ans = 1; long r = 1; // check for all numbers while (r < n) { // compute the number r = ((long)Math.pow(2, m) - 1) * ((long)Math.pow(2, m - 1)); // if less then N if (r < n) ans = r; // increment m to get // the next number m++; } return ans; } // Driver code public static void main(String args[]) { long n = 7; System.out.println(answer(n)); } } // This code is contributed by Sam007 |
Python3
# Python3 program to find # largest number smaller # than equal to n with m # set bits then m-1 0 bits. import math # Returns largest number # with m set bits then # m-1 0 bits. def answer(n): # Start with 2 bits. m = 2; # initial answer is # 1 which meets the # given condition ans = 1; r = 1; # check for all numbers while r < n: # compute the number r = (int)((pow(2, m) - 1) * (pow(2, m - 1))); # if less then N if r < n: ans = r; # increment m to get # the next number m = m + 1; return ans; # Driver Code print(answer(7)); # This code is contributed by mits. |
C#
// C# program to find largest number // smaller than equal to n with m set // bits then m-1 0 bits. using System; class GFG { // Returns largest number with // m set bits then m-1 0 bits. static long answer(long n) { // Start with 2 bits. long m = 2; // initial answer is 1 which // meets the given condition long ans = 1; long r = 1; // check for all numbers while (r < n) { // compute the number r = ((long)Math.Pow(2, m) - 1) * ((long)Math.Pow(2, m - 1)); // if less then N if (r < n) ans = r; // increment m to get // the next number m++; } return ans; } // Driver Code static public void Main () { long n = 7; Console.WriteLine(answer(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find largest number // smaller than equal to n with m set // bits then m-1 0 bits. // Returns largest number with m set // bits then m-1 0 bits. function answer( $n) { // Start with 2 bits. $m = 2; // initial answer is 1 // which meets the // given condition $ans = 1; $r = 1; // check for all numbers while ($r < $n) { // compute the number $r = (pow(2, $m) - 1) * (pow(2, $m - 1)); // if less then N if ($r < $n) $ans = $r; // increment m to get // the next number $m++; } return $ans; } // Driver Code $n = 7; echo answer($n); // This code is contributed by Ajit. ?> |
Output:
6
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
