Largest number with binary representation is m 1's and m-1 0's

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. 
?> 