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. 

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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.
?>

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Output:

6


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

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