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

0 0 9 minutes read

Given n, find the greatest number which is strictly not more than 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 than 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 than 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 than 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 than 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 than 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.
?>

Javascript

<script>
 
// Javascript 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.
        let m = 2;
        
        // initial answer is 1 which
        // meets the given condition
        let ans = 1;
        let r = 1;
        
        // check for all numbers
        while (r < n) {
        
            // compute the number
            r = (Math.pow(2, m) - 1) * 
                (Math.pow(2, m - 1));
        
            // if less than N
            if (r < n)
                ans = r;
        
            // increment m to get 
            // the next number
            m++;
        }
        
        return ans;
    }
 
// Driver code
 
        let n = 7;
        document.write(answer(n));
           
</script>

Output

Time Complexity: O(1)
Auxiliary Space: O(1)

Efficient approach:

This problem can be solved with at most two comparisons, by using the following logic:

1. For a number with n bits, the answer with m set bits followed by m – 1 bits can have a total of n bits, ie, m = ?(n + 1)/2 ? (as m + m – 1 = n).

2. The result can then be computed as 2 ^{(m + m – 1)} – 2 ^{(m – 1)}.

3. However, if the n – bit number is in the range [2^{(m + m – 2)}, 2^{(m + m – 1)} – 2 ^{(m – 1)} – 1], then m must be adjusted to m – 1.

Below is the implementation of the above approach:

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)
{
    // Calculating the number of bits 
    int bits_no = 1 + (int)(log(n)/log(2));
     
    // if m + m - 1 = bits_no, then m = (bits_no + 1) / 2
    int m =  (bits_no + 1) / 2;
     
    // A number with m set bits and m - 1 unset bits can be calculated as
    // (2 ^ (m + m - 1)) - (2 ^ (m - 1))
    int ans = (1 << (m + m - 1)) - (1 << (m - 1));
    if (ans <= n)
        return ans;
     
    // If the result is too large, adjust m by applying decrement operation
    m--;
    return (1 << (m + m - 1)) - (1 << (m - 1));
}
 
// 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.
import java.util.*;
 
class GFG
{
 
  // Returns largest number with m set
  // bits then m-1 0 bits.
  static int answer(int n)
  {
 
    // Calculating the number of bits 
    int bits_no = 1 + (int)(Math.log(n)/Math.log(2));
 
    // if m + m - 1 = bits_no, then m = (bits_no + 1) / 2
    int m =  (bits_no + 1) / 2;
 
    // A number with m set bits and m - 1 
    // unset bits can be calculated as
    // (2 ^ (m + m - 1)) - (2 ^ (m - 1))
    int ans = (1 << (m + m - 1)) - (1 << (m - 1));
    if (ans <= n)
      return ans;
 
    // If the result is too large, adjust m 
    // by applying decrement operation
    m--;
    return (1 << (m + m - 1)) - (1 << (m - 1));
  }
 
  // driver code to check the above condition
  public static void main(String[] args)
  {
    int n = 7;
    System.out.println(answer(n));
  }
}
 
// This code is contributed by phasing17.

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):
 
    # Calculating the number of bits 
    bits_no = 1 + int(math.log(n)/math.log(2));
     
    # if m + m - 1 = bits_no, then m = (bits_no + 1) / 2
    m =  int((bits_no + 1) / 2);
     
    # A number with m set bits and m - 1 unset bits can be calculated as
    # (2 ^ (m + m - 1)) - (2 ^ (m - 1))
    ans = (1 << (m + m - 1)) - (1 << (m - 1));
    if (ans <= n):
        return ans;
     
    # If the result is too large, adjust m by applying decrement operation
    m -= 1;
    return (1 << (m + m - 1)) - (1 << (m - 1));
 
 
# driver code to check the above condition
n = 7;
print(answer(n));
 
# This code is contributed by phasing17.

C#

// C# program to find largest number
// smaller than equal to n with m set
// bits then m-1 0 bits.
using System;
using System.Collections.Generic;
 
class GFG
{
   
  // Returns largest number with m set
  // bits then m-1 0 bits.
  static int answer(int n)
  {
     
    // Calculating the number of bits 
    int bits_no = 1 + (int)(Math.Log(n)/Math.Log(2));
 
    // if m + m - 1 = bits_no, then m = (bits_no + 1) / 2
    int m =  (bits_no + 1) / 2;
 
    // A number with m set bits and m - 1 
    // unset bits can be calculated as
    // (2 ^ (m + m - 1)) - (2 ^ (m - 1))
    int ans = (1 << (m + m - 1)) - (1 << (m - 1));
    if (ans <= n)
      return ans;
 
    // If the result is too large, adjust m 
    // by applying decrement operation
    m--;
    return (1 << (m + m - 1)) - (1 << (m - 1));
  }
 
  // driver code to check the above condition
  public static void Main(string[] args)
  {
    int n = 7;
    Console.WriteLine(answer(n));
  }
}
 
// This code is contributed by phasing17.

Javascript

// JS 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)
{
    // Calculating the number of bits 
    let bits_no = 1 + Math.floor(Math.log(n)/Math.log(2));
     
    // if m + m - 1 = bits_no, then m = (bits_no + 1) / 2
    let m =  (bits_no + 1) / 2;
     
    // A number with m set bits and m - 1 unset bits can be calculated as
    // (2 ^ (m + m - 1)) - (2 ^ (m - 1))
    let ans = (1 << (m + m - 1)) - (1 << (m - 1));
    if (ans <= n)
        return ans;
     
    // If the result is too large, adjust m by applying decrement operation
    m--;
    return (1 << (m + m - 1)) - (1 << (m - 1));
}
 
// driver code to check the above condition
let n = 7;
console.log(answer(n));
 
// This code is contributed by phasing17.

Output

Time Complexity: O(1)
Auxiliary Space: O(1)

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!