Minimum number N such that total set bits of all numbers from 1 to N is at-least X

Given a number X, the task is to find the minimum number N such that total set bits of all numbers from 1 to n is atleast X.

Examples:

Input: x = 5 
Output: 4 
Set bits in 1-> 1
Set bits in 2-> 1
Set bits in 3-> 2 
Set bits in 4-> 1 
Hence first four numbers add upto 5 

Input: x = 20 
Output: 11

Approach: Use binary search to get the minimum most number whose sum of bits till N is atleast X. At start, low is 0 and high is initialized according to the constraint. Check if count of set bits is atleast X, every time it is, change high to mid-1, else change it to mid+1. Everytime we do high = mid-1, store the minimal of answer.

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
#define INF 99999
#define size 10
  
// Function to count sum of set bits
// of all numbers till N
int getSetBitsFromOneToN(int N)
{
    int two = 2, ans = 0;
    int n = N;
  
    while (n) {
        ans += (N / two) * (two >> 1);
  
        if ((N & (two - 1)) > (two >> 1) - 1)
            ans += (N & (two - 1)) - (two >> 1) + 1;
  
        two <<= 1;
        n >>= 1;
    }
    return ans;
}
  
// Function to find the minimum number
int findMinimum(int x)
{
    int low = 0, high = 100000;
  
    int ans = high;
  
    // Binary search for the lowest number
    while (low <= high) {
  
        // Find mid number
        int mid = (low + high) >> 1;
  
        // Check if it is atleast x
        if (getSetBitsFromOneToN(mid) >= x) {
            ans = min(ans, mid);
            high = mid - 1;
        }
        else
            low = mid + 1;
    }
  
    return ans;
}
  
// Driver Code
int main()
{
    int x = 20;
    cout << findMinimum(x);
  
return 0;
}

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// Java implementation of the above approach 


import java.util.*; 


  


class solution 


{ 


static int INF = 99999; 


static int size = 10; 


  


// Function to count sum of set bits 


// of all numbers till N 


static int getSetBitsFromOneToN(int N) 


{ 


    int two = 2, ans = 0; 


    int n = N; 


  


    while (n!=0) { 


        ans += (N / two) * (two >> 1); 


  


        if ((N & (two - 1)) > (two >> 1) - 1) 


            ans += (N & (two - 1)) - (two >> 1) + 1; 


  


        two <<= 1; 


        n >>= 1; 


    } 


    return ans; 


} 


  


// Function to find the minimum number 


static int findMinimum(int x) 


{ 


    int low = 0, high = 100000; 


  


    int ans = high; 


  


    // Binary search for the lowest number 


    while (low <= high) { 


  


        // Find mid number 


        int mid = (low + high) >> 1; 


  


        // Check if it is atleast x 


        if (getSetBitsFromOneToN(mid) >= x) { 


            ans = Math.min(ans, mid); 


            high = mid - 1; 


        } 


        else


            low = mid + 1; 


    } 


  


    return ans; 


} 


  


// Driver Code 


public static void main(String args[]) 


{ 


    int x = 20; 


    System.out.println(findMinimum(x)); 


  


} 


  


} 


  


//This code is contributed by  


// Shashank_Sharma 



















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Python3














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# Python3 implementation of the  


# above approach 


INF = 99999


size = 10


  


# Function to count sum of set bits 


# of all numbers till N 


def getSetBitsFromOneToN(N): 


  


    two, ans = 2, 0


    n = N 


  


    while (n > 0): 


        ans += (N // two) * (two >> 1) 


  


        if ((N & (two - 1)) > (two >> 1) - 1): 


            ans += (N & (two - 1)) - (two >> 1) + 1


  


        two <<= 1


        n >>= 1


    return ans 


  


# Function to find the minimum number 


def findMinimum(x): 


  


    low = 0


    high = 100000


  


    ans = high 


  


    # Binary search for the lowest number 


    while (low <= high):  


  


        # Find mid number 


        mid = (low + high) >> 1


  


        # Check if it is atleast x 


        if (getSetBitsFromOneToN(mid) >= x): 


  


            ans = min(ans, mid) 


            high = mid - 1


        else: 


            low = mid + 1


      


    return ans 


  


# Driver Code 


x = 20


print(findMinimum(x)) 


  


# This code is contributed by 


# Mohit kumar 29 



















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C#














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// Csharp implementation of the above approach  


using System ; 


  


class solution  


{  


static int INF = 99999;  


static int size = 10;  


  


// Function to count sum of set bits  


// of all numbers till N  


static int getSetBitsFromOneToN(int N)  


{  


    int two = 2, ans = 0;  


    int n = N;  


  


    while (n!=0) {  


        ans += (N / two) * (two >> 1);  


  


        if ((N & (two - 1)) > (two >> 1) - 1)  


            ans += (N & (two - 1)) - (two >> 1) + 1;  


  


        two <<= 1;  


        n >>= 1;  


    }  


    return ans;  


}  


  


// Function to find the minimum number  


static int findMinimum(int x)  


{  


    int low = 0, high = 100000;  


  


    int ans = high;  


  


    // Binary search for the lowest number  


    while (low <= high) {  


  


        // Find mid number  


        int mid = (low + high) >> 1;  


  


        // Check if it is atleast x  


        if (getSetBitsFromOneToN(mid) >= x) {  


            ans = Math.Min(ans, mid);  


            high = mid - 1;  


        }  


        else


            low = mid + 1;  


    }  


  


    return ans;  


}  


  


    // Driver Code  


    public static void Main()  


    {  


        int x = 20;  


        Console.WriteLine(findMinimum(x));  


      


    }  


    // This code is contributed by Ryuga 


}  



















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PHP





> 1);


		if (($N & ($two – 1)) > ($two >> 1) – 1)

			$ans += ($N & ($two – 1)) –

			              ($two >> 1) + 1;


		$two <<= 1;
		$n >>= 1;

	}

	return $ans;

}


// Function to find the minimum number

function findMinimum($x)

{

	$low = 0;

	$high = 100000;


	$ans = $high;


	// Binary search for the lowest number

	while ($low <= $high) 
	{

		// Find mid number
		$mid = ($low + $high) >> 1;


		// Check if it is atleast x

		if (getSetBitsFromOneToN($mid) >= $x)

		{

			$ans = min($ans, $mid);

			$high = $mid – 1;

		}

		else

			$low = $mid + 1;

	}


	return $ans;

}


// Driver Code

$x = 20;

echo findMinimum($x);


// This code is contributed

// by Sach_Code

?>





Output:


11





Time Complexity: O(log N * log N)



          
          
          
          

            

    

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