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

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

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++

// 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; 
} 

Java

// 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 

Python3

# 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 

C#

// 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 
}  