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

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  • Difficulty Level : Hard
  • Last Updated : 16 Jun, 2022
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Given a number X, the task is to find the minimum number N such that the total set bits of all numbers from 1 to n is at least 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 at least X. At the start, low is 0, and high is initialized according to the constraint. Check if the count of set bits is at least X, every time it is, change high to mid-1, else change it to mid+1. Every time 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#




// C# 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
}

PHP




<?php
// PHP implementation of the above approach
 
// Function to count sum of set bits
// of all numbers till N
function getSetBitsFromOneToN($N)
{
    $two = 2;
    $ans = 0;
    $n = $N;
 
    while ($n)
    {
        $ans += (int)($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
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
?>

Javascript




<script>
 
// Javascript implementation of the above approach
const INF = 99999;
const size = 10;
 
// Function to count sum of set bits
// of all numbers till N
function getSetBitsFromOneToN(N)
{
    let two = 2, ans = 0;
    let n = N;
 
    while (n)
    {
        ans += parseInt(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
function findMinimum(x)
{
    let low = 0, high = 100000;
 
    let ans = high;
 
    // Binary search for the lowest number
    while (low <= high)
    {
 
        // Find mid number
        let 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
let x = 20;
 
document.write(findMinimum(x));
 
// This code is contributed by subhammahato348
 
</script>

Output: 

11

 

Time Complexity: O(log N * log N), the getSetBitsFromOneToN will take logN time as we are using the bitwise right shift in each traversal which is equivalent to floor division with 2 in each traversal so the cost will be 1+1/2+1/4+….+1/2N which is equivalent to logN. We are using binary search and each time we are calling getSetBitsFromOneToN function. Binary Search also takes logN time as we decrement each time by floor division of 2. So, the effective cost of the program will be O(logN*logN).
 Auxiliary Space: O(1), as we are not using any extra space.


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