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

• Difficulty Level : Hard
• Last Updated : 16 Jun, 2022

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 ``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

 `> 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

 ``

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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