Minimum numbers needed to express every integer below N as a sum

We have an integer N. We need to express N as a sum of K integers such that by adding some(or all) of these integers we can get all the numbers in the range[1, N]. What is the minimum value of K?

Examples: 

Input  : N = 7
Output : 3
Explanation : Three integers are 1, 2, 4. By adding some(or all) of these groups we can get all number in the range 1 to N. 
1; 2; 1+2=3; 4; 1+4=5; 2+4=6; 1+2+4=7

Input  : N = 32
Output : 6
Explanation : Six integers are 1, 2, 4, 8, 16, 1.

1st we solve the problem for small numbers by hand. 
n=1 : 1 
n=2 : 1, 1 
n=3 : 1, 2 
n=4 : 1, 2, 1 
n=5 : 1, 2, 2 
n=6 : 1, 2, 3 
n=7 : 1, 2, 4 
n=8 : 1, 2, 4, 1
If we inspect this closely we can see that if N=2^{m}-1    then the integers are [1, 2, 4..., 2^{m-1}]    . Which is just another way of saying 2^0+2^1+2^2+...+2^{m-1} = \frac{2^{m}-1}{2-1} = 2^{m}-1    .So now we know for 2^{m}-1    minimum value of K is m. 
Now we inspect what happens for 2^{m}    .For 2^{m}    we just add a new integer 1 to our list of integers. Realize that for every number from 2^{m} to 2^{m+1}-1    we can increase the newly added integer by 1 and that will be the optimal list of integers. To verify look at N=4 to N=7, minimum K does not change; only the last integer is increased in each step.

Of course we can implement this in iterative manner in O(log N) time (by inserting successive powers of 2 in the list and the last element will be of the form N-(2^n-1)). But this is exactly same as finding the length of binary expression of N which also can be done in O(log N) time.

// CPP program to find count of integers needed
// to express all numbers from 1 to N.
#include <bits/stdc++.h>
using namespace std;

// function to count length of binary expression of n
int countBits(int n)
{
    int count = 0;
    while (n) {
        count++;
        n >>= 1;
    }
    return count;
}

// Driver code
int main()
{
    int n = 32;
    cout << "Minimum value of K is = " 
         << countBits(n) << endl;
    return 0;
}

// Java  program to find count of integers needed 
// to express all numbers from 1 to N

import java.io.*;

class GFG {
    
// function to count length of binary expression of n 
static int countBits(int n) 
{ 
    int count = 0; 
    while (n>0) { 
        count++; 
        n >>= 1; 
    } 
    return count; 
} 

// Driver code 
    public static void main (String[] args) {
        int n = 32; 
        System.out.println("Minimum value of K is = "+
             countBits(n));
        
    }
}

# Python3 program to find count of integers 
# needed to express all numbers from 1 to N. 

# function to count length of 
# binary expression of n 
def countBits(n):

    count = 0; 
    while (n):
        count += 1; 
        n >>= 1;
        
    return count; 

# Driver code 
n = 32; 
print("Minimum value of K is =", 
                  countBits(n)); 

# This code is contributed by mits

// C# program to find count of 
// integers needed to express all 
// numbers from 1 to N
using System;

class GFG
{
// function to count length of 
// binary expression of n 
static int countBits(int n) 
{ 
    int count = 0; 
    while (n > 0) 
    { 
        count++; 
        n >>= 1; 
    } 
    return count; 
} 

// Driver code
static public void Main ()
{
    int n = 32; 
    Console.WriteLine("Minimum value of K is = "+
                                   countBits(n));
}
}

// This code is contributed
// by Sach_Code

<?php
// PHP program to find count of integers 
// needed to express all numbers from 1 to N. 

// function to count length of 
// binary expression of n 
function countBits($n) 
{ 
    $count = 0; 
    while ($n)
    { 
        $count++; 
        $n >>= 1; 
    } 
    return $count; 
} 

// Driver code 
$n = 32; 
echo "Minimum value of K is = ",
      countBits($n), "\n"; 

// This code is contributed by Sachin
?>

<script>

// Javascript program to find count of 
// integers needed to express all 
// numbers from 1 to N.

// Function to count length of binary
// expression of n
function countBits(n)
{
    var count = 0;
    while (n) 
    {
        count++;
        n >>= 1;
    }
    return count;
}

// Driver code
var n = 32;
document.write("Minimum value of K is = "  + 
               countBits(n));
               
// This code is contributed by rrrtnx

</script>  

output: 

Minimum value of K is = 6

Time Complexity: O(log n)

Auxiliary Space: O(1)

Please see count set bits for more efficient methods to count set bits in an integer.