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.

C++

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// 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;
}

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Java

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// 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));
          
    }
}

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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output:

Minimum value of K is = 6

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

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