Skip to content
Related Articles

Related Articles

Minimum numbers needed to express every integer below N as a sum
  • Difficulty Level : Medium
  • Last Updated : 27 Nov, 2018

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++




// 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




// 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));
          
    }
}


Python 3




# 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#




// 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
// 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
?>



output:

Minimum value of K is = 6

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :