# 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 then the integers are . Which is just another way of saying .So now we know for minimum value of K is m.

Now we inspect what happens for .For we just add a new integer 1 to our list of integers. Realize that for every number from 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  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

 >= 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.

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