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

• Difficulty Level : Medium
• Last Updated : 19 Jul, 2022

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));               } }

## Python3

 # 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 ?>

## Javascript

 

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.

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