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

• Difficulty Level : Medium
• Last Updated : 28 May, 2021

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:

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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 nint countBits(int n){    int count = 0;    while (n) {        count++;        n >>= 1;    }    return count;} // Driver codeint 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 nstatic 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 ndef countBits(n):     count = 0;    while (n):        count += 1;        n >>= 1;             return count; # Driver coden = 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 Nusing System; class GFG{// function to count length of// binary expression of nstatic int countBits(int n){    int count = 0;    while (n > 0)    {        count++;        n >>= 1;    }    return count;} // Driver codestatic 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

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

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