# Summation of floor of harmonic progression

• Last Updated : 13 Sep, 2022

Given an integer N, the task is to find the summation of the harmonic series .

Examples:

Input: N = 5
Output: 10
floor(3/1) + floor(3/2) + floor(3/3) = 3 + 1 + 1 = 5
Input: N = 20
Output: 66

Naive approach: Run a loop from 1 to N and find the summation of the floor values of N / i. Time complexity of this approach will be O(n).
Efficient approach: Use the following formula to calculate the summation of the series:

Now, the loop needs to be run from 1 to sqrt(N) and the time complexity gets reduced to O(sqrt(N))
Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std;   // Function to return the summation of // the given harmonic series long long int getSum(int n) {       // To store the summation     long long int sum = 0;       // Floor of sqrt(n)     int k = sqrt(n);       // Summation of floor(n / i)     for (int i = 1; i <= k; i++) {         sum += floor(n / i);     }       // From the formula     sum *= 2;     sum -= pow(k, 2);       return sum; }   // Driver code int main() {     int n = 5;       cout << getSum(n);       return 0; }

## Java

 // Java implementation of the approach class GFG {           // Function to return the summation of     // the given harmonic series     static long getSum(int n)     {               // To store the summation         long sum = 0;               // Floor of sqrt(n)         int k = (int)Math.sqrt(n);               // Summation of floor(n / i)         for (int i = 1; i <= k; i++)         {             sum += Math.floor(n / i);         }               // From the formula         sum *= 2;         sum -= Math.pow(k, 2);               return sum;     }           // Driver code     public static void main (String[] args)     {         int n = 5;               System.out.println(getSum(n));     } }   // This code is contributed by AnkitRai01

## Python3

 # Python3 implementation of the approach from math import floor, sqrt, ceil   # Function to return the summation of # the given harmonic series def getSum(n):       # To store the summation     summ = 0       # Floor of sqrt(n)     k =(n)**(.5)       # Summation of floor(n / i)     for i in range(1, floor(k) + 1):         summ += floor(n / i)       # From the formula     summ *= 2     summ -= pow(floor(k), 2)       return summ   # Driver code n = 5   print(getSum(n))   # This code is contributed by Mohit Kumar

## C#

 // C# implementation of the approach using System;       class GFG {           // Function to return the summation of     // the given harmonic series     static double getSum(int n)     {               // To store the summation         double sum = 0;               // Floor of sqrt(n)         int k = (int)Math.Sqrt(n);               // Summation of floor(n / i)         for (int i = 1; i <= k; i++)         {             sum += Math.Floor((double)n / i);         }               // From the formula         sum *= 2;         sum -= Math.Pow(k, 2);               return sum;     }           // Driver code     public static void Main (String[] args)     {         int n = 5;               Console.WriteLine(getSum(n));     } }       // This code is contributed by PrinciRaj1992

## Javascript



Output:

10

Time Complexity: O(sqrt(n)), since the for loop runs for sqrt(n) times.
Auxiliary Space: O(1), since no extra space has been taken.

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