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++ implementation of the approach #include <bits/stdc++.h> 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 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 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# 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 |
<script> // Javascript implementation of the approach // Function to return the summation of // the given harmonic series function getSum(n)
{ // To store the summation
let sum = 0;
// Floor of sqrt(n)
let k = parseInt(Math.sqrt(n));
// Summation of floor(n / i)
for (let i = 1; i <= k; i++) {
sum += Math.floor(n / i);
}
// From the formula
sum *= 2;
sum -= Math.pow(k, 2);
return sum;
} // Driver code let n = 5;
document.write(getSum(n));
</script> |
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.