Program to find the Nth Harmonic Number (original) (raw)

Last Updated : 20 Mar, 2023

Given a number N. The task is to find the Nth Harmonic Number.
Let the nth harmonic number be Hn.
The harmonic series is as follows:

H1 = 1
H2 = H1 + 1/2
H3 = H2 + 1/3
H4 = H3 + 1/4
.
.
.
Hn = Hn-1 + 1/n

Examples:

Input : N = 5 Output : 2.45

Input : N = 9 Output : 2.71786

The idea is to traverse from H1 and then consecutively keep finding H2 from H1, H3 from H2 ..... and so on.
Below is the program to find N-th Harmonic Number:

C++ `

// CPP program to find N-th Harmonic Number

#include using namespace std;

// Function to find N-th Harmonic Number double nthHarmonic(int N) { // H1 = 1 float harmonic = 1.00;

// loop to apply the formula 
// Hn = H1 + H2 + H3 ... + Hn-1 + Hn-1 + 1/n
for (int i = 2; i <= N; i++) {
    harmonic += (float)1 / i;
}

return harmonic;

}

// Driver Code int main() { int N = 8;

cout<<nthHarmonic(N);

return 0;

}

Java

// Java program to find N-th Harmonic Number

import java.io.*;

class GFG {

// Function to find N-th Harmonic Number static double nthHarmonic(int N) { // H1 = 1 float harmonic = 1;

// loop to apply the formula 
// Hn = H1 + H2 + H3 ... + Hn-1 + Hn-1 + 1/n
for (int i = 2; i <= N; i++) {
    harmonic += (float)1 / i;
}

return harmonic;

}

// Driver Code

public static void main (String[] args) {
        int N = 8;

System.out.print(nthHarmonic(N));

}

} // This code is contributed // by ajit

Python 3

Python3 program to find

N-th Harmonic Number

Function to find N-th Harmonic Number

def nthHarmonic(N) :

# H1 = 1 
harmonic = 1.00

# loop to apply the formula 
# Hn = H1 + H2 + H3 ... + 
# Hn-1 + Hn-1 + 1/n 
for i in range(2, N + 1) :
    harmonic += 1 / i

return harmonic

Driver code

if name == "main" :

N = 8
print(round(nthHarmonic(N),5))

This code is contributed by ANKITRAI1

C#

// C# program to find N-th Harmonic Number using System;

class GFG {

// Function to find N-th Harmonic Number static double nthHarmonic(int N) { // H1 = 1 float harmonic = 1;

// loop to apply the formula 
// Hn = H1 + H2 + H3 ... + 
// Hn-1 + Hn-1 + 1/n
for (int i = 2; i <= N; i++)
{
    harmonic += (float)1 / i;
}

return harmonic;

}

// Driver Code static public void Main () { int N = 8;

Console.Write(nthHarmonic(N));

} }

// This code is contributed // by Raj

PHP

JavaScript

`

Time Complexity: O(N)

Auxiliary Space: O(1) as using constant space, since no extra space has been taken.

Approach 2: Dynamic Programming:

The DP approach is better than the simple iterative approach because it avoids recomputing the sum from scratch every time. In the simple iterative approach, we add each term of the harmonic series from 1 to N one by one in every iteration. This means that we perform N-1 additions in total, which can be time-consuming for large values of N.

we only need to perform N-1 additions once and store the results in the harmonic vector. Then, we can simply access the N-th Harmonic number from the vector in constant time, which is much faster than recomputing the sum from scratch every time.

C++ `

#include #include using namespace std;

// Function to find N-th Harmonic Number double nthHarmonic(int N) { // H1 = 1 vector harmonic(N+1); harmonic[1] = 1.0; // loop to apply the formula // Hn = H1 + H2 + H3 ... + Hn-1 + Hn-1 + 1/n for (int i = 2; i <= N; i++) { harmonic[i] = harmonic[i-1] + (double)1 / i; }

return harmonic[N]; }

// Driver Code int main() { int N = 8; cout<<nthHarmonic(N);

return 0; }

Java

import java.util.*;

public class HarmonicNumber { // Function to find N-th Harmonic Number public static double nthHarmonic(int N) { // H1 = 1 List harmonic = new ArrayList<>(); harmonic.add(0.0); // Add a dummy value to align with the C++ vector indexing harmonic.add(1.0);

    // loop to apply the formula
    // Hn = H1 + H2 + H3 ... + Hn-1 + Hn-1 + 1/n
    for (int i = 2; i <= N; i++) {
        harmonic.add(harmonic.get(i-1) + 1.0 / i);
    }

    return harmonic.get(N);
}

// Driver Code
public static void main(String[] args) {
    int N = 8;
    System.out.println(nthHarmonic(N));
}

}

Python3

def nthHarmonic(N): # H1 = 1 harmonic = [0.0]*(N+1) harmonic[1] = 1.0

# loop to apply the formula
# Hn = H1 + H2 + H3 ... + Hn-1 + Hn-1 + 1/n
for i in range(2, N+1):
    harmonic[i] = harmonic[i-1] + 1.0 / i

return harmonic[N]

Driver Code

N = 8 print(nthHarmonic(N))

C#

using System;

class HarmonicNumber { // Function to find N-th Harmonic Number static double nthHarmonic(int N) { // H1 = 1 double[] harmonic = new double[N+1]; harmonic[1] = 1.0;

    // loop to apply the formula
    // Hn = H1 + H2 + H3 ... + Hn-1 + Hn-1 + 1/n
    for (int i = 2; i <= N; i++) {
        harmonic[i] = harmonic[i-1] + 1.0 / i;
    }

    return harmonic[N];
}

// Driver Code
static void Main(string[] args) {
    int N = 8;
    Console.WriteLine(nthHarmonic(N));
}

}

JavaScript

// Function to find N-th Harmonic Number function nthHarmonic(N) { // H1 = 1 const harmonic = new Array(N + 1); harmonic[1] = 1.0; // loop to apply the formula // Hn = H1 + H2 + H3 ... + Hn-1 + Hn-1 + 1/n for (let i = 2; i <= N; i++) { harmonic[i] = harmonic[i - 1] + (1 / i); }

return harmonic[N];

}

// Driver Code const N = 8; console.log(nthHarmonic(N));

`

Time Complexity: O(N)

Auxiliary Space: O(1)