Find N fractions that sum upto a given fraction N/D

Given a fraction N/D, the task is to split this fraction into N parts such that their sum is equal to the fraction N/D, i.e., 

Note: Represents the terms in terms of fractions, instead of floating point numbers.

Input: N = 4, D = 2 
Output: 4/5, 1/5, 1/3, 4/6 
Explanation:

Therefore, it is a valid set of fractions such that their sum is 

Input: N = 3, D = 4 
Output: 1/2, 1/10, 3/20 
Explanation:

Therefore, it is a valid set of fractions such that their sum is 

Approach: The key observation in the problem is that the first fraction numerator can be and then further denominators can be using the below recurrence relation.  

Below is the implementation of the above approach: 

C++

// C++ implementation to split the
// fraction into N parts
#include<bits/stdc++.h>
using namespace std;

// Function to split the fraction
// into the N parts
void splitFraction(int n, int d)
{
    int ar[n];
    int first = d + n - 1;
    ar[0] = first;

    // Loop to find the N - 1
    // fraction
    for(int i = 1; i < n; i++)
    {
       int temp = --first;
       first++;

       ar[i] = first * temp;
       --first;
    }

    // Loop to print the Fractions
    for(int i = 0; i < n; i++)
    {
       if (ar[i] % n == 0)
       {
           cout << "1/" << ar[i] / n << ", ";
       }
       else
       {
           cout << n << "/" << ar[i] << ", ";
       }
    }
}

// Driver Code
int main()
{
    int N = 4;
    int D = 2;

    // Function Call
    splitFraction(N, D);
}

// This code is contributed by Bhupendra_Singh

Java

// Java implementation to split the
// fraction into N parts

import java.util.Scanner;

class Solution {

    // Function to split the fraction
    // into the N parts
    public static void
    splitFraction(int n, int d)
    {

        long ar[] = new long[n];
        long first = d + n - 1;
        ar[0] = first;

        // Loop to find the N - 1
        // fraction
        for (int i = 1; i < n; i++) {
            ar[i] = first * (--first);
        }

        // Loop to print the Fractions
        for (int i = 0; i < n; i++) {
            if (ar[i] % n == 0) {
                System.out.print(
                    "1/" + ar[i] / n
                    + ", ");
            }
            else {
                System.out.print(
                    n + "/" + ar[i]
                    + ", ");
            }
        }
    }

    // Driver Code
    public static void main(
        String[] args) throws Exception
    {
        int N = 4;
        int D = 2;

        // Function Call
        splitFraction(N, D);
    }
}

Python3

# Python3 implementation to split the
# fraction into N parts

# Function to split the fraction
# into the N parts
def splitFraction(n, d):

    ar = []
    for i in range(0, n):
        ar.append(0)

    first = d + n - 1
    ar[0] = first

    # Loop to find the N - 1
    # fraction
    for i in range(1, n):
        temp = first - 1
        ar[i] = first * temp
        first -= 1

    # Loop to print the Fractions
    for i in range(0, n):
        if ar[i] % n == 0:
            print("1/", int(ar[i] / n),
                  "," , end = " ")

        else:
            print(n, "/", ar[i], ",", end = " ")

# Driver Code
N = 4
D = 2

# Function Call
splitFraction(N, D)

# This code is contributed by ishayadav181

C#

// C# implementation to split the
// fraction into N parts
using System;

class GFG{

// Function to split the fraction
// into the N parts
public static void splitFraction(int n, int d)
{
    long []ar = new long[n];
    long first = d + n - 1;
    ar[0] = first;

    // Loop to find the N - 1
    // fraction
    for(int i = 1; i < n; i++)
    {
       ar[i] = first * (--first);
    }

    // Loop to print the Fractions
    for(int i = 0; i < n; i++)
    {
       if (ar[i] % n == 0)
       {
           Console.Write("1/" + ar[i] / n + ", ");
       }
       else
       {
           Console.Write(n + "/" + ar[i] + ", ");
       }
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N = 4;
    int D = 2;

    // Function Call
    splitFraction(N, D);
}
}

// This code is contributed by SoumikMondal

Javascript

<script>
      // JavaScript implementation to split the
      // fraction into N parts
      // Function to split the fraction
      // into the N parts
      function splitFraction(n, d) {
        var ar = new Array(n);
        var first = d + n - 1;
        ar[0] = first;

        // Loop to find the N - 1
        // fraction
        for (var i = 1; i < n; i++) {
          ar[i] = first * --first;
        }

        // Loop to print the Fractions
        for (var i = 0; i < n; i++) {
          if (ar[i] % n === 0) {
            document.write("1/" + ar[i] / n + ", ");
          } else {
            document.write(n + "/" + ar[i] + ", ");
          }
        }
      }

      // Driver Code
      var N = 4;
      var D = 2;

      // Function Call
      splitFraction(N, D);
    </script>

Output:

4/5, 1/5, 1/3, 4/6,

Time Complexity: O(n), where n is the given integer.

Auxiliary Space: O(n), where n is the given integer.