Generate a Random Number between 0 and 1

The Task is to generate a random number between 0 and 1.

It is obvious that the number between 0 and 1 will be a floating point number. To generate a random number between 0 and 1, we will make use of the rand() function. The rand() function creates a random number.

Approach: Generating a Random Number between 0 and 1 with RAND_MAX value

RAND_MAX value is basically the maximum value that can be obtained by the rand() function. So, first of all, we will be using the srand() function for creating different values on every run.

Follow the steps mentioned below to implement the idea:

• Start a loop to check whether we are getting different random numbers on every iteration.
• Then, typecast the rand() function value in a double value as the number we want should be a decimal point number. 
• inside the loop, we will output the random value obtained by rand() function divided by the RAND_MAX value.
• As the RAND_MAX value will be greater than the random number obtained by rand() function most of the time, we will get the result as ‘0.n’. 

Below is the implementation of the above approach:

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

int main()
{
    // Using srand() with time(0) to change
    // the random values everytime
    srand(time(0));

    // Loop to check that we get different values
    // on every iteration
    for (int i = 0; i < 10; i++) {

        // Typecasting the random value
        // obtained into a double and
        // then dividing it with RAND_MAX
        cout << ((double)rand()) / RAND_MAX << endl;
    }

    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
import java.util.Random;
import java.util.UUID;

class GFG {
    public static void main (String[] args) {
       for(int i = 0; i < 10; i++)
        {
            Random rand = new Random();
            System.out.println(String.format("%.6f",
                                             (double)Math.abs(rand.nextInt())/Integer.MAX_VALUE));
        }
    }
}

// This code is contributed by ksam24000.

C#

using System;

public class GFG {

    static public void Main()
    {

        // Code
        for (int i = 0; i < 10; i++) {
            Random rand = new Random();
            Console.WriteLine(string.Format(
                "{0:F6}", (double)Math.Abs(rand.Next())
                              / Int32.MaxValue));
        }
    }
}

// This code is contributed by lokeshmvs21.

Javascript

// driver code
for(let i = 0; i < 10; i++)
{
   document.write(Math.random())
}

// This code is contributed by garg28harsh.

Python3

import random
import time

random.seed(time.time())

for i in range(10):
    print(random.random())

Output

0.288481
0.338828
0.512149
0.810278
0.560978
0.683509
0.0417002
0.332409
0.0942754
0.938507

Time Complexity: O(1) 
Auxiliary Space: O(1)

Another Approach:

The same thing can be done using INT_MAX instead of using RAND_MAX.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

int main()
{
    // using srand() with time(0) to change the random values everytime when the program
    // is ran
    srand(time(0));

    // for loop to check that we get different values on every iteration
    for (int i = 1; i < 11; i++) {

        // typecasting the random value obtained into a double and then dividing it
        // with INT_MAX
        cout << ((double)rand()) / INT_MAX << endl;
    }
}

Java

// Java code for above approach
import java.io.*;
import java.util.Random;
import java.util.UUID;

class GFG {
    public static void main (String[] args) {
    // for loop to check that we get different values on every iteration
    for (int i = 1; i < 11; i++)
    {
            Random rand = new Random();
            // typecasting the random value obtained into a double and then dividing it
            // with INT_MAX
            System.out.println(String.format("%.6f",
                                            (double)Math.abs(rand.nextInt())/Integer.MAX_VALUE));
        }
    }
}

// This code is contributed by Pushpesh Raj.

C#

// C# code for above approach
using System;

class Program {
    static void Main(string[] args)
    {
        // for loop to check that we get different values on
        // every iteration
        for (int i = 1; i < 11; i++) {
            Random rand = new Random();
            // typecasting the random value obtained into a
            // double and then dividing it with int.MaxValue
            Console.WriteLine("{0:F6}",
                              (double)Math.Abs(rand.Next())
                                  / int.MaxValue);
        }
    }
}
// This code is contributed by Prajwal Kandekar

Output

0.645138
0.249392
0.313409
0.628442
0.491985
0.902581
0.400526
0.123816
0.397556
0.520698

Time Complexity: O(1) 
Auxiliary Space: O(1)

Approach for generating a random number by defining a range:

The idea of this approach is that we will use the rand() function to get the random numbers between the defined range and then dividing it with a slightly larger number than the largest number in the defined range. 

Follow the steps mentioned below to implement the idea:

• Run a loop and Inside the loop, set the rand() function to create a random number between the defined range. 
• Then take range from 1 to 10⁹. We are taking a big range of numbers as the number should not repeat most of the time. 
  • If the number repeats, we will get the same number between 0 and 1 every time the number repeats. Taking a small range of numbers will cause the program to output the same random values most of the time. 
• After defining the range, we will be dividing the obtained number by n + 1, with n being the largest number in the defined range.

Below is the Implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

int main()
{
    srand(time(0));
    for (int i = 0; i < 10; i++) {
        // defining a range of numbers and then dividing the obtained random number
        // with the largest number in the range + 1
        cout << ((double)(rand() % 100000000) + 1) / 100000001 << endl;
    }
}

Output

0.682537
0.213042
0.429056
0.583094
0.294206
0.327242
0.309042
0.165038
0.665797
0.542586

Time Complexity: O(1), since loop is traversed only from 1 to 10, and rand() method used O(1) time
Auxiliary Space: O(1), since only variables are created.