Program to find the rate percentage from compound interest of consecutive years

Given two integers N1 and N2 which is the Compound Interest of two consecutive years. The task is to calculate the rate percentage.
Examples:

Input: N1 = 660, N2 = 720 
Output: 9.09091 %
Input: N1 = 100, N2 = 120 
Output: 20 % 

Approach: The rate percentage can be calculated with the formula ((N2 – N1) * 100) / N1 where N1 is the compound interest of some year and N2 is the compound interest for the next year.  

Let us consider the 1st Example: 
The difference between the Compound interest in the two consecutive years is because of the interest received on the previous year interest. Therefore, 
–> N2 – N1 = N1 * (Rate / 100) 
–> 720 – 660 = 660 * (Rate / 100) 
–> (60 / 660) * 100 = Rate 
–> Rate = (100 / 11) = 9.09% (Approx)  

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the
// required rate percentage
float Rate(int N1, int N2)
{
    float rate = (N2 - N1) * 100 / float(N1);

    return rate;
}

// Driver code
int main()
{
    int N1 = 100, N2 = 120;

    cout << Rate(N1, N2) << " %";

    return 0;
}

Java

// Java implementation of the approach

class GFG
{

    // Function to return the
    // required rate percentage
    static int Rate(int N1, int N2)
    {
        float rate = (N2 - N1) * 100 / N1;

        return (int)rate;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N1 = 100, N2 = 120;

        System.out.println(Rate(N1, N2) + " %");
    }
}

// This code has been contributed by 29AjayKumar

Python 3

# Python 3 implementation of the approach

# Function to return the
# required rate percentage
def Rate( N1, N2):
    rate = (N2 - N1) * 100 // (N1);

    return rate

# Driver code
if __name__ == "__main__":
    N1 = 100
    N2 = 120

    print(Rate(N1, N2) ," %")

# This code is contributed by ChitraNayal

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the
    // required rate percentage
    static int Rate(int N1, int N2)
    {
        float rate = (N2 - N1) * 100 / N1;

        return (int)rate;
    }

    // Driver code
    static public void Main ()
    {
        int N1 = 100, N2 = 120;

        Console.WriteLine(Rate(N1, N2) + " %");
    }
}

// This code has been contributed by ajit.

PHP

<?php
// PHP implementation of the approach

// Function to return the
// required rate percentage
function Rate($N1, $N2)
{
    $rate = ($N2 - $N1) * 100 / $N1;

    return $rate;
}

// Driver code
$N1 = 100;
$N2 = 120;

echo Rate($N1, $N2), "%";

// This code is contributed by AnkitRai01
?>

Javascript

<script>

// javascript implementation of the approach

    // Function to return the
    // required rate percentage
    function Rate(N1 , N2) {
        var rate = (N2 - N1) * 100 / N1;

        return parseInt( rate);
    }

    // Driver code

        var N1 = 100, N2 = 120;

        document.write(Rate(N1, N2) + " %");

// This code contributed by Rajput-Ji

</script>

Output:

20 %

Time Complexity: O(1), as there is only basic arithmetic operation.

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