Maximum no. of apples that can be kept in a single basket

Given the ‘N’ number of Basket and the total of Green ‘G’ and Red ‘R’ apples. The task is to distribute all the apples in the Basket and tell the maximum number of apples that can be kept in a basket.
Note: None of the basket is empty.

Examples:

Input: N = 2, R = 1, G = 1
Output:  Maximum apple kept is = 1

Input: N = 2, R = 1, G = 2
Output:  Maximum apple kept is = 2

Approach: The idea is to just check the difference between no. of baskets and total no. of apples(red and Green) i.e. first put 1 apple in 1 basket that means the remaining apples will be extra and can be put together in any basket to make the count maximum. As there is already 1 apple in the basket. So, the maximum number of apples will be (No_of_apples – No_of_baskets) + 1. Since it is mentioned that none of the baskets is empty so apples will always be equal to or greater than no. of baskets.

Below is the implementation of the above approach: 

C++

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

// Function that will calculate the probability
int Number(int Basket, int Red, int Green)
{
    return (Green + Red) - Basket + 1;
}

// Driver code
int main()
{
    int Basket = 3, Red = 5, Green = 3;

    cout << "Maximum apple kept is = "
         << Number(Basket, Red, Green);

    return 0;
}

Java

// Java  implementation of above approach

import java.io.*;

class GFG {

    // Function that will calculate the probability
    static int Number(int Basket, int Red, int Green)
    {
        return (Green + Red) - Basket + 1;
    }

    // Driver code
    public static void main(String[] args)
    {

        int Basket = 3, Red = 5, Green = 3;

        System.out.println("Maximum apple kept is = "
                           + Number(Basket, Red, Green));
    }
    // This Code is Contributed by akt_mit
}

Python3

# Python 3 implementation of above approach

# Function that will calculate
# the probability


def Number(Basket, Red, Green):
    return (Green + Red) - Basket + 1


# Driver code
if __name__ == '__main__':
    Basket = 3
    Red = 5
    Green = 3

    print("Maximum apple kept is =",
          Number(Basket, Red, Green))

# This code is contributed by
# Sanjit_Prasad

C#

// C#  implementation of above approach

using System;

public class GFG {

    // Function that will calculate the probability
    static int Number(int Basket, int Red, int Green)
    {
        return (Green + Red) - Basket + 1;
    }

    // Driver code

    static public void Main()
    {
        int Basket = 3, Red = 5, Green = 3;

        Console.WriteLine("Maximum apple kept is = "
                          + Number(Basket, Red, Green));
    }
    // This Code is Contributed by @ajit
}

PHP

<?php
// PHP implementation of above approach

// Function that will calculate
// the probability
function Number($Basket, $Red, $Green)
{
    return ($Green + $Red) - $Basket + 1;
}

// Driver code
$Basket = 3;
$Red = 5 ;
$Green = 3;

echo "Maximum apple kept is = ",
      Number($Basket, $Red, $Green);

// This code is contributed by ANKITRAI1
?>

Javascript

<script>

// Javascript implementation of above approach

// Function that will calculate the probability
function Number(Basket, Red, Green)
{
    return (Green + Red) - Basket + 1;
}

// Driver code
var Basket = 3, Red = 5, Green = 3;

document.write("Maximum apple kept is = " +
               Number(Basket, Red, Green));

// This code is contributed by rutvik_56

</script>

Output:

Maximum apple kept is = 6

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