Find the average of k digits from the beginning and l digits from the end of the given number

Given three integers N, K and L. The task is to find the average of the first K digits and the last L digits of the given number N without any digit overlapping.
Examples:

Input: N = 123456, K = 2, L = 3 
Output: 3.0 
Sum of first K digits will be 1 + 2 = 3 
Sum of last L digits will be 4 + 5 + 6 = 15 
Average = (3 + 15) / (2 + 3) = 18 / 5 = 3
Input: N = 456966, K = 1, L = 1 
Output: 5.0 

Approach: If the count of digits in n is less than (K + L) then it isn’t possible to find the average without digits overlapping and print -1 in that case. If that’s not the case, find the sum of the last L digits of N and store it in a variable say sum1 then find the sum of the first K digits of N and store it in sum2. Now, print the average as (sum1 + sum2) / (K + L).
Below is the implementation of the above approach: 

C++

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

// Function to return the count
// of digits in num
int countDigits(int num)
{
    int cnt = 0;
    while (num > 0)
    {
        cnt++;
        num /= 10;
    }
    return cnt;
}

// Function to return the sum
// of first n digits of num
int sumFromStart(int num, int n, int rem)
{

    // Remove the unnecessary digits
    num /= ((int)pow(10, rem));

    int sum = 0;
    while (num > 0)
    {
        sum += (num % 10);
        num /= 10;
    }
    return sum;
}

// Function to return the sum
// of the last n digits of num
int sumFromEnd(int num, int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum += (num % 10);
        num /= 10;
    }
    return sum;
}

float getAverage(int n, int k, int l)
{

    // If the average can't be calculated without
    // using the same digit more than once
    int totalDigits = countDigits(n);
    if (totalDigits < (k + l))
        return -1;

    // Sum of the last l digits of n
    int sum1 = sumFromEnd(n, l);

    // Sum of the first k digits of n
    // (totalDigits - k) must be removed from the
    // end of the number to get the remaining
    // k digits from the beginning
    int sum2 = sumFromStart(n, k, totalDigits - k);

    // Return the average
    return ((float)(sum1 + sum2) /
            (float)(k + l));
}

// Driver code
int main()
{
    int n = 123456, k = 2, l = 3;
    cout << getAverage(n, k, l);

    return 0;
}

// This code is contributed by PrinciRaj1992

Java

// Java implementation of the approach
class GFG {

    // Function to return the count
    // of digits in num
    public static int countDigits(int num)
    {
        int cnt = 0;
        while (num > 0) {
            cnt++;
            num /= 10;
        }
        return cnt;
    }

    // Function to return the sum
    // of first n digits of num
    public static int sumFromStart(int num, int n, int rem)
    {

        // Remove the unnecessary digits
        num /= ((int)Math.pow(10, rem));

        int sum = 0;
        while (num > 0) {
            sum += (num % 10);
            num /= 10;
        }
        return sum;
    }

    // Function to return the sum
    // of the last n digits of num
    public static int sumFromEnd(int num, int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum += (num % 10);
            num /= 10;
        }
        return sum;
    }

    public static float getAverage(int n, int k, int l)
    {

        // If the average can't be calculated without
        // using the same digit more than once
        int totalDigits = countDigits(n);
        if (totalDigits < (k + l))
            return -1;

        // Sum of the last l digits of n
        int sum1 = sumFromEnd(n, l);

        // Sum of the first k digits of n
        // (totalDigits - k) must be removed from the
        // end of the number to get the remaining
        // k digits from the beginning
        int sum2 = sumFromStart(n, k, totalDigits - k);

        // Return the average
        return ((float)(sum1 + sum2) / (float)(k + l));
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 123456, k = 2, l = 3;
        System.out.print(getAverage(n, k, l));
    }
}

Python3

# implementation of the approach
from math import pow

# Function to return the count
# of digits in num
def countDigits(num):
    cnt = 0
    while (num > 0):
        cnt += 1
        num //= 10
    return cnt

# Function to return the sum
# of first n digits of num
def sumFromStart(num, n, rem):

    # Remove the unnecessary digits
    num //= pow(10, rem)

    sum = 0
    while (num > 0):
        sum += (num % 10)
        num //= 10
    return sum

# Function to return the sum
# of the last n digits of num
def sumFromEnd(num, n):
    sum = 0
    for i in range(n):
        sum += (num % 10)
        num //= 10

    return sum

def getAverage(n, k, l):

    # If the average can't be calculated without
    # using the same digit more than once
    totalDigits = countDigits(n)
    if (totalDigits < (k + l)):
        return -1

    # Sum of the last l digits of n
    sum1 = sumFromEnd(n, l)

    # Sum of the first k digits of n
    # (totalDigits - k) must be removed from the
    # end of the number to get the remaining
    # k digits from the beginning
    sum2 = sumFromStart(n, k, totalDigits - k)

    # Return the average
    return (sum1 + sum2) / (k + l)

# Driver code
if __name__ == '__main__':
    n = 123456
    k = 2
    l = 3
    print(getAverage(n, k, l))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the count
    // of digits in num
    public static int countDigits(int num)
    {
        int cnt = 0;
        while (num > 0)
        {
            cnt++;
            num /= 10;
        }
        return cnt;
    }

    // Function to return the sum
    // of first n digits of num
    public static int sumFromStart(int num,
                                   int n, int rem)
    {

        // Remove the unnecessary digits
        num /= ((int)Math.Pow(10, rem));

        int sum = 0;
        while (num > 0)
        {
            sum += (num % 10);
            num /= 10;
        }
        return sum;
    }

    // Function to return the sum
    // of the last n digits of num
    public static int sumFromEnd(int num, int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += (num % 10);
            num /= 10;
        }
        return sum;
    }

    public static float getAverage(int n, int k, int l)
    {

        // If the average can't be calculated without
        // using the same digit more than once
        int totalDigits = countDigits(n);
        if (totalDigits < (k + l))
            return -1;

        // Sum of the last l digits of n
        int sum1 = sumFromEnd(n, l);

        // Sum of the first k digits of n
        // (totalDigits - k) must be removed from the
        // end of the number to get the remaining
        // k digits from the beginning
        int sum2 = sumFromStart(n, k, totalDigits - k);

        // Return the average
        return ((float)(sum1 + sum2) /
                (float)(k + l));
    }

    // Driver code
    public static void Main(String []args)
    {
        int n = 123456, k = 2, l = 3;
        Console.WriteLine(getAverage(n, k, l));
    }
}

// This code is contributed by Princi Singh

Javascript

<script>
// javascript implementation of the approach

    // Function to return the count
    // of digits in num
    function countDigits(num)
    {
        var cnt = 0;
        while (num > 0)
        {
            cnt++;
            num = parseInt(num/10);
        }
        return cnt;
    }

    // Function to return the sum
    // of first n digits of num
    function sumFromStart(num, n, rem)
    {

        // Remove the unnecessary digits
        num = (parseInt( num/Math.pow(10, rem)));

        var sum = 0;
        while (num > 0)
        {
            sum += (num % 10);
            num = parseInt(num/10);
        }
        return sum;
    }

    // Function to return the sum
    // of the last n digits of num
    function sumFromEnd(num , n)
    {
        var sum = 0;
        for (i = 0; i < n; i++)
        {
            sum += (num % 10);
            num = parseInt(num/10);
        }
        return sum;
    }

    function getAverage(n , k , l) {

        // If the average can't be calculated without
        // using the same digit more than once
        var totalDigits = countDigits(n);
        if (totalDigits < (k + l))
            return -1;

        // Sum of the last l digits of n
        var sum1 = sumFromEnd(n, l);

        // Sum of the first k digits of n
        // (totalDigits - k) must be removed from the
        // end of the number to get the remaining
        // k digits from the beginning
        var sum2 = sumFromStart(n, k, totalDigits - k);

        // Return the average
        return ( (sum1 + sum2) /  (k + l));
    }

    // Driver code
    var n = 123456, k = 2, l = 3;
    document.write(getAverage(n, k, l));

// This code is contributed by Rajput-Ji
</script>

Output:

3.6