# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 `

Output:

```3.6
```

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