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++

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// 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 

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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Output:

3.6


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