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 |
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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)); } } |
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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 |
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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 |
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Output:
3.6
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