Given an array arr[] of size N, the task is to find the average of the array elements without running into overflow.
Examples:
Input: arr[] = { INT_MAX, INT_MAX }
Output:
Average by Standard method: -1.0000000000
Average by Efficient method: 2147483647.0000000000
Explanation:
The average of the two numbers by standard method is (sum / 2).
Since the sum of the two numbers exceed INT_MAX, the obtained output by standard method is incorrect.Input: arr[] = { INT_MAX, 1, 2 }
Output:
Average by Standard method: -715827882.0000000000
Average by Efficient method: 715827883.3333332539
Approach: The given problem can be solved based on the following observations:
- The average of N array elements can be obtained by dividing the sum of the array elements by N. But, calculating sum of the array arr[] may lead to integer overflow, if the array contains large integers.
- Therefore, average of the array can be calculated efficiently by the following steps:
- Traverse the array, using a variable i over the range of indices [0, N – 1]
- Update avg = (avg+ (arr[i] – avg)/(i+1))
Follow the steps below to solve the problem:
- Initialize two variables, say sum as 0 and avg as 0, to store the sum and average of the array elements respectively.
- Traverse the array arr[], update avg = avg + (arr[i] – avg) / (i + 1) and update sum = sum + arr[i].
- After completing the above steps, print the average by the standard method, i.e. sum / N and print the average by the efficient method, i.e. avg
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate average of// an array using standard methoddouble average(int arr[], int N){ // Stores the sum of array int sum = 0; // Find the sum of the array for (int i = 0; i < N; i++) sum += arr[i]; // Return the average return (double)sum / N;}// Function to calculate average of// an array using efficient methoddouble mean(int arr[], int N){ // Store the average of the array double avg = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // Update avg avg += (arr[i] - avg) / (i + 1); } // Return avg return avg;}// Driver Codeint main(){ // Input int arr[] = { INT_MAX, 1, 2 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call cout << "Average by Standard method: " << fixed << setprecision(10) << average(arr, N) << endl; cout << "Average by Efficient method: " << fixed << setprecision(10) << mean(arr, N) << endl; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to calculate average of// an array using standard methodstatic Double average(int arr[], int N){ // Stores the sum of array int sum = 0; // Find the sum of the array for (int i = 0; i < N; i++) sum += arr[i]; // Return the average return Double.valueOf(sum / N);}// Function to calculate average of// an array using efficient methodstatic Double mean(int arr[], int N){ // Store the average of the array Double avg = 0.0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // Update avg avg += Double.valueOf((arr[i] - avg) / (i + 1)); } // Return avg return avg;}// Driver Codepublic static void main(String args[]){ // Input int arr[] = {Integer.MAX_VALUE, 1, 2 }; int N = arr.length; // Function call System.out.println("Average by Standard method: "+ String.format("%.10f", average(arr, N))); System.out.println("Average by Efficient method: "+ String.format("%.10f", mean(arr, N)));}}// This code is contributed by ipg2016107. |
Python3
# Python3 program for the above approachimport sys# Function to calculate average of# an array using standard methoddef average(arr, N): # Stores the sum of array sum = 0 # Find the sum of the array for i in range(N): sum += arr[i] # Return the average return sum // N * 1.0 - 1# Function to calculate average of# an array using efficient methoddef mean(arr, N): # Store the average of the array avg = 0 # Traverse the array arr[] for i in range(N): # Update avg avg += (arr[i] - avg) / (i + 1) # Return avg return round(avg, 7)# Driver Codeif __name__ == '__main__': # Input arr = [2147483647, 1, 2] N = len(arr) # Function call print("Average by Standard method: ","{:.10f}".format( -1.0 * average(arr, N))) print("Average by Efficient method: ","{:.10f}".format( mean(arr, N)))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to calculate average of// an array using standard methodstatic double average(int[] arr, int N){ // Stores the sum of array int sum = 0; // Find the sum of the array for(int i = 0; i < N; i++) sum += arr[i]; // Return the average return (double)(sum / N);}// Function to calculate average of// an array using efficient methodstatic double mean(int[] arr, int N){ // Store the average of the array double avg = 0.0; // Traverse the array arr[] for(int i = 0; i < N; i++) { // Update avg avg += ((double)((arr[i] - avg) / (i + 1))); } // Return avg return avg;}// Driver Codestatic public void Main(){ // Input int[] arr = { Int32.MaxValue, 1, 2 }; int N = arr.Length; // Function call Console.WriteLine("Average by Standard method: " + (average(arr, N)).ToString("F10")); Console.WriteLine("Average by Efficient method: " + (mean(arr, N)).ToString("F10"));}}// This code is contributed by Dharanendra L V. |
Javascript
<script>// Javascript program for the above approach// Function to calculate average of// an array using standard methodfunction average(arr, N){ // Stores the sum of array var sum = 0; // Find the sum of the array for(var i = 0; i < N; i++) sum += arr[i]; if(sum>2147483647) { sum = -2147483647 + (sum - 2147483649) } // Return the average return parseInt(sum / N);}// Function to calculate average of// an array using efficient methodfunction mean(arr, N){ // Store the average of the array var avg = 0; // Traverse the array arr[] for(var i = 0; i < N; i++) { // Update avg avg += parseFloat((arr[i] - avg) / (i + 1)); } // Return avg return avg;}// Driver Code// Inputvar arr = [2147483647, 1, 2 ];var N = arr.length// Function calldocument.write( "Average by Standard method: " + average(arr, N).toFixed(10) + "<br>");document.write( "Average by Efficient method: " + mean(arr, N).toFixed(10)+ "<br>");</script> |
Output:
Average by Standard method: -715827882.0000000000 Average by Efficient method: 715827883.3333332539
Time Complexity: O(N)
Auxiliary Space: O(1)
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