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Find the original Array from given array where ith element is the average of first i elements

  • Difficulty Level : Easy
  • Last Updated : 26 Oct, 2021

Given an array arr[] of length N, the task is to find the original array such that every ith element in the given array(arr[i]) is the average value of the first i elements of the original array.

Examples:

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Input: arr = {4 3 3 3} 
Output: 4 2 3 3
Explanation: (4) / 1 = 1, (4+2) / 2 = 3, (4+2+3) / 3 = 3, (4+2+3+3) / 4 = 3



Input: arr = {2 6 8 10} 
Output: 2 10 12 16
Explanation: (2) / 1 = 2, (2+10) / 2 = 6, (2+10+12) / 3 = 8, (2+10+12+16) / 4 = 10

 

Approach: The given problem can be solved using a mathematical approach. Follow the steps below:

  • Initialize a variable sum to the first element of the array arr
  • Iterate the array arr from 2nd index till the end and at every iteration:
    • Multiply current element arr[i] with current index + 1 (i + 1) and subtract the value of sum from it
    • Add the resulting current element to the variable sum
  • Return the resulting array after modification as it will be the original array

C++




// C++ implementation for the above approach
 
#include <iostream>
using namespace std;
 
// Function to find the original
// array from the modified array
void findOriginal(int arr[], int N)
{
 
    // Initialize the variable sum
    // with the first element of array
    int sum = arr[0];
 
    for (int i = 1; i < N; i++) {
 
        // Calculate original element
        // from average of first i elements
        arr[i] = (i + 1) * arr[i] - sum;
 
        // Add current element to sum
        sum += arr[i];
    }
 
    // Print the array
    for (int i = 0; i < N; i++) {
        cout << arr[i] << " ";
    }
}
 
// Driver function
int main()
{
 
    int arr[] = { 2, 6, 8, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Call the function
    findOriginal(arr, N);
 
    return 0;
}

Java




// Java implementation for the above approach
class GFG
{
   
  // Function to find the original
  // array from the modified array
  static void findOriginal(int arr[], int N)
  {
 
      // Initialize the variable sum
      // with the first element of array
      int sum = arr[0];
 
      for (int i = 1; i < N; i++) {
 
          // Calculate original element
          // from average of first i elements
          arr[i] = (i + 1) * arr[i] - sum;
 
          // Add current element to sum
          sum += arr[i];
      }
 
      // Print the array
      for (int i = 0; i < N; i++) {
          System.out.print(arr[i] + " ");
      }
  }
 
  // Driver function
  public static void main(String [] args)
  {
 
      int [] arr = new int [] { 2, 6, 8, 10 };
      int N = arr.length;
 
      // Call the function
      findOriginal(arr, N);
  }   
}
 
// This code is contributed by ihritik

Python3




# Python implementation for the above approach
 
# Function to find the original
# array from the modified array
def findOriginal(arr, N):
 
    # Initialize the variable sum
    # with the first element of array
    sum = arr[0]
 
    for i in range(1, N):
 
        # Calculate original element
        # from average of first i elements
        arr[i] = (i + 1) * arr[i] - sum
 
        # Add current element to sum
        sum = sum + arr[i]
     
 
    # Print the array
    for i in range (0, N):
        print(arr[i], end=" ")
  
 
# Driver function
 
arr= [ 2, 6, 8, 10 ]
N = len(arr)
 
# Call the function
findOriginal(arr, N)
 
 
# This code is contributed by ihritik

C#




// C# program for above approach
using System;
 
class GFG {
 
    // Function to find the original
    // array from the modified array
    static void findOriginal(int[] arr, int N)
    {
 
        // Initialize the variable sum
        // with the first element of array
        int sum = arr[0];
 
        for (int i = 1; i < N; i++) {
 
            // Calculate original element
            // from average of first i elements
            arr[i] = (i + 1) * arr[i] - sum;
 
            // Add current element to sum
            sum += arr[i];
        }
 
        // Print the array
        for (int i = 0; i < N; i++)
            Console.Write(arr[i] + " ");
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int N = 4;
        int[] arr = { 2, 6, 8, 10 };
        findOriginal(arr, N);
    }
}
// This code is contributed by dwivediyash

Javascript




<script>
// JavaScript implementation for the above approach
 
// Function to find the original
// array from the modified array
function findOriginal(arr, N)
{
 
    // Initialize the variable sum
    // with the first element of array
    var sum = arr[0];
 
    for (var i = 1; i < N; i++) {
 
        // Calculate original element
        // from average of first i elements
        arr[i] = (i + 1) * arr[i] - sum;
 
        // Add current element to sum
        sum += arr[i];
    }
 
    // Print the array
    for (var i = 0; i < N; i++) {
           document.write(arr[i] + " ");
    }
}
 
// Driver code
var arr = [ 2, 6, 8, 10 ];
var N = arr.length;
 
// Call the function
findOriginal(arr, N);
 
// This code is contributed by AnkThon
 
</script>

 
 

Output
2 10 12 16 

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 




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