Find original Array from given Array where each element is sum of prefix and postfix sum

Given an array arr[] of length N, where arr is derived from an array nums[] which is lost. Array arr[] is derived as: 

arr[i] = (nums[0] + nums[1] + … + nums[i]) + (nums[i] + nums[i+1] + … + nums[N-1]). 

The task is to find nums[] array of length N.

Examples:

Input: N = 4, arr[] = {9, 10, 11, 10}
Output: {1, 2, 3, 2}
Explanation: If nums[] = {1, 2, 3, 2}, then according to above definition
arr[0] = (nums[0]) + (nums[0] + nums[1] + nums[2] + nums[3]) = 1 + 1 + 2 + 3 + 2 = 9
arr[1] = (nums[0] + nums[1]) + (nums[1] + nums[2] + nums[3]) = 1 + 2 + 2 + 3 + 2 = 10
arr[2] = (nums[0] + nums[1] + nums[2]) + (nums[2] + nums[3]) = 1 + 2 + 3 + 3 + 2 = 11
arr[3] = (nums[0] + nums[1] + nums[2] + nums[3]) + (nums[3]) = 1 + 2 + 3 + 2 + 2 = 10

Input: N = 2, arr[] = [25, 20]
Output: [10, 5]

Approach: Follow the below idea to solve the problem:

Suppose nums[] contains [a₁, a₂, a₃, …, a_N]
Then, sum = a₁ + a₂ + a₃ + . . . + a_N.
We are given 
b₁ = a₁ + a₁ + a₂ + . . . + a_N = a₁ + sum …..(1)
Similarily,  
b₂ = a₁ + a₂ + a₂ + . . . + a_N = a₂ + sum    …..(2)
. . .  (so on) and in last 
b₁ = a₁ + a₂ + a₃ + . . . + a_N + a_N = a_N + sum …..(N)
where [b₁, b₂, b₃ , . . ., b_N] are elements of arr[] and,  
total = b₁ + b₂ + b₃ + . . . + b_N

Adding all equation (1) + (2) + (3) + …. + (N) we will get

b₁ + b₂ + b₃ + . . . + b_N = (a₁ + sum) + (a₂ + sum) + . . . + (a_N + sum)
total = (a₁ + a₁ + a₂ + . . . + a_N) + (N * sum)
total = (sum) + (N * sum)
total = (N + 1) * sum

Now find the value of sum variable after that simply:
a₁ = (b₁ – sum), a₂ = (b₂ – sum), . . ., a_N = (b_N – sum)

Using the above idea follow the below steps to implement the code:

• First of all, try to store the sum of elements of arr[] in a variable let’s say total
• Using the formula (N + 1) * sum = total, we will get the value of variable sum which denotes the sum of elements present in the nums[] array.
• At last traverse N times to find nums[0] = arr[0] – sum, nums[1] = arr[1] – sum and so on.
• Return the array and print it.

Below is the implementation of the above approach:

C++

// C++ Algorithm for the above approach

#include <iostream>
#include <vector>
using namespace std;

// Function to find the original
// array nums[]
vector<int> findOrgArray(vector<int> arr, int N)
{
    // Total variable stores the sum of
    // elements of arr[]
    int total = 0;
    for (int val : arr)
        total += val;

    // Sum variable stores the sum of
    // elements of nums[]
    int sum = (total / (N + 1));
    vector<int> v;

    // Traversing to find the elements
    // of nums[]
    for (int i = 0; i < N; i++) {
        int val = arr[i] - sum;
        v.push_back(val);
    }

    // Returning nums[]
    return v;
}

int main()
{

    int N = 4;
    vector<int> arr = { 9, 10, 11, 10 };

    vector<int> v = findOrgArray(arr, N);
    for (auto val : v)
        cout << val << " ";
    return 0;
}

Java

// Java algorithm of the above approach

import java.util.*;

class GFG {

    // Driver Code
    public static void main(String[] args)
    {
        int N = 4;
        int[] arr = { 9, 10, 11, 10 };
        List<Integer> nums = findOrgArray(arr, N);
        for (int x : nums)
            System.out.print(x + " ");
    }

    // Function to find the original
    // array nums[]
    public static List<Integer> findOrgArray(int[] arr,
                                             int N)
    {

        // Total variable stores the sum of
        // elements of arr[]
        int total = 0;
        for (int val : arr)
            total += val;

        // Sum variable stores the sum of
        // elements of nums[]
        int sum = (total / (N + 1));
        List<Integer> nums = new ArrayList<>();

        // Traversing to find the elements
        // of nums[]
        for (int i = 0; i < N; i++) {
            int val = arr[i] - sum;
            nums.add(val);
        }

        // Returning nums[]
        return nums;
    }
}

Python3

# python3 Algorithm for the above approach

# Function to find the original
# array nums[]
def findOrgArray(arr, N) :

    # Total variable stores the sum of
    # elements of arr[]
    total = 0
    for i in arr :
        total+= i

    # Sum variable stores the sum of
    # elements of nums[]
    sum = int(total / (N + 1));
    v = []

    # Traversing to find the elements
    # of nums[]
    for i in range (N) :
        val = arr[i] - sum
        v.append(val)

    # Returning nums[]
    return v

# Driver Code
if __name__ == "__main__" :

    N = 4
    arr = [ 9, 10, 11, 10 ]

    v = findOrgArray(arr, N)
    for val in v :
        print(val,end=' ')

# this code is contributed by aditya942003patil

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;

public class GFG
{

  // Function to find the original
  // array nums[]
  public static List<int> findOrgArray(int[] arr,
                                       int N)
  {

    // Total variable stores the sum of
    // elements of arr[]
    int total = 0;
    //for (int x = 0;  x < arr.count; x++)
    foreach (int val in arr)
      total += val;

    // Sum variable stores the sum of
    // elements of nums[]
    int sum = (total / (N + 1));
    List<int> nums = new List<int>();

    // Traversing to find the elements
    // of nums[]
    for (int i = 0; i < N; i++) {
      int val = arr[i] - sum;
      nums.Add(val);
    }

    // Returning nums[]
    return nums;
  }

  // Driver Code
  public static void Main(String []args)
  {
    int N = 4;
    int[] arr = { 9, 10, 11, 10 };
    List<int> nums = findOrgArray(arr, N);
    for (int x = 0;  x < nums.Count; x++)
      Console.Write(nums[x] + " ");
  }
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// Function to find the original
// array nums[]
function findOrgArray(arr, N)
{
    // Total variable stores the sum of
    // elements of arr[]
    let total = 0;
    for (let i = 0; i < N; i++)
        total += arr[i];

    // Sum variable stores the sum of
    // elements of nums[]
    let sum = (total / (N + 1));
    let v= new Array(N);

    // Traversing to find the elements
    // of nums[]
    for (let i = 0; i < N; i++) {
         v[i] = arr[i] - sum;

    }

    // Returning nums[]
    return v;
}

    let N = 4;
    let arr = [ 9, 10, 11, 10 ];

    let v = findOrgArray(arr, N);
    for (let i = 0; i < N; i++)
        document.write(v[i]+ " ");

        // This code is contributed by satwik4409.
    </script>

Output

1 2 3 2 

Time Complexity: O(N)
Auxiliary Space: O(N), to further reduce it to O(1), store the value in the same given array arr[] rather than storing it in a new array.