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

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  • Difficulty Level : Medium
  • Last Updated : 16 Aug, 2022
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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, …, aN] 
Then, sum = a1 + a2 + a3 + . . . + aN.
We are given 
b1 = a1 + a1 + a2 + . . . + aN = a1 + sum …..(1)
Similarily,  
b2 = a1 + a2 + a2 + . . . + aN = a2 + sum    …..(2)
. . .  (so on) and in last 
b1 = a1 + a2 + a3 + . . . + aN + aN = aN + sum …..(N)
where [b1, b2, b3 , . . ., bN] are elements of arr[] and,  
total = b1 + b2 + b3 + . . . + bN

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

b1 + b2 + b3 + . . . + bN = (a1 + sum) + (a2 + sum) + . . . + (aN + sum)
total = (a1 + a1 + a2 + . . . + aN) + (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), . . ., aN = (bN – 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.


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