# Generate Array whose difference of each element with its left yields the given Array

• Last Updated : 05 Mar, 2021

Given an integer N and an arr1[], of (N – 1) integers, the task is to find the sequence arr2[] of N integers in the range [1, N] such that arr1[i] = arr2[i+1] – arr2[i]. The integers in sequence arr1[] lies in range [-N, N].
Examples:

Input: N = 3, arr1[] = {-2, 1}
Output: arr2[] = {3, 1, 2}
Explanation:
arr2[1] – arr2[0] = (1 – 3) = -2 = arr1[0]
arr2[2] – arr2[1] = (2 – 1) = 1 = arr1[1]
Input: N = 5, arr1 = {1, 1, 1, 1, 1}
Output: arr2 = {1, 2, 3, 4, 5}
Explanation:
arr2[1] – arr2[0] = (2 – 1) = 1 = arr1[0]
arr2[2] – arr2[1] = (3 – 2) = 1 = arr1[1]
arr2[3] – arr2[2] = (4 – 3) = 1 = arr1[2]
arr2[4] – arr2[3] = (5 – 4) = 1 = arr1[3]

Approach:
Follow the steps to solve the problem:

1. Assume the first element of arr2[] to be X.
2. The next element will be X + arr1[0].
3. The rest of the elements of arr2[] can be represented, w.r.t X.
4. It is known that the sequence arr2[] can contain integers in the range [1, N]. So the minimum possible integer would be 1.
5. The minimum number of the arr2[] can be found out in terms of X, and equate it with 1 to find the value of X.
6. Finally using the values of X, all the other numbers in arr2[] can be found out.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the sequence``void` `find_seq(``int` `arr[],``              ``int` `m, ``int` `n) {``    ``int` `b[n];``    ``int` `x = 0;` `    ``// initializing 1st element``    ``b[0] = x;` `    ``// Creating sequence in``    ``// terms of x``    ``for` `(``int` `i = 0;``         ``i < n - 1; i++) {` `        ``b[i + 1] = x +``                   ``arr[i] + b[i];``    ``}` `    ``int` `mn = n;` `    ``// Finding min element``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``mn = min(mn, b[i]);``    ``}` `    ``// Finding value of x``    ``x = 1 - mn;` `    ``// Creating original sequence``    ``for` `(``int` `i = 0; i < n; i++) {``        ``b[i] += x;``    ``}` `    ``// Output original sequence``    ``for` `(``int` `i = 0; i < n; i++) {``        ``cout << b[i] << ``" "``;``    ``}``    ``cout << endl;``}` `// Driver function``int` `main()``{``    ``int` `N = 3;``    ``int` `arr[] = { -2, 1 };` `    ``int` `M = ``sizeof``(arr) / ``sizeof``(``int``);``    ``find_seq(arr, M, N);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG{``    ` `// Function to find the sequence``static` `void` `find_seq(``int` `arr[], ``int` `m,``                                ``int` `n)``{``    ``int` `b[] = ``new` `int``[n];``    ``int` `x = ``0``;` `    ``// Initializing 1st element``    ``b[``0``] = x;` `    ``// Creating sequence in``    ``// terms of x``    ``for``(``int` `i = ``0``; i < n - ``1``; i++)``    ``{``       ``b[i + ``1``] = x + arr[i] + b[i];``    ``}` `    ``int` `mn = n;` `    ``// Finding min element``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``       ``mn = Math.min(mn, b[i]);``    ``}` `    ``// Finding value of x``    ``x = ``1` `- mn;` `    ``// Creating original sequence``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``       ``b[i] += x;``    ``}` `    ``// Output original sequence``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``System.out.print(b[i] + ``" "``);``    ``}``    ``System.out.println();``}``    ` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``3``;``    ``int` `arr[] = ``new` `int``[]{ -``2``, ``1` `};``    ``int` `M = arr.length;``    ` `    ``find_seq(arr, M, N);``}``}` `// This code is contributed by Pratima Pandey`

## Python3

 `# Python3 program for the above approach` `# Function to find the sequence``def` `find_seq(arr, m, n):``    ` `    ``b ``=` `[]``    ``x ``=` `0``    ` `    ``# Initializing 1st element``    ``b.append(x)``    ` `    ``# Creating sequence in``    ``# terms of x``    ``for` `i ``in` `range``(n ``-` `1``):``        ``b.append(x ``+` `arr[i] ``+` `b[i])``        ` `    ``mn ``=` `n``    ` `    ``# Finding min element``    ``for` `i ``in` `range``(n):``        ``mn ``=` `min``(mn, b[i])``        ` `    ``# Finding value of x``    ``x ``=` `1` `-` `mn``        ` `    ``# Creating original sequence``    ``for` `i ``in` `range``(n):``        ``b[i] ``+``=` `x``        ` `    ``# Output original sequence``    ``for` `i ``in` `range``(n):``        ``print``(b[i], end ``=` `' '``)``    ` `    ``print``()``    ` `# Driver code``if` `__name__``=``=``'__main__'``:``    ` `    ``N ``=` `3``    ``arr ``=` `[ ``-``2``, ``1` `]``    ``M ``=` `len``(arr)``    ` `    ``find_seq(arr, M, N)` `# This code is contributed by rutvik_56`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG{``    ` `// Function to find the sequence``static` `void` `find_seq(``int` `[]arr, ``int` `m,``                                ``int` `n)``{``    ``int` `[]b = ``new` `int``[n];``    ``int` `x = 0;` `    ``// Initializing 1st element``    ``b[0] = x;` `    ``// Creating sequence in``    ``// terms of x``    ``for``(``int` `i = 0; i < n - 1; i++)``    ``{``       ``b[i + 1] = x + arr[i] + b[i];``    ``}` `    ``int` `mn = n;` `    ``// Finding min element``    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ``mn = Math.Min(mn, b[i]);``    ``}` `    ``// Finding value of x``    ``x = 1 - mn;` `    ``// Creating original sequence``    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ``b[i] += x;``    ``}` `    ``// Output original sequence``    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ``Console.Write(b[i] + ``" "``);``    ``}``    ``Console.WriteLine();``}``    ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 3;``    ``int` `[]arr = ``new` `int``[]{ -2, 1 };``    ``int` `M = arr.Length;``    ` `    ``find_seq(arr, M, N);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

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Output:

`3 1 2`

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

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