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

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 – arr2 = (1 – 3) = -2 = arr1
arr2 – arr2 = (2 – 1) = 1 = arr1

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.
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 = 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 = 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 `

Output:

```3 1 2
```

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

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