# Reconstruct the array by replacing arr[i] with (arr[i-1]+1) % M

Given an array of N elements and an integer M. Now, the array is modified by replacing some of the array elements with -1. The task is to print the original array.

The elements in the orginal array are related as, for every index i, a[i] = (a[i-1]+1)% M.

It is guaranteed that there is one non zero value in the array.

Examples:

```Input: arr[] = {5, -1, -1, 1, 2, 3}, M = 7
Output: 5 6 0 1 2 3
M = 7, so value at index 2 should be (5+1) % 7 = 6
value at index 3 should be (6+1) % 7 = 0

Input: arr[] = {5, -1, 7, -1, 9, 0}, M = 10
Output: 5 6 7 8 9 0
```

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

Approach: First find the index of the non negative value index i. Then simply go in two directions i.e. From i-1 to 0 and i+1 to n.

• For index i-1 the value can be calculated by (a[i+1]-1+m)%m because (a – b) mod p = ((a mod p – b mod p) + p) mod p.
• For indexes i+1 the values can be calculated by (a[i-1]+1)%m.

Below is the implementation of the above approach:

## C++

 `// C++ implemenatation of above approach  ` `#include   ` `using` `namespace` `std;  ` ` `  `void` `construct(``int` `n, ``int` `m, ``int` `a[]) ` `{ ` `    ``int` `ind = 0; ` ` `  `    ``// Finding the index which is not -1 ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(a[i] != -1)  ` `        ``{ ` `            ``ind = i; ` `            ``break``; ` `        ``} ` `    ``} ` `     `  `    ``// Calculating the values of  ` `    ``// the indexes ind-1 to 0 ` `    ``for` `(``int` `i = ind - 1; i > -1; i--) ` `    ``{ ` `        ``if` `(a[i] == -1) ` `            ``a[i] = (a[i + 1] - 1 + m) % m; ` `    ``} ` `     `  `    ``// Calculating the values of ` `    ``// the indexes ind + 1 to n ` `    ``for` `(``int` `i = ind + 1; i < n; i++)  ` `    ``{ ` `        ``if` `(a[i] == -1) ` `            ``a[i] = (a[i - 1] + 1) % m; ` `    ``} ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``cout<< a[i] << ``" "``; ` `    ``} ` ` `  `} ` ` `  `// Driver code ` `int` `main()  ` `{  ` ` `  `    ``int` `n = 6, m = 7; ` `    ``int` `a[] = { 5, -1, -1, 1, 2, 3 }; ` `    ``construct(n, m, a); ` `    ``return` `0;  ` `}  ` ` `  `// This code is contributed by 29AjayKumar `

## Java

 `// Java implementation of the above approach ` `class` `GFG  ` `{ ` `    ``static` `void` `construct(``int` `n, ``int` `m, ``int``[] a) ` `    ``{ ` `        ``int` `ind = ``0``; ` ` `  `        ``// Finding the index which is not -1 ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``if` `(a[i] != -``1``)  ` `            ``{ ` `                ``ind = i; ` `                ``break``; ` `            ``} ` `        ``} ` `         `  `        ``// Calculating the values of  ` `        ``// the indexes ind-1 to 0 ` `        ``for` `(``int` `i = ind - ``1``; i > -``1``; i--) ` `        ``{ ` `            ``if` `(a[i] == -``1``) ` `                ``a[i] = (a[i + ``1``] - ``1` `+ m) % m; ` `        ``} ` `         `  `        ``// Calculating the values of ` `        ``// the indexes ind + 1 to n ` `        ``for` `(``int` `i = ind + ``1``; i < n; i++)  ` `        ``{ ` `            ``if` `(a[i] == -``1``) ` `                ``a[i] = (a[i - ``1``] + ``1``) % m; ` `        ``} ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``System.out.print(a[i] + ``" "``); ` `        ``} ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `n = ``6``, m = ``7``; ` `        ``int``[] a = { ``5``, -``1``, -``1``, ``1``, ``2``, ``3` `}; ` `        ``construct(n, m, a); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python implementation of the above approach ` `def` `construct(n, m, a): ` `    ``ind ``=` `0` ` `  `    ``# Finding the index which is not -1 ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(a[i]!``=``-``1``): ` `            ``ind ``=` `i ` `            ``break` ` `  `    ``# Calculating the values of the indexes ind-1 to 0 ` `    ``for` `i ``in` `range``(ind``-``1``, ``-``1``, ``-``1``): ` `        ``if` `(a[i]``=``=``-``1``): ` `            ``a[i]``=``(a[i ``+` `1``]``-``1` `+` `m)``%` `m ` ` `  `    ``# Calculating the values of the indexes ind + 1 to n ` `    ``for` `i ``in` `range``(ind ``+` `1``, n): ` `        ``if``(a[i]``=``=``-``1``): ` `            ``a[i]``=``(a[i``-``1``]``+``1``)``%` `m ` `    ``print``(``*``a) ` ` `  `# Driver code ` `n, m ``=` `6``, ``7` `a ``=``[``5``, ``-``1``, ``-``1``, ``1``, ``2``, ``3``] ` `construct(n, m, a) `

## C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``static` `void` `construct(``int` `n, ``int` `m, ``int``[] a) ` `    ``{ ` `        ``int` `ind = 0; ` ` `  `        ``// Finding the index which is not -1 ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `(a[i] != -1)  ` `            ``{ ` `                ``ind = i; ` `                ``break``; ` `            ``} ` `        ``} ` `         `  `        ``// Calculating the values of  ` `        ``// the indexes ind-1 to 0 ` `        ``for` `(``int` `i = ind - 1; i > -1; i--) ` `        ``{ ` `            ``if` `(a[i] == -1) ` `                ``a[i] = (a[i + 1] - 1 + m) % m; ` `        ``} ` `         `  `        ``// Calculating the values of ` `        ``// the indexes ind + 1 to n ` `        ``for` `(``int` `i = ind + 1; i < n; i++)  ` `        ``{ ` `            ``if` `(a[i] == -1) ` `                ``a[i] = (a[i - 1] + 1) % m; ` `        ``} ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``Console.Write(a[i] + ``" "``); ` `        ``} ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``int` `n = 6, m = 7; ` `        ``int``[] a = { 5, -1, -1, 1, 2, 3 }; ` `        ``construct(n, m, a); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```5 6 0 1 2 3
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : 29AjayKumar

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.