Skip to content
Related Articles

Related Articles

Improve Article

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

  • Difficulty Level : Basic
  • Last Updated : 25 Aug, 2021

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 original 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 

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. 

Below is the implementation of the above approach: 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
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

Javascript




<script>
 
// Javascript implementation of above approach
 
function construct(n, m, a)
{
    var ind = 0;
 
    // Finding the index which is not -1
    for (var i = 0; i < n; i++)
    {
        if (a[i] != -1)
        {
            ind = i;
            break;
        }
    }
     
    // Calculating the values of
    // the indexes ind-1 to 0
    for (var 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 (var i = ind + 1; i < n; i++)
    {
        if (a[i] == -1)
            a[i] = (a[i - 1] + 1) % m;
    }
    for (var i = 0; i < n; i++)
    {
        document.write( a[i] + " ");
    }
 
}
 
// Driver code
var n = 6, m = 7;
var a = [5, -1, -1, 1, 2, 3];
construct(n, m, a);
 
</script>
Output: 
5 6 0 1 2 3

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :