Skip to content
Related Articles

Related Articles

Reconstruct the array by replacing arr[i] with (arr[i-1]+1) % M
  • Difficulty Level : Basic
  • Last Updated : 17 Feb, 2020

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

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 <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
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.

My Personal Notes arrow_drop_up
Recommended Articles
Page :