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

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  • Difficulty Level : Basic
  • Last Updated : 12 Sep, 2022
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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 

Complexity Analysis:

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

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