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 import java.io.*;
public 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)