- Navigate the numbers from 0 to n-1.
- Now navigate through array.
- If (i==a[j]) , then replace the element at i position with a[j] position.
- If there is any element in which -1 is used instead of the number then it will be replaced automatically.
- Now, iterate through the array and check if (a[i]!=i) , if it s true then replace a[i] with -1.
Below is the implementation for the above approach:
- C++
// C++ program for above approach
#include <iostream>
using namespace std;
// Function to tranform the array
void fixArray( int ar[], int n)
{
`` int i, j, temp;
`` // Iterate over the array
`` for (i = 0; i < n; i++)
`` {
`` for (j = 0; j < n; j++)
`` {
`` // Checf is any ar[j]
`` // exists such that
`` // ar[j] is equal to i
`` if (ar[j] == i) {
`` temp = ar[j];
`` ar[j] = ar[i];
`` ar[i] = temp;
`` break ;
`` }
`` }
`` }
`` // Iterate over array
`` for (i = 0; i < n; i++)
`` {
`` // If not present
`` if (ar[i] != i)
`` {
`` ar[i] = -1;
`` }
`` }
`` // Print the output
`` cout << "Array after Rearranging" << endl;
`` for (i = 0; i < n; i++) {
`` cout << ar[i] << " " ;
`` }
}
// Driver Code
int main()
{
`` int n, ar[] = { -1, -1, 6, 1, 9, 3, 2, -1, 4, -1 };
`` n = sizeof (ar) / sizeof (ar[0]);
`` // Function Call
`` fixArray(ar, n);
}
Output
Array after Rearranging -1 1 2 3 4 -1 6 -1 -1 9
Time Complexity: O(n2)
- JAVA
// Java program for above approach
class GFG{
// Function to tranform the array
public static void fixArray( int ar[], int n)
{
int i, j, temp;
// Iterate over the array
for (i = 0 ; i < n; i++)
{
for (j = 0 ; j < n; j++)
{
// Checf is any ar[j]
// exists such that
// ar[j] is equal to i
if (ar[j] == i)
{
temp = ar[j];
ar[j] = ar[i];
ar[i] = temp;
break ;
}
}
}
// Iterate over array
for (i = 0 ; i < n; i++)
{
// If not present
if (ar[i] != i)
{
ar[i] = - 1 ;
}
}
// Print the output
System.out.println( "Array after Rearranging" );
for (i = 0 ; i < n; i++)
{
System.out.print(ar[i] + " " );
}
}
// Driver Code
public static void main(String[] args)
{
int n, ar[] = { - 1 , - 1 , 6 , 1 , 9 ,
3 , 2 , - 1 , 4 , - 1 };
n = ar.length;
// Function Call
fixArray(ar, n);
}
}
Output
Array after Rearranging -1 1 2 3 4 -1 6 -1 -1 9
Time Complexity: O(n2)