Rearrange an array such that arr[i] = i in C

Problem statement
Consider an array having N elements. Rearrange the array such that A[i] = i and if i is not present, replace it with -1 at that place. As not all elements in between 0 to N-1 will be present.
Naive Approach
Here we have used two for loops. One for traversing through numbers O to N-1 and the other for traversing through the array.

/*C code to Rearrange an array such that arr[i] = i*/
#include<stdio.h>

void reArrangeArray(int arr[], int n)
{
	int i, j, temp;


	for (i = 0; i < n; i++)
	{
		for (j = 0; j < n; j++)
		{
			if (arr[j] == i) {
				temp = arr[j];
				arr[j] = arr[i];
				arr[i] = temp;
				break;
			}
		}
	}

	for (i = 0; i < n; i++)
	{
		if (arr[i] != i)
		{
			arr[i] = -1;
		}
	}

	printf("\nArray after rearranging:\n");
	for (i = 0; i < n; i++) {
		printf("%d\t",arr[i]);
	}
}

int main()
{
	int n, arr[] = { -1, -1, 5, 1, 8, 3, 2, -1, 4, -1 };
	n = sizeof(arr) / sizeof(arr[0]);
	
	printf("Array is:\n");
	for (int i = 0; i < n; i++) {
		printf("%d\t",arr[i]);
	}
	
	reArrangeArray(arr, n);
}

Output

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