Rearrange an array such that arr[i] = i

Question - 

Given an array of elements of length N, ranging from 0 to N – 1. All elements may not be present in the array. If the element is not present then there will be -1 present in the array. Rearrange the array such that A[i] = i and if i is not present, display -1 at that place.

Example - 

nput : arr = {-1, -1, 6, 1, 9, 3, 2, -1, 4, -1}
Output : [-1, 1, 2, 3, 4, -1, 6, -1, -1, 9]

Input : arr = {19, 7, 0, 3, 18, 15, 12, 6, 1, 8,
              11, 10, 9, 5, 13, 16, 2, 14, 17, 4}
Output : [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 
         11, 12, 13, 14, 15, 16, 17, 18, 19]

Approach - 

we check each index is present in array as element or not. if present then print index else print -1.

Implementation - 

#include<iostream>

using namespace std;

int binarysearch(int arr[],int n,int num)

{

    for(int i=0;i<n;i++){

        if(arr[i]==num)

        return num;

    }

    return -1;

}

int main()

{

    int n;

    cout<<"enter the no of elements : ";cin>>n;

    int arr[n];

    cout<<"Enter elements : ";

    for(int i=0;i<n;i++)

        cin>>arr[i];

    //print resultant

    for(int i=0;i<n;i++){

    int a=binarysearch(arr,n,i);

        

            cout<<a<<", ";

    }

        

        return 0;

}

Code on compiler

Time complexity - O(n*n)