# Convert an array to reduced form | Set 2 (Using vector of pairs)

Given an array with n distinct elements, convert the given array to a form where all elements are in range from 0 to n-1. The order of elements is same, i.e., 0 is placed in place of smallest element, 1 is placed for second smallest element, … n-1 is placed for largest element.

```Input:  arr[] = {10, 40, 20}
Output: arr[] = {0, 2, 1}

Input:  arr[] = {5, 10, 40, 30, 20}
Output: arr[] = {0, 1, 4, 3, 2}
```

We have discussed simple and hashing based solutions.

In this post, a new solution is discussed. The idea is to create a vector of pairs. Every element of pair contains element and index. We sort vector by array values. After sorting, we copy indexes to original array.

```// C++ program to convert an array in reduced
// form
#include <bits/stdc++.h>
using namespace std;

// Converts arr[0..n-1] to reduced form.
void convert(int arr[], int n)
{
// A vector of pairs. Every element of
// pair contains array element and its
// index
vector <pair<int, int> > v;

// Put all elements and their index in
// the vector
for (int i = 0; i < n; i++)
v.push_back(make_pair(arr[i], i));

// Sort the vector by array values
sort(v.begin(), v.end());

// Put indexes of modified vector in arr[]
for (int i=0; i<n; i++)
arr[i] = v[i].second;
}

// Utility function to print an array.
void printArr(int arr[], int n)
{
for (int i=0; i<n; i++)
cout << arr[i] << " ";
}

// Driver program to test above method
int main()
{
int arr[] = {10, 20, 15, 12, 11, 50};
int n = sizeof(arr)/sizeof(arr[0]);

cout << "Given Array is \n";
printArr(arr, n);

convert(arr , n);

cout << "\n\nConverted Array is \n";
printArr(arr, n);

return 0;
}
```

Output :

```Given Array is
10 20 15 12 11 50

Converted Array is
0 4 3 2 1 5
```

Time Complexity : O(n Log n)
Auxiliary Space : O(n)

This article is contributed by Arpit Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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