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

• Difficulty Level : Medium
• Last Updated : 07 Jun, 2018

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 ``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 > 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

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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