# 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 ` `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)

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Improved By : aganjali10

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