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)

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