# Convert an array to reduced form | Set 1 (Simple and Hashing)

• Difficulty Level : Easy
• Last Updated : 30 Jun, 2022

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

The expected time complexity is O(n Log n).

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Method 1 (Simple):

A simple solution is to first find the minimum element, replace it with 0, consider the remaining array and find the minimum in the remaining array and replace it with 1, and so on. The time complexity of this solution is O(n2)

Method 2 (Efficient):

The idea is to use hashing and sorting. Below are the steps.

1. Create a temp array and copy the contents of the given array to temp[]. This takes O(n) time.
2. Sort temp[] in ascending order. This takes O(n Log n) time.
3. Create an empty hash table. This takes O(1) time.
4. Traverse temp[] from left to right and store mapping of numbers and their values (in converted array) in the hash table. This takes O(n) time on average.
5. Traverse given array and change elements to their positions using a hash table. This takes O(n) time on average.

Below are implementations of the above idea.

## C++

 // C++ program to convert an array in reduced// form#include using namespace std; void convert(int arr[], int n){    // Create a temp array and copy contents    // of arr[] to temp    int temp[n];    memcpy(temp, arr, n*sizeof(int));     // Sort temp array    sort(temp, temp + n);     // Create a hash table. Refer    unordered_map umap;     // One by one insert elements of sorted    // temp[] and assign them values from 0    // to n-1    int val = 0;    for (int i = 0; i < n; i++)        umap[temp[i]] = val++;     // Convert array by taking positions from    // umap    for (int i = 0; i < n; i++)        arr[i] = umap[arr[i]];} void printArr(int arr[], int n){    for (int i=0; i

## Java

 // Java Program to convert an Array// to reduced formimport java.util.*; class GFG{    public static void convert(int arr[], int n)    {        // Create a temp array and copy contents        // of arr[] to temp        int temp[] = arr.clone();         // Sort temp array        Arrays.sort(temp);         // Create a hash table.        HashMap umap = new HashMap<>();         // One by one insert elements of sorted        // temp[] and assign them values from 0        // to n-1        int val = 0;        for (int i = 0; i < n; i++)            umap.put(temp[i], val++);         // Convert array by taking positions from        // umap        for (int i = 0; i < n; i++)            arr[i] = umap.get(arr[i]);    }     public static void printArr(int arr[], int n)    {        for (int i = 0; i < n; i++)            System.out.print(arr[i] + " ");    }     // Driver code    public static void main(String[] args)    {         int arr[] = {10, 20, 15, 12, 11, 50};        int n = arr.length;         System.out.println("Given Array is ");        printArr(arr, n);         convert(arr , n);         System.out.println("\n\nConverted Array is ");        printArr(arr, n);     }} // This code is contributed by Abhishek Panwar

## Python3

 # Python3 program to convert an array# in reduced formdef convert(arr, n):    # Create a temp array and copy contents    # of arr[] to temp    temp = [arr[i] for i in range (n) ]         # Sort temp array    temp.sort()         # create a map    umap = {}              # One by one insert elements of sorted    # temp[] and assign them values from 0    # to n-1    val = 0    for i in range (n):        umap[temp[i]] = val        val += 1         # Convert array by taking positions from umap    for i in range (n):        arr[i] = umap[arr[i]]     def printArr(arr, n):    for i in range(n):        print(arr[i], end = " ") # Driver Codeif __name__ == "__main__":    arr = [10, 20, 15, 12, 11, 50]    n = len(arr)    print("Given Array is ")    printArr(arr, n)    convert(arr , n)    print("\n\nConverted Array is ")    printArr(arr, n) # This code is contributed by Abhishek Gupta

## C#

 // C# Program to convert an Array// to reduced formusing System;using System.Collections.Generic;using System.Linq; class GFG{    public static void convert(int []arr, int n)    {        // Create a temp array and copy contents        // of []arr to temp        int []temp = new int[arr.Length];        Array.Copy(arr, 0, temp, 0, arr.Length);         // Sort temp array        Array.Sort(temp);         // Create a hash table.        Dictionary umap =            new Dictionary();         // One by one insert elements of sorted        // []temp and assign them values from 0        // to n - 1        int val = 0;        for (int i = 0; i < n; i++)            if(umap.ContainsKey(temp[i]))                umap[temp[i]] = val++;            else                umap.Add(temp[i], val++);         // Convert array by taking positions from        // umap        for (int i = 0; i < n; i++)            arr[i] = umap[arr[i]];    }     public static void printArr(int []arr, int n)    {        for (int i = 0; i < n; i++)            Console.Write(arr[i] + " ");    }     // Driver code    public static void Main(String[] args)    {         int []arr = {10, 20, 15, 12, 11, 50};        int n = arr.Length;         Console.WriteLine("Given Array is ");        printArr(arr, n);         convert(arr , n);         Console.WriteLine("\n\nConverted Array is ");        printArr(arr, n);    }} // This code is contributed by PrinciRaj1992

## Javascript



Output

Given Array is
10 20 15 12 11 50

Converted Array is
0 4 3 2 1 5

Time complexity: O(n)
Auxiliary Space: O(n)

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