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

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}

Expected time complexity is O(n Log n).

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

Method 2 (Efficient)
The idea is to use hashing and sorting. Below are steps.
1) Create a temp array and copy contents of 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[] form left to right and store mapping of numbers and their values (in converted array) in hash table. This takes O(n) time on average.
5) Traverse given array and change elements to their positions using hash table. This takes O(n) time on average.

Overall time complexity of this solution is O(n Log n).

Below are implementations of 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 form import 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 form def 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 Code if __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

Output :

Given Array is
10 20 15 12 11 50

Converted Array is
0 4 3 2 1 5

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

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