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++

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// C++ program to convert an array in reduced
// form
#include <bits/stdc++.h>
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<int, int> 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<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;
}

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Java

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// 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<Integer, Integer> 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

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Python3

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# 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

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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)

This article is contributed by Dheeraj Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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