Skip to content
Related Articles

Related Articles

Improve Article

Convert an array to reduced form | Set 3 (Binary Search)

  • Last Updated : 24 May, 2021
Geek Week

Given an array arr[] consisting of N distinct integers, the task is to convert the given array into a sequence of first N non-negative integers, i.e. [0, N – 1] such that the order of the elements is the same, i.e. 0 is placed at the index of the smallest array element, 1 at the index of the second smallest element, and so on.

Examples:

Input: arr[] = {10, 40, 20}
Output: 0 2 1

Input: arr[] = {5, 10, 40, 30, 20}
Output: 0 1 4 3 2

 

Hashing-Based Approach: Please refer to the Set 1 post of this article for the hashing-based approach. 
Time Complexity: O(N* log N)
Auxiliary Space: O(N)

Vector Of Pairs Based Approach: Please refer to the Set 2 post of this article for the approach using the vector of pairs
Time Complexity: O(N* log N)
Auxiliary Space: O(N)

Binary Search-based Approach: Follow the steps to solve the problem:



Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the reduced form
// of the given array arr[]
void convert(int arr[], int n)
{
    // Stores the sorted form of the
    // the given array arr[]
    int brr[n];
 
    for (int i = 0; i < n; i++)
        brr[i] = arr[i];
 
    // Sort the array brr[]
    sort(brr, brr + n);
 
    // Traverse the given array arr[]
    for (int i = 0; i < n; i++) {
 
        int l = 0, r = n - 1, mid;
 
        // Perform the Binary Search
        while (l <= r) {
 
            // Calculate the value of
            // mid
            mid = (l + r) / 2;
 
            if (brr[mid] == arr[i]) {
 
                // Print the current
                // index and break
                cout << mid << ' ';
                break;
            }
 
            // Update the value of l
            else if (brr[mid] < arr[i]) {
                l = mid + 1;
            }
 
            // Update the value of r
            else {
                r = mid - 1;
            }
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 10, 20, 15, 12, 11, 50 };
    int N = sizeof(arr) / sizeof(arr[0]);
    convert(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG
{
 
    // Function to find the reduced form
    // of the given array arr[]
    static void convert(int arr[], int n)
    {
       
        // Stores the sorted form of the
        // the given array arr[]
        int brr[] = new int[n];
 
        for (int i = 0; i < n; i++)
            brr[i] = arr[i];
 
        // Sort the array brr[]
        Arrays.sort(brr);
 
        // Traverse the given array arr[]
        for (int i = 0; i < n; i++) {
 
            int l = 0, r = n - 1, mid;
 
            // Perform the Binary Search
            while (l <= r) {
 
                // Calculate the value of
                // mid
                mid = (l + r) / 2;
 
                if (brr[mid] == arr[i]) {
 
                    // Print the current
                    // index and break
                    System.out.print(mid + " ");
                    break;
                }
 
                // Update the value of l
                else if (brr[mid] < arr[i]) {
                    l = mid + 1;
                }
 
                // Update the value of r
                else {
                    r = mid - 1;
                }
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 10, 20, 15, 12, 11, 50 };
        int N = arr.length;
        convert(arr, N);
    }
}
 
// This code is contributed by Kingash.

Python3




# Python3 program for the above approach
 
# Function to find the reduced form
# of the given array arr[]
def convert(arr, n):
   
    # Stores the sorted form of the
    # the given array arr[]
    brr = [i for i in arr]
 
    # Sort the array brr[]
    brr = sorted(brr)
 
    # Traverse the given array arr[]
    for i in range(n):
 
        l, r, mid = 0, n - 1, 0
 
        # Perform the Binary Search
        while (l <= r):
 
            # Calculate the value of
            # mid
            mid = (l + r) // 2
 
            if (brr[mid] == arr[i]):
 
                # Prthe current
                # index and break
                print(mid,end=" ")
                break
            # Update the value of l
            elif (brr[mid] < arr[i]):
                l = mid + 1
            # Update the value of r
            else:
                r = mid - 1
 
# Driver Code
if __name__ == '__main__':
    arr=[10, 20, 15, 12, 11, 50]
    N = len(arr)
    convert(arr, N)
 
    # This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the reduced form
// of the given array arr[]
static void convert(int[] arr, int n)
{
     
    // Stores the sorted form of the
    // the given array arr[]
    int[] brr = new int[n];
 
    for(int i = 0; i < n; i++)
        brr[i] = arr[i];
 
    // Sort the array brr[]
    Array.Sort(brr);
 
    // Traverse the given array arr[]
    for(int i = 0; i < n; i++)
    {
        int l = 0, r = n - 1, mid;
 
        // Perform the Binary Search
        while (l <= r)
        {
             
            // Calculate the value of
            // mid
            mid = (l + r) / 2;
 
            if (brr[mid] == arr[i])
            {
                 
                // Print the current
                // index and break
                Console.Write(mid + " ");
                break;
            }
 
            // Update the value of l
            else if (brr[mid] < arr[i])
            {
                l = mid + 1;
            }
 
            // Update the value of r
            else
            {
                r = mid - 1;
            }
        }
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = { 10, 20, 15, 12, 11, 50 };
    int N = arr.Length;
     
    convert(arr, N);
}
}
 
// This code is contributed by ukasp

Javascript




<script>
 
// javascript program for the above approach
 
// Function to find the reduced form
// of the given array arr[]
function convert(arr, n)
{
    // Stores the sorted form of the
    // the given array arr[]
    var brr = Array(n).fill(0);
    var i;
    for (i = 0; i < n; i++)
        brr[i] = arr[i];
 
    // Sort the array brr[]
    brr.sort();
 
    // Traverse the given array arr[]
    for (i = 0; i < n; i++) {
 
        var l = 0, r = n - 1, mid;
 
        // Perform the Binary Search
        while (l <= r) {
 
            // Calculate the value of
            // mid
            mid = parseInt((l + r) / 2,10);
 
            if (brr[mid] == arr[i]) {
 
                // Print the current
                // index and break
                document.write(mid + ' ');
                break;
            }
 
            // Update the value of l
            else if (brr[mid] < arr[i]) {
                l = mid + 1;
            }
 
            // Update the value of r
            else {
                r = mid - 1;
            }
        }
    }
}
 
// Driver Code
    var arr = [10, 20, 15, 12, 11, 50];
    var N = arr.length;
    convert(arr, N);
 
// This code is contributed by SURENDRA_GANGWAR.
</script>
Output: 
0 4 3 2 1 5

 

Time Complexity: O(N * log N)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :