Convert an Array to reduced for using Binary Search
Last Updated :
28 Nov, 2022
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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 1Input: 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:
- Initialize an array, say brr[] and store all the elements of the array arr[] into brr[].
- Sort the array brr[] in ascending order.
- Traverse the given array arr[] and for each element i.e., arr[i] find the lower_bound of arr[i] in the array brr[] and print the index of is as the result for the current element.
Below is the implementation of the above approach:
// 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 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 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]):
# Print the 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# 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
<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)
