Convert an Array to reduced for using Binary Search
Last Updated :
28 Nov, 2022
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++
#include <bits/stdc++.h>
using namespace std;
void convert( int arr[], int n)
{
int brr[n];
for ( int i = 0; i < n; i++)
brr[i] = arr[i];
sort(brr, brr + n);
for ( int i = 0; i < n; i++) {
int l = 0, r = n - 1, mid;
while (l <= r) {
mid = (l + r) / 2;
if (brr[mid] == arr[i]) {
cout << mid << ' ' ;
break ;
}
else if (brr[mid] < arr[i]) {
l = mid + 1;
}
else {
r = mid - 1;
}
}
}
}
int main()
{
int arr[] = { 10, 20, 15, 12, 11, 50 };
int N = sizeof (arr) / sizeof (arr[0]);
convert(arr, N);
return 0;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG
{
static void convert( int arr[], int n)
{
int brr[] = new int [n];
for ( int i = 0 ; i < n; i++)
brr[i] = arr[i];
Arrays.sort(brr);
for ( int i = 0 ; i < n; i++) {
int l = 0 , r = n - 1 , mid;
while (l <= r) {
mid = (l + r) / 2 ;
if (brr[mid] == arr[i]) {
System.out.print(mid + " " );
break ;
}
else if (brr[mid] < arr[i]) {
l = mid + 1 ;
}
else {
r = mid - 1 ;
}
}
}
}
public static void main(String[] args)
{
int arr[] = { 10 , 20 , 15 , 12 , 11 , 50 };
int N = arr.length;
convert(arr, N);
}
}
|
Python3
def convert(arr, n):
brr = [i for i in arr]
brr = sorted (brr)
for i in range (n):
l, r, mid = 0 , n - 1 , 0
while (l < = r):
mid = (l + r) / / 2
if (brr[mid] = = arr[i]):
print (mid,end = " " )
break
elif (brr[mid] < arr[i]):
l = mid + 1
else :
r = mid - 1
if __name__ = = '__main__' :
arr = [ 10 , 20 , 15 , 12 , 11 , 50 ]
N = len (arr)
convert(arr, N)
|
C#
using System;
class GFG{
static void convert( int [] arr, int n)
{
int [] brr = new int [n];
for ( int i = 0; i < n; i++)
brr[i] = arr[i];
Array.Sort(brr);
for ( int i = 0; i < n; i++)
{
int l = 0, r = n - 1, mid;
while (l <= r)
{
mid = (l + r) / 2;
if (brr[mid] == arr[i])
{
Console.Write(mid + " " );
break ;
}
else if (brr[mid] < arr[i])
{
l = mid + 1;
}
else
{
r = mid - 1;
}
}
}
}
public static void Main( string [] args)
{
int [] arr = { 10, 20, 15, 12, 11, 50 };
int N = arr.Length;
convert(arr, N);
}
}
|
Javascript
<script>
function convert(arr, n)
{
var brr = Array(n).fill(0);
var i;
for (i = 0; i < n; i++)
brr[i] = arr[i];
brr.sort();
for (i = 0; i < n; i++) {
var l = 0, r = n - 1, mid;
while (l <= r) {
mid = parseInt((l + r) / 2,10);
if (brr[mid] == arr[i]) {
document.write(mid + ' ' );
break ;
}
else if (brr[mid] < arr[i]) {
l = mid + 1;
}
else {
r = mid - 1;
}
}
}
}
var arr = [10, 20, 15, 12, 11, 50];
var N = arr.length;
convert(arr, N);
</script>
|
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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