Given an array of N distinct elements and a number x, arrange array elements according to the absolute difference with x, i. e., an element having minimum difference comes first, and so on.
Note: If two or more elements are at equal distances arrange them in the same sequence as in the given array.
Examples:
Input: x = 7, arr[] = {10, 5, 3, 9, 2}
Output: arr[] = {5, 9, 10, 3, 2}
Explanation:
7 – 10 = 3(abs)
7 – 5 = 2
7 – 3 = 4
7 – 9 = 2(abs)
7 – 2 = 5
So according to the difference with X,
elements are arranged as 5, 9, 10, 3, 2.
Input: x = 6, arr[] = {1, 2, 3, 4, 5}
Output: arr[] = {5, 4, 3, 2, 1}
Sort an array according to absolute difference with a given value using self-balancing BST:
To solve the problem follow the below idea:
The idea is to use a self-balancing binary search tree. We traverse the input array and for every element, we find its difference with x and store the difference as key and element as the value in a self-balancing binary search tree. Finally, we traverse the tree and print its inorder traversal which is the required output.
Approach for C++ Implementation:
In C++, self-balancing-binary-search-tree is implemented by set, map, and multimap. We can’t use set here as we have key-value pairs (not only keys). We also can’t directly use map also as a single key can belong to multiple values and map allows a single value for a key. So we use multimap which stores key-value pairs and can have multiple values for a key.
Follow the below steps to solve the problem:
- Store the values in the multimap with the difference with X as key.
- In multimap, the values will be already in sorted order according to key i.e. difference with X because it implements self-balancing-binary-search-tree internally.
- Update all the values of an array with the values of the map so that the array has the required output.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void rearrange( int arr[], int n, int x)
{
multimap< int , int > m;
multimap< int , int >::iterator it;
for ( int i = 0; i < n; i++)
m.insert(make_pair( abs (x - arr[i]), arr[i]));
int i = 0;
for (it = m.begin(); it != m.end(); it++)
arr[i++] = (*it).second;
}
void printArray( int arr[], int n)
{
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
}
int main()
{
int arr[] = { 10, 5, 3, 9, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 7;
rearrange(arr, n, x);
printArray(arr, n);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static void rearrange( int [] arr, int n, int x)
{
TreeMap<Integer, ArrayList<Integer> > m
= new TreeMap<>();
for ( int i = 0 ; i < n; i++) {
int diff = Math.abs(x - arr[i]);
if (m.containsKey(diff)) {
ArrayList<Integer> al = m.get(diff);
al.add(arr[i]);
m.put(diff, al);
}
else {
ArrayList<Integer> al = new ArrayList<>();
al.add(arr[i]);
m.put(diff, al);
}
}
int index = 0 ;
for (Map.Entry entry : m.entrySet()) {
ArrayList<Integer> al = m.get(entry.getKey());
for ( int i = 0 ; i < al.size(); i++)
arr[index++] = al.get(i);
}
}
static void printArray( int [] arr, int n)
{
for ( int i = 0 ; i < n; i++)
System.out.print(arr[i] + " " );
}
public static void main(String args[])
{
int [] arr = { 10 , 5 , 3 , 9 , 2 };
int n = arr.length;
int x = 7 ;
rearrange(arr, n, x);
printArray(arr, n);
}
}
|
Python3
def rearrange(arr, n, x):
m = {}
for i in range (n):
m[arr[i]] = abs (x - arr[i])
m = {k: v for k, v in sorted (m.items(),
key = lambda item: item[ 1 ])}
i = 0
for it in m.keys():
arr[i] = it
i + = 1
def printArray(arr, n):
for i in range (n):
print (arr[i], end = " " )
if __name__ = = "__main__" :
arr = [ 10 , 5 , 3 , 9 , 2 ]
n = len (arr)
x = 7
rearrange(arr, n, x)
printArray(arr, n)
|
C#
using System;
using System.Collections.Generic;
public class GFG {
static void rearrange( int [] arr, int n, int x)
{
SortedDictionary< int , List< int > > m
= new SortedDictionary< int , List< int > >();
for ( int i = 0; i < n; i++) {
int diff = Math.Abs(x - arr[i]);
if (m.ContainsKey(diff)) {
List< int > al = m;
al.Add(arr[i]);
m = al;
}
else {
List< int > al = new List< int >();
al.Add(arr[i]);
m.Add(diff, al);
}
}
int index = 0;
foreach ( int entry in m.Keys)
{
List< int > al = m[entry];
for ( int i = 0; i < al.Count; i++)
arr[index++] = al[i];
}
}
static void printArray( int [] arr, int n)
{
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
}
static public void Main()
{
int [] arr = { 10, 5, 3, 9, 2 };
int n = arr.Length;
int x = 7;
rearrange(arr, n, x);
printArray(arr, n);
}
}
|
Javascript
function rearrange(arr,n,x)
{
let m = new Map();
for (let i = 0; i < n; i++)
{
m.set(arr[i],Math.abs(x-arr[i]));
}
let m1 = new Map([...m.entries()].sort((a, b) =>
a[1] - b[1]));
let index = 0;
for (let [key, value] of m1.entries())
{
arr[index++] =key
}
}
function printArray(arr,n)
{
for (let i = 0; i < n; i++)
document.write(arr[i] + " " );
}
let arr=[10, 5, 3, 9 ,2];
let n = arr.length;
let x = 7;
rearrange(arr, n, x);
printArray(arr, n);
|
Time Complexity: O(N Log N)
Auxiliary Space: O(N)
This article is contributed by Sahil Chhabra.
Sort an array according to absolute difference with a given value using C++ STL:
In C++, we can use stable_sort(), and write a lambda expression for the custom comparator function. This solution is elegant and far easier to understand. The only challenge in writing the lambda expression is to send the value ‘x’ into the lambda expression to be able to use it inside the expression. This can be achieved either by operator overloading with the help of a class or using a much simpler capture.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void rearrange( int arr[], int n, int x)
{
stable_sort(arr, arr + n, [x]( int a, int b) {
if ( abs (a - x) < abs (b - x))
return true ;
else
return false ;
});
}
int main()
{
int arr[] = { 10, 5, 3, 9, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 7;
rearrange(arr, n, x);
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
}
|
Java
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
class GFG {
static List<Integer> rearrange(List<Integer> arr, int n,
int x)
{
Collections.sort(
arr,
(a, b) -> Math.abs(x - a) - Math.abs(x - b));
return arr;
}
public static void main(String[] args)
{
List<Integer> arr = new ArrayList<>();
arr.add( 10 );
arr.add( 5 );
arr.add( 3 );
arr.add( 9 );
arr.add( 2 );
int n = arr.size();
int x = 7 ;
arr = rearrange(arr, n, x);
for ( int num : arr) {
System.out.print(num + " " );
}
}
}
|
Python3
def rearrange(arr, n, x):
arr.sort(key = lambda a: abs (a - x) )
arr = [ 10 , 5 , 3 , 9 , 2 ];
n = len (arr)
x = 7 ;
rearrange(arr, n, x);
print ( * arr)
|
C#
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
static List< int > rearrange(List< int > arr, int n, int x)
{
return arr.OrderBy(a => Math.Abs(x - a)).ToList();
}
public static void Main( string [] args)
{
List< int > arr = new List< int >{10, 5, 3, 9, 2};
int n = arr.Count;
int x = 7;
arr = rearrange(arr, n, x);
foreach ( int num in arr)
Console.Write(num + " " );
}
}
|
Javascript
function rearrange(arr, n, x)
{
arr.sort( function (a, b)
{
return Math.abs(a - x) - Math.abs(b - x)
})
}
let arr = [ 10, 5, 3, 9, 2 ];
let n = arr.length
let x = 7;
rearrange(arr, n, x);
console.log(arr.join( " " ))
|
Time Complexity: O(N log N)
Auxiliary Space: O(1)
This article is contributed by D. Mohit Varsha. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.