Find k closest numbers in an unsorted array

11 Feb, 2021
Given an unsorted array and two numbers x and k, find k closest values to x.Examples:  
Input : arr[] = {10, 2, 14, 4, 7, 6}, x = 5, k = 3 
Output : 4 6 7
Three closest values of x are 4, 6 and 7.

Input : arr[] = {-10, -50, 20, 17, 80}, x = 20, k = 2
Output : 17, 20
 
A simple solution is to sort the array. Then apply the method discussed to k closest values in a sorted array.Time Complexity : O(n Log n)A better solution is to use Heap Data Structure 1) Make a max heap of differences with first k elements. 2) For every element starting from (k+1)-th element, do following. .....a) Find difference of current element with x. .....b) If difference is more than root of heap, ignore current element. .....c) Else insert the current element to the heap after removing the root. 3) Finally the heap has k closest elements.  
C++
Java
Python3
C#
// C++ program to find k closest elements#include <bits/stdc++.h>using namespace std; void printKclosest(int arr[], int n, int x,                   int k){    // Make a max heap of difference with    // first k elements.    priority_queue<pair<int, int> > pq;    for (int i = 0; i < k; i++)        pq.push({ abs(arr[i] - x), i });     // Now process remaining elements.    for (int i = k; i < n; i++) {         int diff = abs(arr[i] - x);         // If difference with current        // element is more than root,        // then ignore it.        if (diff > pq.top().first)            continue;         // Else remove root and insert        pq.pop();        pq.push({ diff, i });    }     // Print contents of heap.    while (pq.empty() == false) {        cout << arr[pq.top().second] << " ";        pq.pop();    }} // Driver program to test above functionsint main(){    int arr[] = { -10, -50, 20, 17, 80 };    int x = 20, k = 2;    int n = sizeof(arr) / sizeof(arr[0]);    printKclosest(arr, n, x, k);    return 0;}
//Java program to find k closest elementsimport java.util.Comparator;import java.util.PriorityQueue; class Pair{    Integer key;    Integer value;         public Pair(Integer key, Integer value)    {        this.key = key;        this.value = value;    }    public Integer getKey()    {        return key;    }    public void setKey(Integer key)    {        this.key = key;    }    public Integer getValue()    {        return value;    }    public void setValue(Integer value)    {        this.value = value;    }} class GFG{     public static void printKclosest(int[] arr, int n,                                 int x, int k){     // Make a max heap.    PriorityQueue<Pair> pq = new PriorityQueue<>(                             new Comparator<Pair>()    {        public int compare(Pair p1, Pair p2)        {            return p2.getValue().compareTo(                   p1.getValue());        }    });         // Build heap of difference with    // first k elements    for(int i = 0; i < k; i++)    {        pq.offer(new Pair(arr[i],                 Math.abs(arr[i] - x)));    }         // Now process remaining elements.    for(int i = k; i < n; i++)    {        int diff = Math.abs(arr[i] - x);                 // If difference with current        // element is more than root,        // then ignore it.        if(diff > pq.peek().getValue()) continue;                 // Else remove root and insert        pq.poll();        pq.offer(new Pair(arr[i], diff));    }         // Print contents of heap.    while(!pq.isEmpty())    {        System.out.print(pq.poll().getKey() + " ");    }} // Driver codepublic static void main(String[] args){    int arr[] = { -10, -50, 20, 17, 80 };    int x = 20, k = 2;    int n = arr.length;         printKclosest(arr, n, x, k);}} // This code is contributed by Ashok Borra
# Python3 program to find k closest elementsimport mathimport sysfrom queue import PriorityQueuedef printKclosest(arr,n,x,k):     # Make a max heap of difference with    # first k elements.    pq = PriorityQueue()    for i in range(k):        pq.put((-abs(arr[i]-x),i))     # Now process remaining elements    for i in range(k,n):        diff = abs(arr[i]-x)        p,pi = pq.get()        curr = -p         # If difference with current        # element is more than root,        # then put it back.        if diff>curr:            pq.put((-curr,pi))            continue        else:             # Else remove root and insert            pq.put((-diff,i))                 # Print contents of heap.    while(not pq.empty()):        p,q = pq.get()        print("{} ".format(arr[q]),end = "") # Driver program to test above functionsif __name__=='__main__':    arr = [-10,-50,20,17,80]    x,k = 20,2    n = len(arr)    printKclosest(arr, n, x, k) # This code is contributed by Vikash Kumar 37
// C# program to find k closest elementsusing System;using System.Collections.Generic;class GFG {   static void printKclosest(int[] arr, int n, int x, int k)  {    // Make a max heap of difference with    // first k elements.    List<Tuple<int, int>> pq = new List<Tuple<int, int>>();    for (int i = 0; i < k; i++)    {      pq.Add(new Tuple<int,int>(Math.Abs(arr[i] - x), i));    }     pq.Sort();    pq.Reverse();     // Now process remaining elements.    for (int i = k; i < n; i++)    {       int diff = Math.Abs(arr[i] - x);       // If difference with current      // element is more than root,      // then ignore it.      if (diff > pq[0].Item1)        continue;       // Else remove root and insert      pq.RemoveAt(0);      pq.Add(new Tuple<int,int>(diff, i));      pq.Sort();      pq.Reverse();    }     // Print contents of heap.    while (pq.Count > 0)    {      Console.Write(arr[pq[0].Item2] + " ");      pq.RemoveAt(0);    }  }   // Driver code  static void Main()  {    int[] arr = { -10, -50, 20, 17, 80 };    int x = 20, k = 2;    int n = arr.Length;    printKclosest(arr, n, x, k);  }} // This code is contributed by divyesh072019.
17 20
 
Time Complexity : O(n Log k) 
Vikash Kumar 37
nidhi_biet
Dheeraj37
coded9
divyesh072019
Amazon
cpp-pair
cpp-priority-queue
STL
Arrays
Heap
Amazon
Arrays
Heap
STL
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