Find k closest elements to a given value

Given a sorted array arr[] and a value X, find the k closest elements to X in arr[]. 

Examples:

Input: K = 4, X = 35
       arr[] = {12, 16, 22, 30, 35, 39, 42, 
               45, 48, 50, 53, 55, 56}
Output: 30 39 42 45

Note that if the element is present in array, then it should not be in output, only the other closest elements are required.

In the following solutions, it is assumed that all elements of array are distinct.

A simple solution is to do linear search for k closest elements. 

1. Start from the first element and search for the crossover point (The point before which elements are smaller than or equal to X and after which elements are greater). This step takes O(n) time. 
2. Once we find the crossover point, we can compare elements on both sides of crossover point to print k closest elements. This step takes O(k) time.

The time complexity of the above solution is O(n).

An Optimized Solution is to find k elements in O(Logn + k) time. The idea is to use Binary Search to find the crossover point. Once we find index of crossover point, we can print k closest elements in O(k) time.  

C++

#include<stdio.h>

/* Function to find the cross over point (the point before
which elements are smaller than or equal to x and after
which greater than x)*/
int findCrossOver(int arr[], int low, int high, int x)
{
// Base cases
if (arr[high] <= x) // x is greater than all
    return high;
if (arr[low] > x) // x is smaller than all
    return low;

// Find the middle point
int mid = (low + high)/2; /* low + (high - low)/2 */

/* If x is same as middle element, then return mid */
if (arr[mid] <= x && arr[mid+1] > x)
    return mid;

/* If x is greater than arr[mid], then either arr[mid + 1]
    is ceiling of x or ceiling lies in arr[mid+1...high] */
if(arr[mid] < x)
    return findCrossOver(arr, mid+1, high, x);

return findCrossOver(arr, low, mid - 1, x);
}

// This function prints k closest elements to x in arr[].
// n is the number of elements in arr[]
void printKclosest(int arr[], int x, int k, int n)
{
    // Find the crossover point
    int l = findCrossOver(arr, 0, n-1, x);
    int r = l+1; // Right index to search
    int count = 0; // To keep track of count of elements already printed

    // If x is present in arr[], then reduce left index
    // Assumption: all elements in arr[] are distinct
    if (arr[l] == x) l--;

    // Compare elements on left and right of crossover
    // point to find the k closest elements
    while (l >= 0 && r < n && count < k)
    {
        if (x - arr[l] < arr[r] - x)
            printf("%d ", arr[l--]);
        else
            printf("%d ", arr[r++]);
        count++;
    }

    // If there are no more elements on right side, then
    // print left elements
    while (count < k && l >= 0)
        printf("%d ", arr[l--]), count++;

    // If there are no more elements on left side, then
    // print right elements
    while (count < k && r < n)
        printf("%d ", arr[r++]), count++;
}

/* Driver program to check above functions */
int main()
{
int arr[] ={12, 16, 22, 30, 35, 39, 42,
            45, 48, 50, 53, 55, 56};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 35, k = 4;
printKclosest(arr, x, 4, n);
return 0;
}

Java

// Java program to find k closest elements to a given value
class KClosest
{
    /* Function to find the cross over point (the point before
       which elements are smaller than or equal to x and after
       which greater than x)*/
    int findCrossOver(int arr[], int low, int high, int x)
    {
        // Base cases
        if (arr[high] <= x) // x is greater than all
            return high;
        if (arr[low] > x)  // x is smaller than all
            return low;

        // Find the middle point
        int mid = (low + high)/2;  /* low + (high - low)/2 */

        /* If x is same as middle element, then return mid */
        if (arr[mid] <= x && arr[mid+1] > x)
            return mid;

        /* If x is greater than arr[mid], then either arr[mid + 1]
          is ceiling of x or ceiling lies in arr[mid+1...high] */
        if(arr[mid] < x)
            return findCrossOver(arr, mid+1, high, x);

        return findCrossOver(arr, low, mid - 1, x);
    }

    // This function prints k closest elements to x in arr[].
    // n is the number of elements in arr[]
    void printKclosest(int arr[], int x, int k, int n)
    {
        // Find the crossover point
        int l = findCrossOver(arr, 0, n-1, x);
        int r = l+1;   // Right index to search
        int count = 0; // To keep track of count of elements
                       // already printed

        // If x is present in arr[], then reduce left index
        // Assumption: all elements in arr[] are distinct
        if (arr[l] == x) l--;

        // Compare elements on left and right of crossover
        // point to find the k closest elements
        while (l >= 0 && r < n && count < k)
        {
            if (x - arr[l] < arr[r] - x)
                System.out.print(arr[l--]+" ");
            else
                System.out.print(arr[r++]+" ");
            count++;
        }

        // If there are no more elements on right side, then
        // print left elements
        while (count < k && l >= 0)
        {
            System.out.print(arr[l--]+" ");
            count++;
        }


        // If there are no more elements on left side, then
        // print right elements
        while (count < k && r < n)
        {
            System.out.print(arr[r++]+" ");
            count++;
        }
    }

    /* Driver program to check above functions */
    public static void main(String args[])
    {
        KClosest ob = new KClosest();
        int arr[] = {12, 16, 22, 30, 35, 39, 42,
                     45, 48, 50, 53, 55, 56
                    };
        int n = arr.length;
        int x = 35, k = 4;
        ob.printKclosest(arr, x, 4, n);
    }
}
/* This code is contributed by Rajat Mishra */

Python3

# Function to find the cross over point
# (the point before which elements are
# smaller than or equal to x and after
# which greater than x)
def findCrossOver(arr, low, high, x) :

    # Base cases
    if (arr[high] <= x) : # x is greater than all
        return high

    if (arr[low] > x) : # x is smaller than all
        return low

    # Find the middle point
    mid = (low + high) // 2 # low + (high - low)// 2

    # If x is same as middle element,
    # then return mid
    if (arr[mid] <= x and arr[mid + 1] > x) :
        return mid

    # If x is greater than arr[mid], then
    # either arr[mid + 1] is ceiling of x
    # or ceiling lies in arr[mid+1...high]
    if(arr[mid] < x) :
        return findCrossOver(arr, mid + 1, high, x)

    return findCrossOver(arr, low, mid - 1, x)

# This function prints k closest elements to x
# in arr[]. n is the number of elements in arr[]
def printKclosest(arr, x, k, n) :

    # Find the crossover point
    l = findCrossOver(arr, 0, n - 1, x)
    r = l + 1 # Right index to search
    count = 0 # To keep track of count of
              # elements already printed

    # If x is present in arr[], then reduce
    # left index. Assumption: all elements
    # in arr[] are distinct
    if (arr[l] == x) :
        l -= 1

    # Compare elements on left and right of crossover
    # point to find the k closest elements
    while (l >= 0 and r < n and count < k) :

        if (x - arr[l] < arr[r] - x) :
            print(arr[l], end = " ")
            l -= 1
        else :
            print(arr[r], end = " ")
            r += 1
        count += 1

    # If there are no more elements on right
    # side, then print left elements
    while (count < k and l >= 0) :
        print(arr[l], end = " ")
        l -= 1
        count += 1

    # If there are no more elements on left
    # side, then print right elements
    while (count < k and r < n) :
        print(arr[r], end = " ")
        r += 1
        count += 1

# Driver Code
if __name__ == "__main__" :

    arr =[12, 16, 22, 30, 35, 39, 42,
              45, 48, 50, 53, 55, 56]

    n = len(arr)
    x = 35
    k = 4

    printKclosest(arr, x, 4, n)

# This code is contributed by Ryuga

C#

// C# program to find k closest elements to
// a given value
using System;

class GFG {

    /* Function to find the cross over point
    (the point before which elements are
    smaller than or equal to x and after which
    greater than x)*/
    static int findCrossOver(int []arr, int low,
                                int high, int x)
    {

        // Base cases
        // x is greater than all
        if (arr[high] <= x)
            return high;

        // x is smaller than all
        if (arr[low] > x)
            return low;

        // Find the middle point
        /* low + (high - low)/2 */
        int mid = (low + high)/2;

        /* If x is same as middle element, then
        return mid */
        if (arr[mid] <= x && arr[mid+1] > x)
            return mid;

        /* If x is greater than arr[mid], then
        either arr[mid + 1] is ceiling of x or
        ceiling lies in arr[mid+1...high] */
        if(arr[mid] < x)
            return findCrossOver(arr, mid+1,
                                      high, x);

        return findCrossOver(arr, low, mid - 1, x);
    }

    // This function prints k closest elements
    // to x in arr[]. n is the number of
    // elements in arr[]
    static void printKclosest(int []arr, int x,
                                  int k, int n)
    {

        // Find the crossover point
        int l = findCrossOver(arr, 0, n-1, x);

        // Right index to search
        int r = l + 1;

        // To keep track of count of elements
        int count = 0;

        // If x is present in arr[], then reduce
        // left index Assumption: all elements in
        // arr[] are distinct
        if (arr[l] == x) l--;

        // Compare elements on left and right of
        // crossover point to find the k closest
        // elements
        while (l >= 0 && r < n && count < k)
        {
            if (x - arr[l] < arr[r] - x)
                Console.Write(arr[l--]+" ");
            else
                Console.Write(arr[r++]+" ");
            count++;
        }

        // If there are no more elements on right
        // side, then print left elements
        while (count < k && l >= 0)
        {
            Console.Write(arr[l--]+" ");
            count++;
        }

        // If there are no more elements on left
        // side, then print right elements
        while (count < k && r < n)
        {
            Console.Write(arr[r++] + " ");
            count++;
        }
    }

    /* Driver program to check above functions */
    public static void Main()
    {
        int []arr = {12, 16, 22, 30, 35, 39, 42,
                         45, 48, 50, 53, 55, 56};
        int n = arr.Length;
        int x = 35;
        printKclosest(arr, x, 4, n);
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP Program to Find k closest
// elements to a given value

/* Function to find the cross
   over point (the point before
   which elements are smaller
   than or equal to x and after
   which greater than x) */
function findCrossOver($arr, $low,
                       $high, $x)
{

    // Base cases

    // x is greater than all
    if ($arr[$high] <= $x)
        return $high;

    // x is smaller than all
    if ($arr[$low] > $x)
        return $low;

    // Find the middle point
    /* low + (high - low)/2 */
    $mid = ($low + $high)/2;

    /* If x is same as middle
       element, then return mid */
    if ($arr[$mid] <= $x and
        $arr[$mid + 1] > $x)
        return $mid;

    /* If x is greater than arr[mid],
       then either arr[mid + 1] is
       ceiling of x or ceiling lies
       in arr[mid+1...high] */
    if($arr[$mid] < $x)
        return findCrossOver($arr, $mid + 1,
                                 $high, $x);

    return findCrossOver($arr, $low,
                      $mid - 1, $x);
}

// This function prints k
// closest elements to x in arr[].
// n is the number of elements
// in arr[]
function printKclosest($arr, $x, $k, $n)
{

    // Find the crossover point
    $l = findCrossOver($arr, 0, $n - 1, $x);

    // Right index to search
    $r = $l + 1;

    // To keep track of count of
    // elements already printed
    $count = 0;

    // If x is present in arr[],
    // then reduce left index
    // Assumption: all elements
    // in arr[] are distinct
    if ($arr[$l] == $x) $l--;

    // Compare elements on left
    // and right of crossover
    // point to find the k
    // closest elements
    while ($l >= 0 and $r < $n
              and $count < $k)
    {
        if ($x - $arr[$l] < $arr[$r] - $x)
            echo $arr[$l--]," ";
        else
            echo $arr[$r++]," ";
        $count++;
    }

    // If there are no more
    // elements on right side,
    // then print left elements
    while ($count < $k and $l >= 0)
        echo $arr[$l--]," "; $count++;

    // If there are no more
    // elements on left side,
    // then print right elements
    while ($count < $k and $r < $n)
        echo $arr[$r++]; $count++;
}

// Driver Code
$arr =array(12, 16, 22, 30, 35, 39, 42,
                45, 48, 50, 53, 55, 56);
$n = count($arr);
$x = 35; $k = 4;

printKclosest($arr, $x, 4, $n);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// JavaScript program to find k
// closest elements to a given value

// Function to find the cross over point
// (the point before which elements are
// smaller than or equal to x and after
// which greater than x)
function findCrossOver(arr, low, high, x)
{

    // Base cases
    if (arr[high] <= x)  // x is greater than all
        return high

    if (arr[low] > x)  // x is smaller than all
        return low

    // Find the middle point
    var mid = (low + high) // 2 // low + (high - low)// 2

    // If x is same as middle element,
    // then return mid
    if (arr[mid] <= x && arr[mid + 1] > x)
        return mid

    // If x is greater than arr[mid], then
    // either arr[mid + 1] is ceiling of x
    // or ceiling lies in arr[mid+1...high]
    if (arr[mid] < x)
        return findCrossOver(arr, mid + 1, high, x)

    return findCrossOver(arr, low, mid - 1, x)

}

// This function prints k closest elements to x
// in arr[]. n is the number of elements in arr[]
function printKclosest(arr, x, k, n)
{

    // Find the crossover point
    var l = findCrossOver(arr, 0, n - 1, x)
    var r = l + 1 // Right index to search
    var count = 0 // To keep track of count of
                  // elements already printed

    // If x is present in arr[], then reduce
    // left index. Assumption: all elements
    // in arr[] are distinct
    if (arr[l] == x)
        l -= 1

    // Compare elements on left and right of crossover
    // point to find the k closest elements
    while (l >= 0 && r < n && count < k)
    {

        if (x - arr[l] < arr[r] - x)
        {
            document.write(arr[l] + " ")
            l -= 1
        }
        else
        {
            document.write(arr[r] + " ")
            r += 1
        }
        count += 1
    }

    // If there are no more elements on right
    // side, then print left elements
    while (count < k && l >= 0)
    {
        print(arr[l])
        l -= 1
        count += 1
    }

    // If there are no more elements on left
    // side, then print right elements
    while (count < k && r < n)
    {
        print(arr[r])
        r += 1
        count += 1
    }
}

// Driver Code
var arr = [ 12, 16, 22, 30, 35, 39, 42,
            45, 48, 50, 53, 55, 56 ]

var n = arr.length
var x = 35
var k = 4

printKclosest(arr, x, 4, n)

// This code is contributed by AnkThon

</script>

Output

39 30 42 45 

Time complexity: O(Logn + k).
Auxiliary Space: O(1), since no extra space has been used.

Approach 2: Using Priority Queue

This approach uses a priority queue (max heap) to maintain the k closest numbers to x. It iterates over the elements in the array and calculates their absolute differences from x. The pairs of absolute differences and negative values are pushed into the max heap. If the size of the max heap exceeds k, the element with the maximum absolute difference is removed. Finally, the top k elements from the max heap are extracted and stored in a result vector. The vector is then reversed to obtain the closest numbers in ascending order before being returned as the result.

Below is the implementation:

C++

#include <bits/stdc++.h>
using namespace std;

vector<int> findClosestElements(vector<int>& arr, int k,
                                int x)
{
    // Create a max heap to store the pairs of absolute
    // differences and negative values
    priority_queue<pair<int, int> > maxH;

      int n = arr.size();

    for (int i = 0; i < n; i++) {
        // Skip if the element is equal to x
        if (arr[i] == x)
            continue;

        // Calculate the absolute difference and add the
        // pair to the max heap
        maxH.push({ abs(arr[i] - x), -arr[i] });

        // If the size of the max heap exceeds k, remove the
        // element with the maximum absolute difference
        if (maxH.size() > k)
            maxH.pop();
    }

    // Store the result in a vector
    vector<int> result;

    // Retrieve the top k elements from the max heap
    while (!maxH.empty()) {
        // Get the top element from the max heap
        auto p = maxH.top();
        maxH.pop();

        // Add the negative value to the result vector
        result.push_back(-p.second);
    }

    // Reverse the result vector to get the closest numbers
    // in ascending order
    reverse(result.begin(), result.end());

    return result;
}

int main()
{
    vector<int> arr = { 12, 16, 22, 30, 35, 39, 42,
                        45, 48, 50, 53, 55, 56 };
    int k = 4, x = 35;
    vector<int> res = findClosestElements(arr, k, x);
    for (int i = 0; i < res.size(); i++) {
        cout << res[i] << " ";
    }
    cout << endl;
    return 0;
}

Output

39 30 42 45 

Time Complexity: O(n log k), where n is the size of the array and k is the number of elements to be returned. The priority queue takes O(log k) time to insert an element and O(log k) time to remove the top element. Therefore, traversing through the array and inserting elements into the priority queue takes O(n log k) time. Popping elements from the priority queue and pushing them into the result vector takes O(k log k) time. Therefore, the total time complexity is O(n log k + k log k) which is equivalent to O(n log k).
Auxiliary Space: O(k), as we are using a priority queue of size k+1 and a vector of size k to store the result.

Exercise: Extend the optimized solution to work for duplicates also, i.e., to work for arrays where elements don’t have to be distinct.