Closest K Elements in a Sorted Array
Last Updated :
11 Jun, 2025
Improve
Try it on GfG Practice 
You are given a sorted array arr[] containing unique integers, a number k, and a target value x. Your goal is to return exactly k elements from the array that are closest to x, excluding x itself if it is present in the array.
An element a is closer to x than b if:
|a - x| < |b - x|, or|a - x| == |b - x|anda > b(i.e., prefer the larger element if tied)
Examples:
Input: arr[] = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56], k = 4, x = 35
Output: 39 30 42 45
Explanation: First closest element to 35 is 39.
Second closest element to 35 is 30.
Third closest element to 35 is 42.
And fourth closest element to 35 is 45.Input: arr[] = [1, 3, 4, 10, 12], k = 2, x = 4
Output: 3 1
Explanation:4is excluded, Closest elements to4are:3 (1),1 (3). So, the 2 closest elements are:3 1
Table of Content
[Naive Aprroach] Using Absolute difference and Custom Sorting O(n*log(n)) Time and O(k) Space
The idea is to use the absolute difference between each element and the target value to measure how close they are. Then, we apply custom sorting: elements with smaller differences come first, and in case of a tie, the larger element is preferred. we take the k element from get from custrom sorted array then return it.
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
// Function to return k closest elements to x
vector<int> printKClosest(vector<int> &arr, int k, int x) {
// Custom comparator using absolute difference and tie-breaker
sort(arr.begin(), arr.end(), [x](int a, int b) {
int diffA = abs(a - x);
int diffB = abs(b - x);
// If differences are equal, prefer the larger element
if (diffA == diffB)
return a > b;
return diffA < diffB;
});
vector<int> result;
int count = 0;
// Pick first k elements which are not equal to x
for (int num : arr) {
// skip if element is equal to x
if (num == x) continue;
result.push_back(num);
count++;
if (count == k)
break;
}
return result;
}
// Main function with example
int main() {
vector<int> arr = {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};
int k = 4;
int x = 35;
vector<int> closest = printKClosest(arr, k, x);
for (int num : closest) {
cout << num << " ";
}
return 0;
}
import java.util.*;
public class Main {
// Function to return k closest elements to x
public static int[] printKClosest(int[] arr, int k, int x) {
// Convert array to list for easier sorting with custom comparator
List<Integer> list = new ArrayList<>();
for (int num : arr) {
list.add(num);
}
// Custom sort using absolute difference and tie-breaking
list.sort((a, b) -> {
int diffA = Math.abs(a - x);
int diffB = Math.abs(b - x);
if (diffA == diffB) {
// prefer larger element
return b - a;
}
return diffA - diffB;
});
List<Integer> result = new ArrayList<>();
int count = 0;
// Pick first k elements not equal to x
for (int num : list) {
// skip if element is equal to x
if (num == x) continue;
result.add(num);
count++;
if (count == k) break;
}
// Convert list to array
int[] closest = new int[k];
for (int i = 0; i < k; i++) {
closest[i] = result.get(i);
}
return closest;
}
// Main function with example
public static void main(String[] args) {
int[] arr = {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};
int k = 4;
int x = 35;
int[] closest = printKClosest(arr, k, x);
for (int num : closest) {
System.out.print(num + " ");
}
}
}
def printKClosest(arr, k, x):
# Custom sort using absolute difference and tie-breaking
# -a to prefer larger on tie
arr.sort(key=lambda a: (abs(a - x), -a))
result = []
count = 0
# Pick first k elements not equal to x
for num in arr:
if num == x:
continue # skip if element is equal to x
result.append(num)
count += 1
if count == k:
break
return result
if __name__ == "__main__":
arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]
k = 4
x = 35
closest = printKClosest(arr, k, x)
print( *closest)
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
// Function to return k closest elements to x
public static int[] printKClosest(int[] arr, int k, int x){
// Convert to list for sorting with custom comparator
List<int> list = arr.ToList();
// Custom sort using absolute difference and tie-breaking
list.Sort((a, b) => {
int diffA = Math.Abs(a - x);
int diffB = Math.Abs(b - x);
if (diffA == diffB)
// prefer larger element
return b - a;
return diffA - diffB;
});
List<int> result = new List<int>();
int count = 0;
// Pick first k elements not equal to x
foreach (int num in list){
// skip if element is equal to x
if (num == x) continue;
result.Add(num);
count++;
if (count == k) break;
}
return result.ToArray();
}
// Main function with example
static void Main(){
int[] arr = { 12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56 };
int k = 4;
int x = 35;
int[] closest = printKClosest(arr, k, x);
foreach (int num in closest){
Console.Write(num + " ");
}
}
}
function printKClosest(arr, k, x) {
// Custom sort using absolute difference and tie-breaking
arr.sort((a, b) => {
let diffA = Math.abs(a - x);
let diffB = Math.abs(b - x);
if (diffA === diffB) return b - a; // prefer larger element
return diffA - diffB;
});
let result = [];
let count = 0;
// Pick first k elements not equal to x
for (let num of arr) {
if (num === x) continue;
result.push(num);
count++;
if (count === k) break;
}
return result;
}
// Cotume code
let arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56];
let k = 4;
let x = 35;
let closest = printKClosest(arr, k, x);
console.log(...closest);
Output
39 30 42 45
[Better Approach] Using Linear Search- O(n) time and O(k) space
The idea is to first go through the array to find the last element that is less than or equal to the target value, skipping the target if it's present. Then, we use two pointers to choose the k closest elements by comparing their differences, while following the tie-breaking rules.
// C++ program to find k closest elements to a given value
#include <bits/stdc++.h>
using namespace std;
// Function to find k closest elements to a given value
vector<int> printKClosest(vector<int> &arr, int k, int x) {
int n = arr.size();
int i = 0;
// Find index of element just less than x
while (i < n && arr[i] < x) i++;
int left = i - 1, right = i;
// If value at right index is x, increment
if (arr[right] == x) right++;
vector<int> res;
while (left >=0 && right < n && res.size() < k) {
int leftDiff = abs(arr[left] - x);
int rightDiff = abs(arr[right] - x);
if (leftDiff < rightDiff) {
res.push_back(arr[left]);
left--;
}
else {
res.push_back(arr[right]);
right++;
}
}
// If k elements are not filled
while (left >=0 && res.size() < k) {
res.push_back(arr[left]);
left--;
}
while (right < n && res.size() < k) {
res.push_back(arr[right]);
right++;
}
return res;
}
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 = printKClosest(arr, k, x);
for (int val: res) cout << val << " ";
cout << endl;
return 0;
}
// Java program to find k closest elements to a given value
class GfG {
// Function to find k closest elements to a given value
static int[] printKClosest(int[] arr, int k, int x) {
int n = arr.length;
int i = 0;
// Find index of element just less than x
while (i < n && arr[i] < x) i++;
int left = i - 1, right = i;
// If value at right index is x, increment
if (right < n && arr[right] == x) right++;
int[] res = new int[k];
int count = 0;
while (left >= 0 && right < n && count < k) {
int leftDiff = Math.abs(arr[left] - x);
int rightDiff = Math.abs(arr[right] - x);
if (leftDiff < rightDiff) {
res[count++] = arr[left];
left--;
}
else {
res[count++] = arr[right];
right++;
}
}
// If k elements are not filled
while (left >= 0 && count < k) {
res[count++] = arr[left];
left--;
}
while (right < n && count < k) {
res[count++] = arr[right];
right++;
}
return res;
}
public static void main(String[] args) {
int[] arr =
{12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};
int k = 4, x = 35;
int[] res = printKClosest(arr, k, x);
for (int val : res) System.out.print(val + " ");
System.out.println();
}
}
# Python program to find k closest elements to a given value
# Function to find k closest elements to a given value
def printKClosest(arr, k, x):
n = len(arr)
i = 0
# Find index of element just less than x
while i < n and arr[i] < x:
i += 1
left = i - 1
right = i
# If value at right index is x, increment
if right < n and arr[right] == x:
right += 1
res = []
while left >= 0 and right < n and len(res) < k:
leftDiff = abs(arr[left] - x)
rightDiff = abs(arr[right] - x)
if leftDiff < rightDiff:
res.append(arr[left])
left -= 1
else:
res.append(arr[right])
right += 1
# If k elements are not filled
while left >= 0 and len(res) < k:
res.append(arr[left])
left -= 1
while right < n and len(res) < k:
res.append(arr[right])
right += 1
return res
if __name__ == "__main__":
arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]
k = 4
x = 35
res = printKClosest(arr, k, x)
print(' '.join(map(str, res)))
// C# program to find k closest elements to a given value
using System;
class GfG {
// Function to find k closest elements to a given value
static int[] printKClosest(int[] arr, int k, int x) {
int n = arr.Length;
int i = 0;
// Find index of element just less than x
while (i < n && arr[i] < x) i++;
int left = i - 1, right = i;
// If value at right index is x, increment
if (right < n && arr[right] == x) right++;
int[] res = new int[k];
int count = 0;
while (left >= 0 && right < n && count < k) {
int leftDiff = Math.Abs(arr[left] - x);
int rightDiff = Math.Abs(arr[right] - x);
if (leftDiff < rightDiff) {
res[count++] = arr[left];
left--;
}
else {
res[count++] = arr[right];
right++;
}
}
// If k elements are not filled
while (left >= 0 && count < k) {
res[count++] = arr[left];
left--;
}
while (right < n && count < k) {
res[count++] = arr[right];
right++;
}
return res;
}
static void Main() {
int[] arr =
{12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};
int k = 4, x = 35;
int[] res = printKClosest(arr, k, x);
foreach (int val in res) Console.Write(val + " ");
Console.WriteLine();
}
}
// JavaScript program to find k closest elements to a given value
// Function to find k closest elements to a given value
function printKClosest(arr, k, x) {
let n = arr.length;
let i = 0;
// Find index of element just less than x
while (i < n && arr[i] < x) i++;
let left = i - 1, right = i;
// If value at right index is x, increment
if (right < n && arr[right] === x) right++;
let res = [];
while (left >= 0 && right < n && res.length < k) {
let leftDiff = Math.abs(arr[left] - x);
let rightDiff = Math.abs(arr[right] - x);
if (leftDiff < rightDiff) {
res.push(arr[left]);
left--;
}
else {
res.push(arr[right]);
right++;
}
}
// If k elements are not filled
while (left >= 0 && res.length < k) {
res.push(arr[left]);
left--;
}
while (right < n && res.length < k) {
res.push(arr[right]);
right++;
}
return res;
}
let arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56];
let k = 4, x = 35;
let res = printKClosest(arr, k, x);
console.log(res.join(' '));
Output
39 30 42 45
[Expected Approach] Using Binary Search - O(k + log n) time and O(k) space
The idea is to first use binary search to quickly find the last element in the array that is less than or equal to the target (skipping the target itself if it exists). Then, we use a two-pointer approach to select the k closest elements, following the tie-breaking rules.
// C++ program to find k closest elements to a given value
#include <bits/stdc++.h>
using namespace std;
// Function to find k closest elements to a given value
vector<int> printKClosest(vector<int> &arr, int k, int x) {
int n = arr.size();
int i = 0;
int low = 0, high = n - 1, pos = -1;
// Binary search to find last element less than x
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] < x) {
pos = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
int left = pos, right = pos + 1;
// If value at right index is x, increment
if (arr[right] == x) right++;
vector<int> res;
// Use two-pointer technique to pick k closest elements
while (left >=0 && right < n && res.size() < k) {
int leftDiff = abs(arr[left] - x);
int rightDiff = abs(arr[right] - x);
if (leftDiff < rightDiff) {
res.push_back(arr[left]);
left--;
}
else {
res.push_back(arr[right]);
right++;
}
}
// If k elements are not filled
while (left >=0 && res.size() < k) {
res.push_back(arr[left]);
left--;
}
while (right < n && res.size() < k) {
res.push_back(arr[right]);
right++;
}
return res;
}
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 = printKClosest(arr, k, x);
for (int val: res) cout << val << " ";
cout << endl;
return 0;
}
// Java program to find k closest elements to a given value
class GfG {
// Function to find k closest elements to a given value
static int[] printKClosest(int[] arr, int k, int x) {
int n = arr.length;
int low = 0, high = n - 1, pos = -1;
// Binary search to find last element less than x
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] < x) {
pos = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
int left = pos, right = pos + 1;
// If value at right index is x, increment
if (right < n && arr[right] == x) right++;
int[] res = new int[k];
int count = 0;
// Use two-pointer technique to pick k closest elements
while (left >= 0 && right < n && count < k) {
int leftDiff = Math.abs(arr[left] - x);
int rightDiff = Math.abs(arr[right] - x);
if (leftDiff < rightDiff) {
res[count++] = arr[left];
left--;
}
else {
res[count++] = arr[right];
right++;
}
}
// If k elements are not filled
while (left >= 0 && count < k) {
res[count++] = arr[left];
left--;
}
while (right < n && count < k) {
res[count++] = arr[right];
right++;
}
return res;
}
public static void main(String[] args) {
int[] arr =
{12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};
int k = 4, x = 35;
int[] res = printKClosest(arr, k, x);
for (int val : res) System.out.print(val + " ");
System.out.println();
}
}
# Python program to find k closest elements to a given value
# Function to find k closest elements to a given value
def printKClosest(arr, k, x):
n = len(arr)
low, high, pos = 0, n - 1, -1
# Binary search to find last element less than x
while low <= high:
mid = (low + high) // 2
if arr[mid] < x:
pos = mid
low = mid + 1
else:
high = mid - 1
left, right = pos, pos + 1
# If value at right index is x, increment
if right < n and arr[right] == x:
right += 1
res = []
# Use two-pointer technique to pick k closest elements
while left >= 0 and right < n and len(res) < k:
leftDiff = abs(arr[left] - x)
rightDiff = abs(arr[right] - x)
if leftDiff < rightDiff:
res.append(arr[left])
left -= 1
else:
res.append(arr[right])
right += 1
# If k elements are not filled
while left >= 0 and len(res) < k:
res.append(arr[left])
left -= 1
while right < n and len(res) < k:
res.append(arr[right])
right += 1
return res
if __name__ == "__main__":
arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]
k = 4
x = 35
res = printKClosest(arr, k, x)
print(' '.join(map(str, res)))
// C# program to find k closest elements to a given value
using System;
class GfG {
// Function to find k closest elements to a given value
static int[] printKClosest(int[] arr, int k, int x) {
int n = arr.Length;
int low = 0, high = n - 1, pos = -1;
// Binary search to find last element less than x
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] < x) {
pos = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
int left = pos, right = pos + 1;
// If value at right index is x, increment
if (right < n && arr[right] == x) right++;
int[] res = new int[k];
int count = 0;
// Use two-pointer technique to pick k closest elements
while (left >= 0 && right < n && count < k) {
int leftDiff = Math.Abs(arr[left] - x);
int rightDiff = Math.Abs(arr[right] - x);
if (leftDiff < rightDiff) {
res[count++] = arr[left];
left--;
}
else {
res[count++] = arr[right];
right++;
}
}
// If k elements are not filled
while (left >= 0 && count < k) {
res[count++] = arr[left];
left--;
}
while (right < n && count < k) {
res[count++] = arr[right];
right++;
}
return res;
}
static void Main() {
int[] arr =
{12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};
int k = 4, x = 35;
int[] res = printKClosest(arr, k, x);
foreach (int val in res) Console.Write(val + " ");
Console.WriteLine();
}
}
// JavaScript program to find k closest elements to a given value
// Function to find k closest elements to a given value
function printKClosest(arr, k, x) {
let n = arr.length;
let low = 0, high = n - 1, pos = -1;
// Binary search to find last element less than x
while (low <= high) {
let mid = Math.floor((low + high) / 2);
if (arr[mid] < x) {
pos = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
let left = pos, right = pos + 1;
// If value at right index is x, increment
if (right < n && arr[right] === x) right++;
let res = [];
// Use two-pointer technique to pick k closest elements
while (left >= 0 && right < n && res.length < k) {
let leftDiff = Math.abs(arr[left] - x);
let rightDiff = Math.abs(arr[right] - x);
if (leftDiff < rightDiff) {
res.push(arr[left]);
left--;
}
else {
res.push(arr[right]);
right++;
}
}
// If k elements are not filled
while (left >= 0 && res.length < k) {
res.push(arr[left]);
left--;
}
while (right < n && res.length < k) {
res.push(arr[right]);
right++;
}
return res;
}
let arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56];
let k = 4, x = 35;
let res = printKClosest(arr, k, x);
console.log(res.join(' '));
Output
39 30 42 45
