Find k smallest elements in an array
Last Updated :
23 Jul, 2025
Given an array arr[] and an integer k, the task is to find k smallest elements in the given array. Elements in the output array can be in any order.
Examples:
Input: arr[] = [1, 23, 12, 9, 30, 2, 50], k = 3
Output: [1, 2, 9]
Input: arr[] = [11, 5, 12, 9, 44, 17, 2], k = 2
Output: [2, 5]
[Approach - 1] Using Sorting
The idea is to sort the input array in ascending order, so the first k elements in the array will be the k smallest elements.
C++
// C++ program to find k smallest elements in an
// array using sorting
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> kSmallest(vector<int> &arr, int k) {
// sort the given array in ascending order
sort(arr.begin(), arr.end());
// store the first k element in result array
vector<int> res(arr.begin(), arr.begin() + k);
return res;
}
int main() {
vector<int>arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
vector<int> res = kSmallest(arr, k);
for(int ele : res)
cout << ele << " ";
return 0;
}
// Java program to find k smallest elements in an
// array using sorting
import java.util.Arrays;
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> kSmallest(int[] arr, int k) {
// Sort the given array in ascending order
Arrays.sort(arr);
// Store the first k elements in result ArrayList
ArrayList<Integer> res = new ArrayList<>();
for (int i = 0; i < k; i++)
res.add(arr[i]);
return res;
}
public static void main(String[] args) {
int[] arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
ArrayList<Integer> res = kSmallest(arr, k);
for (int ele : res)
System.out.print(ele + " ");
}
}
# Python program to find k smallest elements in an
# array using sorting
def kSmallest(arr, k):
# Sort the given array in ascending order
arr.sort()
# Store the first k elements in result array
res = arr[:k]
return res
if __name__ == "__main__":
arr = [1, 23, 12, 9, 30, 2, 50]
k = 3
res = kSmallest(arr, k)
for ele in res:
print(ele, end=" ")
// C# program to find k smallest elements in an
// array using sorting
using System;
using System.Collections.Generic;
class GfG {
static List<int> kSmallest(int[] arr, int k) {
// Sort the given array in ascending order
Array.Sort(arr);
// Store the first k elements in result List
List<int> res = new List<int>();
for (int i = 0; i < k; i++)
res.Add(arr[i]);
return res;
}
static void Main() {
int[] arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
List<int> res = kSmallest(arr, k);
foreach (int ele in res)
Console.Write(ele + " ");
}
}
// JavaScript program to find k smallest elements in an
// array using sorting
function kSmallest(arr, k) {
// Sort the given array in ascending order
arr.sort((a, b) => a - b);
// Store the first k elements in result array
let res = arr.slice(0, k);
return res;
}
const arr = [1, 23, 12, 9, 30, 2, 50];
const k = 3;
const res = kSmallest(arr, k);
console.log(res.join(' '));
Time complexity: O(n * logn)
Auxiliary Space: O(1)
[Approach - 2] Using Quick Sort Partitioning Step (Quick Select Algorithm)
The idea is to use the partitioning step of QuickSort to find the k smallest elements in the array, without sorting the entire array. In the partitioning step, we rearrange the elements in a way that all elements smaller than or equal to a chosen pivot (usually the last element) are placed on its left, and all elements greater than the pivot are on its right. And pivot element in its correct sorted position.
After each partition, we compare the number of elements in the left part of the array (which contains all elements smaller than or equal to the pivot) with k:
- Number of elements in the left = k, it means all elements in the left part (including pivot) are the k smallest elements.
- Number of elements in the left > k, it means that k smallest elements exist in the left subarray only, so we recursively search in the left subarray.
- Number of elements in the left < k, it means that the k smallest elements include the entire left part of the array along with some elements from the right part. Therefore we reduce k by the number of elements already covered on the left side and search in the right subarray.
C++
// C++ program to find the k smallest elements in the array
// using partitioning step of quick sort
#include <iostream>
#include <vector>
using namespace std;
// Function to partition the array around a pivot
int partition(vector<int> &arr, int left, int right) {
// Last element is chosen as a pivot.
int pivot = arr[right];
int i = left;
for (int j = left; j < right; j++) {
if (arr[j] <= pivot) {
swap(arr[i], arr[j]);
i++;
}
}
swap(arr[i], arr[right]);
// The correct sorted position of the pivot
return i;
}
void quickSelect(vector<int> &arr, int left, int right, int k) {
if (left <= right) {
int pivotIdx = partition(arr, left, right);
// Count of all elements in the left part
int leftCnt = pivotIdx - left + 1;
// If leftCnt is equal to k, then we have
// found the k smallest element
if (leftCnt == k)
return;
// Search in the left subarray
if (leftCnt > k)
quickSelect(arr, left, pivotIdx - 1, k);
// Reduce the k by number of elements already covered
// and search in the right subarray
else
quickSelect(arr, pivotIdx + 1, right, k - leftCnt);
}
}
vector<int> kSmallest(vector<int> &arr, int k) {
int n = arr.size();
quickSelect(arr, 0, n - 1, k);
// First k elements of the array, will be the smllest
vector<int> res(arr.begin(), arr.begin() + k);
return res;
}
int main() {
vector<int> arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
vector<int> res = kSmallest(arr, k);
for (int ele : res)
cout << ele << " ";
return 0;
}
// Java program to find the k smallest elements in the array
// using partitioning step of quick sort
import java.util.ArrayList;
class GfG {
// Function to partition the array around a pivot
static int partition(int[] arr, int left, int right) {
// Last element is chosen as a pivot.
int pivot = arr[right];
int i = left;
for (int j = left; j < right; j++) {
if (arr[j] <= pivot) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
}
}
int temp = arr[i];
arr[i] = arr[right];
arr[right] = temp;
// The correct sorted position of the pivot
return i;
}
static void quickSelect(int[] arr, int left, int right, int k) {
if (left <= right) {
int pivotIdx = partition(arr, left, right);
// Count of all elements in the left part
int leftCnt = pivotIdx - left + 1;
// If leftCnt is equal to k, then we have
// found the k smallest element
if (leftCnt == k)
return;
// Search in the left subarray
if (leftCnt > k)
quickSelect(arr, left, pivotIdx - 1, k);
// Reduce the k by number of elements already covered
// and search in the right subarray
else
quickSelect(arr, pivotIdx + 1, right, k - leftCnt);
}
}
static ArrayList<Integer> kSmallest(int[] arr, int k) {
quickSelect(arr, 0, arr.length - 1, k);
ArrayList<Integer> res = new ArrayList<>();
// First k elements of the array, will be the smallest
for(int i = 0; i < k; i++)
res.add(arr[i]);
return res;
}
public static void main(String[] args) {
int[] arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
ArrayList<Integer> res = kSmallest(arr, k);
for (int ele : res)
System.out.print(ele + " ");
}
}
# Python program to find the k smallest elements in the array
# using partitioning step of quick sort
# Function to partition the array around a pivot
def partition(arr, left, right):
# Last element is chosen as a pivot.
pivot = arr[right]
i = left
for j in range(left, right):
if arr[j] <= pivot:
arr[i], arr[j] = arr[j], arr[i]
i += 1
arr[i], arr[right] = arr[right], arr[i]
# The correct sorted position of the pivot
return i
def quickSelect(arr, left, right, k):
if left <= right:
pivotIdx = partition(arr, left, right)
# Count of all elements in the left part
leftCnt = pivotIdx - left + 1
# If leftCnt is equal to k, then we have
# found the k smallest element
if leftCnt == k:
return
# Search in the left subarray
if leftCnt > k:
quickSelect(arr, left, pivotIdx - 1, k)
# Reduce the k by number of elements already covered
# and search in the right subarray
else:
quickSelect(arr, pivotIdx + 1, right, k - leftCnt)
def kSmallest(arr, k):
quickSelect(arr, 0, len(arr) - 1, k)
# First k elements of the array, will be the smallest
return arr[:k]
if __name__ == "__main__":
arr = [1, 23, 12, 9, 30, 2, 50]
k = 3
res = kSmallest(arr, k)
print(" ".join(map(str, res)))
// C# program to find the k smallest elements in the array
// using partitioning step of quick sort
using System;
using System.Collections.Generic;
class GfG {
// Function to partition the array around a pivot
static int partition(int[] arr, int left, int right) {
// Last element is chosen as a pivot.
int pivot = arr[right];
int i = left;
for (int j = left; j < right; j++) {
if (arr[j] <= pivot) {
swap(arr, i, j);
i++;
}
}
swap(arr, i, right);
// The correct sorted position of the pivot
return i;
}
static void quickSelect(int[] arr, int left, int right, int k) {
if (left <= right) {
int pivotIdx = partition(arr, left, right);
// Count of all elements in the left part
int leftCnt = pivotIdx - left + 1;
// If leftCnt is equal to k, then we have
// found the k smallest element
if (leftCnt == k)
return;
// Search in the left subarray
if (leftCnt > k)
quickSelect(arr, left, pivotIdx - 1, k);
// Reduce the k by number of elements already covered
// and search in the right subarray
else
quickSelect(arr, pivotIdx + 1, right, k - leftCnt);
}
}
static List<int> kSmallest(int[] arr, int k) {
quickSelect(arr, 0, arr.Length - 1, k);
// First k elements of the array, will be the smallest
List<int> res = new List<int>();
for (int i = 0; i < k; i++) {
res.Add(arr[i]);
}
return res;
}
static void swap(int[] arr, int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void Main() {
int[] arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
List<int> res = kSmallest(arr, k);
foreach (int ele in res)
Console.Write(ele + " ");
}
}
// JavaScript program to find the k smallest elements in the array
// using partitioning step of quick sort
// Function to partition the array around a pivot
function partition(arr, left, right) {
// Last element is chosen as a pivot.
let pivot = arr[right];
let i = left;
for (let j = left; j < right; j++) {
if (arr[j] <= pivot) {
[arr[i], arr[j]] = [arr[j], arr[i]];
i++;
}
}
[arr[i], arr[right]] = [arr[right], arr[i]];
// The correct sorted position of the pivot
return i;
}
function quickSelect(arr, left, right, k) {
if (left <= right) {
let pivotIdx = partition(arr, left, right);
// Count of all elements in the left part
let leftCnt = pivotIdx - left + 1;
// If leftCnt is equal to k, then the first
// k element of the array will be smallest
if (leftCnt === k)
return;
// Search in the left subarray
if (leftCnt > k)
quickSelect(arr, left, pivotIdx - 1, k);
// Reduce the k by number of elements already covered
// and search in the right subarray
else
quickSelect(arr, pivotIdx + 1, right, k - leftCnt);
}
}
function kSmallest(arr, k) {
quickSelect(arr, 0, arr.length - 1, k);
// First k elements of the array, will be the smallest
return arr.slice(0, k);
}
const arr = [1, 23, 12, 9, 30, 2, 50];
const k = 3;
const res = kSmallest(arr, k);
console.log(res.join(' '));
Time Complexity: O(n*n) in worst case(O(n) on average)
Auxiliary Space: O(n)
[Approach - 3] Using Priority Queue(Max-Heap)
The idea is, as we iterate through the array, we keep track of the k smallest elements at each step. To achieve this, we use a max-heap. First, we insert the initial k elements into the max-heap. After that, for each next element, we compare it with the top of the heap. Since the top element of the max-heap is the largest among the k elements, if the current element is smaller than the top, it means the top element is no longer one of the k smallest elements. In this case, we remove the top and insert the smaller element. After completing the entire traversal, the heap will contain exactly the k smallest elements of the array.
C++
// C++ program to find the k smallest elements in the
// array using max heap
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
vector<int> kSmallest(vector<int> &arr, int k) {
// Max Priority Queue (Max-Heap)
priority_queue<int> maxH;
for (int i = 0; i < arr.size(); i++) {
// Insert initial k elements in the max heap
if (maxH.size() < k)
maxH.push(arr[i]);
// If the top of heap is greater than the arr[i]
// then remove this and insert arr[i]
else if(maxH.top() > arr[i]) {
maxH.pop();
maxH.push(arr[i]);
}
}
vector<int> res;
// Max heap will contain only k
// smallest element
while (!maxH.empty()) {
res.push_back(maxH.top());
maxH.pop();
}
return res;
}
int main() {
vector<int> arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
vector<int> res = kSmallest(arr, k);
for(int ele : res)
cout << ele << " ";
return 0;
}
// Java program to find the k smallest elements in the array
// using max heap
import java.util.PriorityQueue;
import java.util.ArrayList;
import java.util.Collections;
class GfG {
static ArrayList<Integer> kSmallest(int[] arr, int k) {
// Max Priority Queue (Max-Heap)
PriorityQueue<Integer> maxH =
new PriorityQueue<>(Collections.reverseOrder());
for (int i = 0; i < arr.length; i++) {
// Insert initial k elements in the max heap
if (maxH.size() < k)
maxH.add(arr[i]);
// If the top of heap is greater than arr[i]
// then remove this and insert arr[i]
else if (maxH.peek() > arr[i]) {
maxH.poll();
maxH.add(arr[i]);
}
}
// Max heap will contain all k smallest elements
ArrayList<Integer> res = new ArrayList<>();
while (!maxH.isEmpty()) {
res.add(maxH.poll());
}
return res;
}
public static void main(String[] args) {
int[] arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
ArrayList<Integer> res = kSmallest(arr, k);
for (int ele : res) {
System.out.print(ele + " ");
}
}
}
# Python program to find the k smallest elements in the
# array using max heap
import heapq
def kSmallest(arr, k):
# Max Priority Queue (Max-Heap)
maxH = []
for num in arr:
# Insert initial k elements in the max heap
if len(maxH) < k:
heapq.heappush(maxH, -num)
# If the top of heap is greater than
# arr[i], remove this and insert arr[i]
elif -maxH[0] > num:
heapq.heappop(maxH)
heapq.heappush(maxH, -num)
# Max heap will contain all k smallest elements
res = [-x for x in maxH]
return res
if __name__ == "__main__":
arr = [1, 23, 12, 9, 30, 2, 50]
k = 3
res = kSmallest(arr, k)
for ele in res:
print(ele, end=" ")
// C# program to find the k smallest elements in the
// array using max heap
using System;
using System.Collections.Generic;
class GfG {
static List<int> kSmallest(int[] arr, int k) {
// Max Priority Queue (Max-Heap)
SortedSet<int> maxH =
new SortedSet<int>(Comparer<int>.Create((a, b) => b.CompareTo(a)));
for (int i = 0; i < arr.Length; i++) {
// Insert initial k elements in the max heap
if (maxH.Count < k)
maxH.Add(arr[i]);
// If the top of heap is greater than arr[i]
// then remove this and insert arr[i]
else if (maxH.Min > arr[i]) {
maxH.Remove(maxH.Min);
maxH.Add(arr[i]);
}
}
// Max heap will contain all k smallest elements
List<int> res = new List<int>(maxH);
return res;
}
static void Main() {
int[] arr = {1, 23, 12, 9, 30, 2, 50};
int k = 3;
List<int> res = kSmallest(arr, k);
foreach (int ele in res)
Console.Write(ele + " ");
}
}
// JavaScript program to find the k smallest elements in the
// array using max heap
function kSmallest(arr, k) {
// Max Priority Queue (Max-Heap)
let maxH = [];
for (let i = 0; i < arr.length; i++) {
// Insert initial k elements in the max heap
if (maxH.length < k) {
maxH.push(arr[i]);
maxH.sort((a, b) => b - a);
}
// If the top of heap is greater than arr[i]
// then remove this and insert arr[i]
else if (maxH[0] > arr[i]) {
maxH.shift();
maxH.push(arr[i]);
maxH.sort((a, b) => b - a);
}
}
// Max heap will contain only k smallest elements
return maxH;
}
const arr = [1, 23, 12, 9, 30, 2, 50];
const k = 3;
const res = kSmallest(arr, k);
console.log(res.join(' '));
Time Complexity: O(n * log(k))
Auxiliary Space: O(k)
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