Find Kth smallest element in array

Kth smallest element in array

Given an array of integers which is non sorted, find kth smallest element in that array. For example: if input array is A = [3,5,1,2,6,9,7], 4th smallest element in array A is 5, because if you sort the array A, it looks like A = [1,2,3,5,6,7,9] and now you can easily see that 4th element is 5.

This problem is commonly asked in Microsoft and Amazon interviews as it has multiple layers and there is some many things that can be measured with this one problem.

Kth smallest element : Line of thought

First of all, in any interview, try to come up with brute force solution. Brute force solution to find Kth smallest element in array of integers would be to sort array and return A[k-1] element (K-1 as array is zero base indexed).

What is the complexity of brute force solution? It’s O(n2)? Well, we have sort algorithms like merge sort and heap sort which work in O(n log n) complexity. Problem with both searches is that they use additional space. Quick sort is another sort algorithm. It has problem that it’s worst case complexity will be O(n2), which happens when input is completely sorted.
In our case, input is given as unsorted already, so we can expect that quick sort will function with O(n log n) complexity which is it’s average case complexity. Advantage of using quick sort is that there is no additional space complexity.

Optimising quick sort

Let’s see how quicksort works and see if we can optimize solution further?
Idea behind quicksort is to find the correct place for the selected pivot. Once the pivot is at the correct position, all the elements on the left side of the pivot are smaller and on the right side of the pivot are greater than the pivot. This step is partitioning.

If after partitioning, pivot is at position j, can we say that pivot is actually jth smallest element of the array? What if j is equal to k? Well problem solved, we found the kth smallest element.

If j is less than k, left subarray is less than k, we need to include more elements from right subarray, therefore kth smallest element is in right subarray somewhere. We have already found j smallest elements, all we need to find is k-j elements from right subarray.

What if j is greater than k? In this case, we have to drop some elements from left subarray, so our search space would be left subarray after partition.

Theoretically, this algorithm still has complexity of O(n log n), but practically, you do not need to sort the entire array before you find k smallest elements.

Algorithm to find K smallest elements in array

  1. Select a pivot and partition the array with pivot at correct position j
  2. If position of pivot, j, is equal to k, return A[j].
  3. If j is less than k, discard array from start to j, and look for (k-j)th smallest element in right sub array, go to step 1.
  4. If j is greater than k, discard array from j to end and look for kth element in left subarray, go to step 1

Let’s take an example and see if this algorithm works? A =  [4, 2, 1, 7, 5, 3, 8, 10, 9, 6 ], and we have to find fifth smallest element in array A.

Kth smallest element in array

Start with pivot as first index of array, so pivot = 0, partition the array into two parts around pivot such that all elements on left side of pivot element, i.e. A[pivot] are smaller and all elements on right side are greater than A[pivot].

Start with pivot as first index of array, so pivot = 0, partition the array into two parts around pivot such that all elements on left side of pivot element, i.e. A[pivot] are smaller and all elements on right side are greater than A[pivot].

In our example, array A will look like below after pivot has found it’s correct position.

k smallest element
After partition, correct position of pivot is index 3

If pivot == k-1 (array is represented as zero base index), then A[pivot] is kth smallest element. Since pivot (3) is less than k-1 (4), look for kth smallest element on right side of the pivot.

k remains as it is as opposed to k-j mentioned in algorithm as pivot is given w.r.t entire array and not w.r.t subarray.

In second iteration, pivot = 4 (index and not element). After second execution of quick sort array A will be like

After partition of right subarray, correct position of pivot is index 4

pivot(4) which is equal to k-1(5-1). 5th smallest element in array A is 5.

Kth smallest element : Implementation


	* Created by sangar on 30.9.18.
public class KthSmallest {
	private void swap(int[] a, int i, int j){
		int temp = a[i];
		a[i] = a[j];
		a[j] = temp;
	private int partition(int[] a, int start, int end){
		int pivot = a[start];
		int i  = start+1;
		int j  = end;

		while(i <= j){
			while(a[i] < pivot) i++;
			while(a[j] > pivot) j--;

			if(i < j) {
				swap(a, i, j);
		swap(a, start, j);
		return j;

	public int findKthSmallestElement(int a[], int start, 
				int end, int k){
		if(start <= end){
		int p = partition(a, start, end);
		if(p == k-1){
			return a[p];
		if(p > k-1)
			return findKthSmallestElement(a, start, p, k);
		if(p < k-1)
			return findKthSmallestElement(a, p+1, end, k);
		return -1;
package test;

import org.junit.jupiter.api.Test;

import static org.junit.jupiter.api.Assertions.assertEquals;

 * Created by sangar on 28.8.18.
public class KthSmallestTest {

	KthSmallest tester = new KthSmallest();
	private int[] a = {4, 2, 1, 7, 5, 3, 8, 10, 9};
	public void kthSmallest() {
		assertEquals(7, tester.findKthSmallestElement(a,0,8,6));

	public void firstSmallest() {
		assertEquals(1, tester.findKthSmallestElement(a,0,8,1));

	public void lastSmallest() {
		assertEquals(10, tester.findKthSmallestElement(a,0,8,9));

	public void kGreaterThanSize() {
		assertEquals(-1, tester.findKthSmallestElement(a,0,8,15));
	public void emptyArray() {
		int[] a = {};
		assertEquals(-1, tester.findKthSmallestElement(a,0,0,1));

	public void nullArray() {
		assertEquals(-1, tester.findKthSmallestElement(null,0,0,1));

Complexity of using quick sort algorithm to find kth smallest element in array of integers in still O(n log n).

Kth smallest element using heaps

Imagine a case where there are a billion integers in the array and you have to find 5 smallest elements from that array. The complexity of O(n log n) is too costly for that use case. Above algorithm using quick sort does not take into consideration disparity between k and n.

We want top k elements, how about we chose those k elements randomly, call it set A and then go through all other n-k elements, call it set B, check if element from set B (n-k elements) can displace element in set A (k elements)?

What will be the condition for an element from set B to replace an element in set A? Well, if the new element is less than maximum in set A than the maximum in set A cannot be in the set of k smallest elements right?  Maximum element in set A would be replaced by the new element from set B.

Now, the problem is how to quickly find the maximum out of set A. Heap is the best data structure there. What kind of heap: min heap or max heap? Max heap as it store the maximum of the set at the root of it.

Let’s defined concrete steps to find k smallest elements using max heap. 

  1. Create a max heap of size k from first k elements of array.
  2. Scan all elements in array one by one.
    1.  If current element is less than max on heap, add current element to heap and heapify.
    2. If not, then go to next element.
  3. At the end, max heap will contain k smallest elements of array and root will be kth smallest element.

Let’s take an example and see if this algorithm works? The input array is shown below and we have to find the 6th smallest element in this array.

kth smallest element using heaps
input array

Step 1 : Create a max heap with first 6 elements of array.

Create a max heap with set A

Step 2: Take next element from set B and check if it is less than the root of max heap. In this case, yes it is. Remove the root and insert the new element into max heap.

Element from set B removes root from max heap and added to max heap

Step 2: It continues to 10, nothing happens as the new element is greater than the root of max heap. Same for 9.  At 6, again the root of max heap is greater than 6. So remove the root and add 6 to max heap.

Again, new element from set B is less than root of max heap. Root is removed and new element is added.

Array scan is finished, so just return the root of the max heap, 6 which is the sixth smallest element in given array.

	public int findKthSmallestElementUsingHeap(int a[], int k){

	PriorityQueue<Integer>  maxHeap =
			new PriorityQueue<>(k, Collections.reverseOrder());

		if(a == null || k > a.length) return -1;
		//Create max with first k elements
		for(int i=0; i<k; i++){

		/*Keep updating max heap based on new element
		If new element is less than root, 
		remove root and add new element

		for(int i=k; i<a.length; i++){
			if(maxHeap.peek() > a[i]){
		return maxHeap.peek();

Can you calculate the complexity of above algorithm? heapify() has complexity of log(k) with k elements on heap. In worst case, we have to do heapify() for all elements in array, which is n, so overall complexity of algorithm becomes O(n log k). Also, there is additional space complexity of O(k) to store heap.
When is very small as compared to n, this algorithm again depends on the size of array.

We want k smallest elements, if we pick first k elements from a min heap, will it solve the problem? I think so. Create a min heap of n elements in place from the given array, and then pick first k elements.
Creation of heap has complexity of O(n), do more reading on it. All we need to do is delete k times from this heap, each time there will be heapify(). It will have complexity of O(log n) for n element heap. So, overall complexity would be O(n + k log n).

Depending on what you want to optimize, select correct method to find kth smallest element in array.

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Count sort : Sorting in linear time

Count sort : Sorting in linear time

Are there any sorting algorithm which has worst case complexity of O(n)? There are a few like count sort and decision tree algorithms. In this post we would discuss about count sort and couple of problems where this counting sort algorithm can be applied.

Counting sort was invented by Harold H. Seward
To apply counting sort, we need to keep in mind following assumptions:

  1. There should be duplicate values in the input
  2. There should be at most K different type of values in input.
  3. The input data ranges in 0 to K

Count sort goes for O(n2) if K is very close to n i.e. a very few elements are duplicate and rest all are unique. So above three conditions are necessary to have this algorithm work in linear time.

Count Sort -Approach
Let’s see how does it work? There are three steps involved in the algorithm:

  • Sampling the data, counting the frequencies of each element in the input set.
  • Accumulating frequencies to find out relative positions of each element.
  • Third step distributes each element to its appropriate place.
  • Now let’s take one example and go through the above three steps:
    Input is  an array A = [1,2,3,2,2,3,3,1,1,1,1,3,2,2]. Here we can see that K=3. Now let’s perform the first step of the method, i.e. count the frequency of each element. We keep a hash and keep track of each element’s count. Since values are upper bound by K, we need at most K sized hash. We initialize this hash as zero. A is input array and n is the size of the array.

    int char count [K];
    for(i=0; i<K;i++){

    Count looks likes count =[5,5,4]. The complexity of this step is O(n). The second step involves the accumulation of frequencies where we add frequencies of previous elements to the current element.
    F(j) = summation of f[i] for all i=0

    F(j) = summation of f[i] for all i<=j and i>=0
       Count[i] +=count[i-1];

    The second step runs for K times hence the complexity of O(K). The third step involves distribution. Let’s create an output array called as B of size n For every element in A we check the corresponding count in count array and place the element at that position. So, first 1 will be at position 4 (there are total 5 and array is an index from 0), the first two will be placed at 9 and the first three will be at 13. Subsequently, lower positions will be filled as we encounter them in the input array.

    for(i=0; i<n;i++){
           B[--count[a[i]]] = a[i];

    This step has complexity of O(n)

    void count_sort(int a[], int n, int K){
            int count [K];
            int i,j;
            int b[n];
            for(i=0; i<=K;i++){
                    count[i] +=count[i-1];
            for(i=0; i<n;i++){
                    b[count[a[i]]] = a[i];
    int main(){
            int a[]= {0,1,1,1,0,0,0,3,3,3,4,4};
            int n  =  sizeof(a)/sizeof(a[0]);
    Iter 0 :0  0  0  0  1  0  0  0  0  0  0  0  0  0  
    Iter 1 :0  0  0  0  1  0  0  0  0  2  0  0  0  0  
    Iter 2 :0  0  0  0  1  0  0  0  0  2  0  0  0  3  
    Iter 3 :0  0  0  0  1  0  0  0  2  2  0  0  0  3  
    Iter 4 :0  0  0  0  1  0  0  2  2  2  0  0  0  3  
    Iter 5 :0  0  0  0  1  0  0  2  2  2  0  0  3  3  
    Iter 6 :0  0  0  0  1  0  0  2  2  2  0  3  3  3  
    Iter 7 :0  0  0  1  1  0  0  2  2  2  0  3  3  3  
    Iter 8 :0  0  1  1  1  0  0  2  2  2  0  3  3  3  
    Iter 9 :0  1  1  1  1  0  0  2  2  2  0  3  3  3  
    Iter 10 :1  1  1  1  1  0  0  2  2  2  0  3  3  3  
    Iter 11 :1  1  1  1  1  0  0  2  2  2  3  3  3  3  
    Iter 12 :1  1  1  1  1  0  2  2  2  2  3  3  3  3  
    Iter 13 :1  1  1  1  1  2  2  2  2  2  3  3  3  3  
    Final O/P :1  1  1  1  1  2  2  2  2  2  3  3  3  3

    Total complexity of the algorithm being O(K)+ O(n) + O(K) +O(n) = O(K+n). Please refer to the master theorem, how this is derived. Now if we see, K+n remains in order of n till the time K<n, if K goes on to n^2, the complexity of algorithm goes for polynomial time instead of linear time.

    There is immediate application of this algorithm in following problem:
    Let’s there is an array which contains Black, White and Red balls, we need to arrange these balls in such a way that all black balls come first, white second and Red at last. Hint: assign black 0, white 1 and Red 2 and see. 🙂

    In the next post, we would discuss how extra space can be saved, how initial positions of elements can be maintained. We would go through an interesting problem to discuss the above optimization.

    Why this filling from the end of the span? Because this step makes count sort stable sort. Here we have seen a sorting algorithm which can give us o/p linear time given some particular conditions met on the i/p data.