Kth smallest element in two sorted arrays

We have already solved a problem to find kth smallest element in an array using quicksort modification and min-heap. Today, our problem is to find Kth smallest element in two sorted arrays.
For example if A = [10, 20, 40, 60] and B =[15, 35, 50, 70, 100] and K = 4 then solution should be 35 because union of above arrays will be C = [10,15,20,35,40,50,60,70,100] and fourth smallest element is 35.

Kth smallest element in two sorted arrays

As I always insist on doing: what will be the most brute force solution for this problem. Simple, merge two sorted arrays into one and find the kth smallest element in the array. There are different ways to merge two sorted arrays, merge part of merge sort will be most efficient when two parts are already sorted. The time complexity of merge is O(n+m) and additional space complexity of O(m+n) where m and m are sizes of two sorted arrays.

If you are preparing for an interview, you can signup for a free session to find a coach to help you with your preparation.

Now that we already know brute force solution, can we do better than this? The k-th smallest element can be present in any one of the two arrays. How many elements can we include from first array A, with size m, it can be max(k-1, m).

Let’s suppose we took i elements from the first array, how many maximum numbers of elements we can take from array B? Since, we have zero-based indexing of array,

i+j = k-1

kth smallest element in two sorted arrays

Now, we need to find combination of i and j such that all elements from 0 to i are less than B[j] and all elements from 0 to j should be smaller than A[i], and min(A[i], B[j]) will be the kth smallest element in the combined array.

k-th smallest element in two sorted array

For A[i], all elements from index 0 to i-1 are smaller, all we need to check is that all elements in array B from index 0 to index j-1 are smaller too. Therefor A[i] >= B[j-1]

Similarly for B[j], it must satisfy the condition B[j] >= A[i-1].

kth smallest element in array

In order to be kth smallest element, index i and j have to satisfy two conditions:

  1. A[i] >= B[j-1] and B[j] >= A[i-1]
  2. i+j  =  k-1 or j = k-1-i

What if A[i] < B[j-1]  as we saw in the first picture?  That means i is too small, we need to increase i and when we increase i and A[i], j and B[j] is decreased automatically, which makes it possible to satisfy A[i] >= B[j-1].

In the same vain when B[j] < A[i-1],  i is too big and it’s a good choice to decrease it.

This is where binary search comes into the picture. We can start i as mid of array A, j = k-i, and see if this i satisfies the condition. There can be three possible outcomes for the condition.

1. A[i-1] >= B[j] and B[j-1] <=A[i] is true, we return the index min(A[i], B[j])
2. IfB[j-1] > A[i], in this case, A[i] is too small. How can we increase it? by moving towards right. If i is increased, value A[i] is bound to increase, and also it will decrease j. In this case, B[j-1] will decrease and A[i] will increase which will make B[j-1] >=A[i] is true. So, limit search space for i to  mid+1 to min(k, m)
3. A[i-1] > B[j], means A[i-1] is too big. And we must decrease i to get A[i-1] >=B[j]. Limit search space for i from 0 to i-1.

Since, i is bound by 0 to min(k-1,m), j will never go below 0. When i or j is 0, return min(A[i], B[j]).

Let’s take an example and see how it works. Below are the two sorted arrays and K = 7.

The first thing to notice is that k is greater than m, size of array A. Hence the range of i will be 0 to m i.e. 5. Find mid, i = 2 and j = k – 2-1 = 4

Now, A[i] < B[j-1], it is evident that we chose i too small. Search space is now from i+1 to m.

Again we find mid which is index 4 of array A. Index j = 2. At this point condition A[i-1] <= B[j] is false. This means we chose i too big and hence we decrease range till i-1 which now 3.

New i and j are shown below. Condition A[i-1] <= B[j] and B[j-1] <= A[j] is true. Return min(A[i], B[j]) as kth smallest element which is 6.

Implemmentation to find kth smallest element

package com.company;

 * Created by sangar on 19.4.18.
public class KthSmallestElement {

    public static double findKthSmallestElement(int[] A, int[] B, int k){

        int[] temp;

        int lenA = A.length;
        int lenB = B.length;

        if(lenA + lenB < k) return -1;

        int iMin = 0;
        int iMax = Integer.min(A.length, k-1);

        int i = 0;
        int j = 0;

        while (iMin <= iMax) {
            i = (iMin + iMax) / 2;
            j = k - 1 - i; // because of zero based index
            if (B[j - 1] > A[i]) {
                // i is too small, must increase it
                iMin = i + 1;
            } else if (i > 0 && A[i - 1] > B[j]) {
                // i is too big, must decrease it
                iMax = i - 1;
            } else {
                // i is perfect
               return Integer.min(A[i], B[j]);
        return -1;

    public static void main(String[] args){
        int[] a = {1,3,5,6,7,8,9,11};
        int[] b = {1,4,6,8,12,14,15,17};

        double smallest = findKthSmallestElement(a,b, 9);
        System.out.println("Kth smallest element is : " + smallest);

Complexity of algorithm to find Kth smallest element two sorted arrays is O(log(N+M)).

Please share if there is something wrong or missing. we would be glad to hear from you. If you are preparing for an interview and want to understand how to approach it, please book a free coaching session.

Find 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.

Companies asked in

This problem is commonly asked in Microsoft and Amazon interviews as it has multiple layers and there are some many things that can be tested 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 the 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(nlogn) complexity.

The problem with both searches is that they use additional space. Quick sort is another sorting algorithm. It has problem that it’s worst-case complexity will be O(n2), which happens when input is completely sorted.
In our case, the input is given as unsorted already, so we can expect that quicksort will function with O(n log n) complexity which is its average-case complexity. Advantage of using quicksort 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 the complexity of O(n log n), but practically, you do not need to sort the entire array before you find k smallest elements.

If you are preparing for a technical interview and need personal coaching along with mock interviews, book a free demo session with us

Algorithm to find Kth smallest element 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 the correct position.

kth 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

kth smallest element
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.


package com.company;

	* 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;
Test cases
package test;

import com.company.KthSmallest;
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 quicksort algorithm to find the kth smallest element in the array of integers is still O(n logn).

Kth smallest element using heaps

Before going into details of this problem, I strongly recommend reading heap fundamentals.

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 quicksort 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 a 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
input array

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

Create a max heap with set A

Step 2: Take the 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.

kth largest element
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. Remove the root and add 6 to max heap.

nth smallest number in an integer array
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 a 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(nlogk). 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 + klogn).

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

Please share if there is something wrong or missing. If you are interested in taking coaching sessions from our experienced teachers, please reach out to us at [email protected]