Find Kth smallest element in binary search tree

Kth smallest element in a binary a search tree

Given a binary search tree, find kth smallest element in the binary search tree. For example, 5th smallest element in below binary search tree would be 14, if store the tree in sorted order 5,7,9,10,14,15,19; 14 is the fifth smallest element in that order.

kth smallest element in binary search tree
Kth smallest element in binary search tree

Kth smallest element in binary search tree: thoughts

As mentioned earlier in a lot of posts like delete a binary tree or mirror a binary tree, first try to find the traversal order required to solve this problem. One hint we already got is that we want all the nodes on BST traversed in sorted order. What kind of traversal gives us a sorted order of nodes? Of course, it is inorder traversal.

So idea is to do an inorder traversal of the binary search tree and store all the nodes in an array. Once traversal is finished, find the kth smallest element in the sorted array.

This approach, however, scans the entire tree and also has space complexity of O(n) because we store all the nodes of tree in an array. Can we avoid scanning the whole tree and storing nodes?

If we keep count of how many nodes are traversed during inorder traversal, we can actually stop traversal as soon as we see k nodes are visited. In this case, we do not store nodes, just a counter. 

Kth smallest element in binary tree: implementation

package com.company.BST;

import java.util.Stack;

/**
 * Created by sangar on 9.11.18.
 */
public class FindKthSmallestInBST {
    private int counter;

    public FindKthSmallestInBST(){
        counter = 0;
    }
    public TreeNode findKthSmallest(TreeNode root, int k){
        if(root == null) return root;

        //Traverse left subtree first
        TreeNode left = findKthSmallest(root.getLeft(),k);
        //If we found kth node on left subtree
        if(left != null) return left;
        //If k becomes zero, that means we have traversed k nodes.
        if(++counter == k) return root;

        return findKthSmallest(root.getRight(),k);
    }
}

Test cases

package test;

import com.company.BST.BinarySearchTree;
import com.company.BST.FindKthSmallestInBST;
import com.company.BST.TreeNode;
import org.junit.jupiter.api.Test;

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

/**
 * Created by sangar on 23.9.18.
 */
public class KthSmallestElementTest {

    FindKthSmallestInBST tester = new FindKthSmallestInBST();

    @Test
    public void kthSmallestElementTest() {

        BinarySearchTree<Integer> binarySearchTree =
			new BinarySearchTree<>();
        binarySearchTree.insert(10);
        binarySearchTree.insert(8);
        binarySearchTree.insert(15);
        binarySearchTree.insert(12);
        binarySearchTree.insert(6);
        binarySearchTree.insert(9);

        TreeNode kthNode =
			tester.findKthSmallest(binarySearchTree.getRoot(),1);
        assertEquals(6, kthNode.getValue());
    }

    @Test
    public void kthSmallestElementOnRightSubtreeTest() {

        BinarySearchTree<Integer> binarySearchTree =
				new BinarySearchTree<>();
        binarySearchTree.insert(10);
        binarySearchTree.insert(8);
        binarySearchTree.insert(15);
        binarySearchTree.insert(12);
        binarySearchTree.insert(6);
        binarySearchTree.insert(9);

        TreeNode kthNode =
			tester.findKthSmallest(binarySearchTree.getRoot(),5);
        assertEquals(12, kthNode.getValue());
    }

    @Test
    public void kthSmallestElementAbsentSubtreeTest() {

        BinarySearchTree<Integer> binarySearchTree =
				new BinarySearchTree<>();
        binarySearchTree.insert(10);
        binarySearchTree.insert(8);
        binarySearchTree.insert(15);
        binarySearchTree.insert(12);
        binarySearchTree.insert(6);
        binarySearchTree.insert(9);

        TreeNode kthNode =
			tester.findKthSmallest(binarySearchTree.getRoot(),10);
        assertEquals(null, kthNode);
    }

    @Test
    public void kthSmallestElementNulltreeTest() {

        BinarySearchTree<Integer> binarySearchTree =
				new BinarySearchTree<>();
        binarySearchTree.insert(10);
        binarySearchTree.insert(8);
        binarySearchTree.insert(15);
        binarySearchTree.insert(12);
        binarySearchTree.insert(6);
        binarySearchTree.insert(9);

        TreeNode kthNode = tester.findKthSmallest(null,10);
        assertEquals(null, kthNode);
    }
}

The complexity of this algorithm to find kth smallest element is O(k) as we traverse only k nodes on binary search tree.

There is hidden space complexity here. Recursive function requires call stack memory, which is limited to Operation System default. More deep you go in recursion, more space we use on stack. If tree is completely skewed, there are more chances of stack overflow. Also recursive function is very difficult to debug in production environments. Below is the non-recursive solution for the same problem.

Non-recursive way to find kth smallest element in BST

public int kthSmallest(TreeNode root, int k) {
        Stack<TreeNode> s = new Stack<TreeNode>();

        TreeNode current = root;
        int result = 0;

        while(!s.isEmpty() || current!=null){
            if(current!=null){
                s.push(current);
                current = current.getLeft();
            }else{
                TreeNode  temp = s.pop();
                k--;
                if(k==0)
                    result = (int)temp.getValue();
                current = temp.getRight();
            }
        }

        return result;
    }

Please share if there is something wrong or missing. If you are preparing for an interview and need help, please reach out to us at communications@algorithmsandme.com

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

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;
	}
}
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};
	@Test
	public void kthSmallest() {
		assertEquals(7, tester.findKthSmallestElement(a,0,8,6));
	}

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

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

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

	@Test
	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){
	//https://stackoverflow.com/questions/11003155/change-priorityqueue-to-max-priorityqueue

	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++){
			maxHeap.add(a[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]){
				maxHeap.remove();
				maxHeap.add(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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