#### Move all **elements** larger than the pivot near the end of the current segment 4. Place the pivot on the remaining position in between 5. enter no. of nodes 5 enter the node values 8 3 10 1 6 enter value of k 3 the **kth smallest element** is : 6. We follow below steps to find **kth smallest element** using MIN heap: 1. Construct MIN Heap. 2. Then remove the root **element** k times to get the **kth smallest element**. 3. For this we can use priority queue from C++ STD lib. 4. Given an array and a positive integer k, write a program to find the **kth smallest element** in the array. This is an excellent. For the sake of example, our input would be. var input = [100, 300, 2, 3, 1000, 67, 89, 10, 764, 1, 546] At the start, input will be our input array, with low = 0 and high = 10 and value of k = 3 that is we want to find **kth smallest element**. Please note that this is an in-place algorithm. Which means as it finds the **element**, it also reorders.

**elements**larger than the pivot near the end of the current segment 4. Place the pivot on the remaining position in between 5. enter no. of nodes 5 enter the node values 8 3 10 1 6 enter value of k 3 the

**kth smallest element**is : 6.

**Kth Smallest**in an Unsorted Array in Java. In the given array, the task is to find the

**kth smallest element**of the array, where k is always less than the size of the given array. Examples: Input: arr[] = {56, 34, 7, 9, 0, 48, 41, 8} k = 3. Output: The 3 rd

**smallest element**of the array is 8. Given an array arr[ ] and a number K where K is smaller than the size of the array, the task is to find the

**Kth smallest element**in the given array. It is given that all array

**elements**are distinct. See original problem statement here. Solution Approach : Introduction : Idea is to create a min-heap then extract the first (k-1)

**elements**.