**Objective: **Find Kth Smallest or Largest element in an Array

**Example:**

int[] arrA = { 2, 3, 11, 16, 27, 4, 15, 9, 8 }; Output: The 4th smallest element is : 8

**Approach:**

**Naive Approach: Use Sorting**

Sort the array and return kth element. Time Complexity – O(nlogn).

**Better Approach: Use Quick Sort Technique **

*Here we are finding the kth smallest element in array. Same approach you can apply to find the kth largest element.*

- In this technique we select a pivot element and after a one round of operation the pivot element takes its correct place in the array.
- Once that is done we check if the pivot element is the kth element in array, if yes then return it.
- But if pivot element is less than the kth element, that clearly means that the kth element is on the right side of the pivot. So make a recursive call from pivot+1 to end.
- Similarly if pivot element is greater than the kth element, that clearly means that the kth element is on the left side of the pivot. So make a recursive call from start to pivot-1.

**Average Time Complexity : O(n)**

**NOTE: **Instead of printing the kth element, you can print all elements from index 0 to k and this will the solution of the problem** “ Print First K smallest or largest elements in an array”.**

**Complete Code:**

**Output:**

The 4th smallest element is : 8

after line swap(arrA, pivot, right), element at index ‘right’ will be at correct position not element at pivot index.

Code need little modification as mentioned below.

if (right == k)

return arrA[right];

else if (right < k)

return kSmall(arrA, right + 1, end, k);

else

return kSmall(arrA, start, right – 1, k);