Find Kth Smallest or Largest element in an Array.

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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:

public class KthSmallestElement {
public int findkthSmallestElement(int[] arrA, int k) {
k = k – 1; // since array index starts with 0
return kSmall(arrA, 0, arrA.length – 1, k);
}
public int kSmall(int[] arrA, int start, int end, int k) {
int left = start;
int right = end;
int pivot = start;
while (left <= right) {
while (left <= right && arrA[left] <= arrA[pivot])
left++;
while (left <= right && arrA[right] >= arrA[pivot])
right—;
if (left < right) {
swap(arrA, left, right);
left++;
right—;
}
}
swap(arrA, pivot, right);
if (pivot == k)
return arrA[pivot];// if pivot is kth element , return
else if (pivot < k)
// if pivot is less than k, then kth element will be on the right
return kSmall(arrA, pivot + 1, end, k);
else
// if pivot is greater than k, then kth element will be on the right
return kSmall(arrA, start, pivot – 1, k);
}
public void swap(int[] arrA, int a, int b) {
int x = arrA[a];
arrA[a] = arrA[b];
arrA[b] = x;
}
public static void main(String args[]) {
int[] arrA = { 2, 3, 11, 16, 27, 4, 15, 9, 8 };
KthSmallestElement k = new KthSmallestElement();
int a = 4;
int x = k.findkthSmallestElement(arrA, a);
System.out.print("The " + a + "th smallest element is : " + x);
}
}

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KthSmallestElement.java
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Output:

The 4th smallest element is : 8

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