Given an array of positive integers **nums**, return the smallest **k** values, in any order you want.

**Example:**

Input:nums = [5, 9, 3, 6, 2, 1, 3, 2, 7, 5], k = 4Output:[1, 2, 2, 3]Explanation:Smallest number is 1, 2nd smallest is 2, 3rd smallest is 2, 4th smallest is 3

The result can be in any order, [2, 1, 3, 2] is also a correct answer.

For this lesson, your algorithm should run in **O(n log n)** time and use **O(1)** extra space.

(There are faster solutions which we will discuss in future lessons)

The core challenge of this problem is to find the smallest k integers from an array of positive integers. This problem is significant in various applications such as data analysis, where you might need to find the smallest values in a dataset. A common pitfall is to use a naive approach that does not meet the time complexity requirements.

To solve this problem, we can consider the following approaches:

A naive solution would be to sort the entire array and then return the first k elements. While this approach is simple, it is not optimal because it sorts the entire array, which is unnecessary.

An optimized solution involves using a sorting algorithm that runs in O(n log n) time. We can sort the array and then return the first k elements. This approach meets the time complexity requirement and is straightforward to implement.

Here is a step-by-step breakdown of the optimized algorithm:

- Sort the array using a sorting algorithm like QuickSort or MergeSort, which runs in O(n log n) time.
- Return the first k elements of the sorted array.

```
import java.util.Arrays;
public class SmallestKIntegers {
public static int[] findSmallestK(int[] nums, int k) {
// Step 1: Sort the array
Arrays.sort(nums);
// Step 2: Create a result array to store the smallest k elements
int[] result = new int[k];
// Step 3: Copy the first k elements from the sorted array to the result array
for (int i = 0; i < k; i++) {
result[i] = nums[i];
}
return result;
}
public static void main(String[] args) {
int[] nums = {5, 9, 3, 6, 2, 1, 3, 2, 7, 5};
int k = 4;
int[] smallestK = findSmallestK(nums, k);
System.out.println(Arrays.toString(smallestK)); // Output: [1, 2, 2, 3]
}
}
```

The time complexity of this approach is O(n log n) due to the sorting step. The space complexity is O(1) extra space, as we are not using any additional data structures that grow with the input size.

Potential edge cases include:

- k is greater than the length of the array: This should be handled by either returning the entire array or throwing an error.
- k is zero: The function should return an empty array.
- Array contains duplicate values: The function should still return the smallest k values, including duplicates.

To test the solution comprehensively, consider the following test cases:

- Simple case: nums = [5, 9, 3, 6, 2, 1, 3, 2, 7, 5], k = 4
- Edge case: nums = [1, 2, 3], k = 5 (k greater than array length)
- Edge case: nums = [1, 2, 3], k = 0 (k is zero)
- Array with duplicates: nums = [1, 1, 1, 1], k = 2

When approaching such problems, consider the following tips:

- Understand the problem constraints and requirements.
- Think about the most efficient way to solve the problem within the given constraints.
- Break down the problem into smaller steps and solve each step methodically.
- Practice solving similar problems to improve your problem-solving skills.

In this blog post, we discussed how to find the smallest k integers from an array of positive integers using an optimized approach that runs in O(n log n) time. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving your algorithmic thinking and coding skills.

For further reading and practice, consider the following resources:

- LeetCode - Practice coding problems
- GeeksforGeeks - Tutorials and coding challenges
- Coursera - Online courses on algorithms and data structures