Smallest K in Java

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 = 4
Output: [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.

Note:

For this lesson, your algorithm should run in O(n log k) time and use O(k) extra space.
(There are faster solutions which we will discuss in future lessons)

Understanding the Problem

The core challenge of this problem is to efficiently find the smallest k integers from an array of positive integers. This problem is significant in scenarios where we need to filter out the smallest elements from a large dataset, such as finding the top k smallest scores in a competition or the least expensive items in a list.

Potential pitfalls include misunderstanding the requirement to return the smallest k values in any order and not optimizing the solution to meet the O(n log k) time complexity constraint.

Approach

To solve this problem, we can use a max-heap (priority queue) to keep track of the smallest k elements. The idea is to maintain a heap of size k. As we iterate through the array, we add elements to the heap. If the heap size exceeds k, we remove the largest element from the heap. This ensures that the heap always contains the k smallest elements.

Here is a step-by-step approach:

Initialize a max-heap (priority queue) to store the smallest k elements.
Iterate through each element in the array.
Add the current element to the heap.
If the heap size exceeds k, remove the largest element from the heap.
After processing all elements, the heap will contain the k smallest elements.

Algorithm

Let's break down the algorithm step-by-step:

Create a max-heap to store the smallest k elements.
For each element in the array:
- Add the element to the heap.
- If the heap size exceeds k, remove the largest element from the heap.
Convert the heap to a list and return it as the result.

Code Implementation

import java.util.PriorityQueue;
import java.util.Collections;
import java.util.List;
import java.util.ArrayList;

public class SmallestKIntegers {
    public static List<Integer> findSmallestK(int[] nums, int k) {
        // Max-heap to store the smallest k elements
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
        
        // Iterate through each element in the array
        for (int num : nums) {
            // Add the current element to the heap
            maxHeap.add(num);
            
            // If the heap size exceeds k, remove the largest element
            if (maxHeap.size() > k) {
                maxHeap.poll();
            }
        }
        
        // Convert the heap to a list and return it
        return new ArrayList<>(maxHeap);
    }

    public static void main(String[] args) {
        int[] nums = {5, 9, 3, 6, 2, 1, 3, 2, 7, 5};
        int k = 4;
        List<Integer> result = findSmallestK(nums, k);
        System.out.println(result); // Output can be [1, 2, 2, 3] in any order
    }
}

Complexity Analysis

The time complexity of this approach is O(n log k) because we perform log k operations for each of the n elements in the array. The space complexity is O(k) because we store at most k elements in the heap.

Edge Cases

Potential edge cases include:

k is greater than the length of the array: In this case, the result should be the entire array.
k is 0: The result should be an empty list.
All elements in the array are the same: The result should contain k instances of that element.

Testing

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

nums = [5, 9, 3, 6, 2, 1, 3, 2, 7, 5], k = 4 (Expected output: [1, 2, 2, 3])
nums = [1, 2, 3, 4, 5], k = 3 (Expected output: [1, 2, 3])
nums = [5, 5, 5, 5, 5], k = 2 (Expected output: [5, 5])
nums = [10, 20, 30], k = 0 (Expected output: [])
nums = [1, 2, 3], k = 5 (Expected output: [1, 2, 3])

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Understand the problem requirements and constraints thoroughly.
Think about different data structures and their properties to find the most efficient solution.
Break down the problem into smaller steps and solve each step incrementally.
Practice similar problems to improve problem-solving skills and familiarity with different algorithms.

Conclusion

In this blog post, we discussed how to find the smallest k integers from an array using a max-heap to achieve O(n log k) time complexity. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving algorithmic thinking and problem-solving skills.

Additional Resources

For further reading and practice, consider the following resources:

Smallest K Integers II in Java with O(n log k) Time Complexity