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.
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)
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 in data analysis, statistics, and competitive programming.
Potential pitfalls include misunderstanding the requirement to return the smallest k values in any order and not optimizing the solution to meet the required time and space complexity.
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 exceeds size k, we remove the largest element (the root of the max-heap). This ensures that the heap always contains the k smallest elements.
Here is a step-by-step approach:
Let's break down the algorithm step-by-step:
#include <iostream>
#include <vector>
#include <queue>
#include <functional>
std::vector<int> smallestKIntegers(const std::vector<int>& nums, int k) {
// Max-heap to store the smallest k elements
std::priority_queue<int> maxHeap;
// Iterate through each element in the array
for (int num : nums) {
// Add the current element to the heap
maxHeap.push(num);
// If the heap size exceeds k, remove the largest element
if (maxHeap.size() > k) {
maxHeap.pop();
}
}
// Extract the elements from the heap
std::vector<int> result;
while (!maxHeap.empty()) {
result.push_back(maxHeap.top());
maxHeap.pop();
}
return result;
}
int main() {
std::vector<int> nums = {5, 9, 3, 6, 2, 1, 3, 2, 7, 5};
int k = 4;
std::vector<int> result = smallestKIntegers(nums, k);
std::cout << "The smallest " << k << " integers are: ";
for (int num : result) {
std::cout << num << " ";
}
std::cout << std::endl;
return 0;
}
The time complexity of this approach is O(n log k), where n is the number of elements in the array. This is because we perform log k operations for each of the n elements when maintaining the heap of size k. The space complexity is O(k) since we store only k elements in the heap.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
When approaching such problems, it's essential to:
In this blog post, we discussed how to find the smallest k integers from an array using a max-heap. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing efficient algorithms and improving problem-solving skills.
For further reading and practice, consider the following resources: