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 maxVal) time and use O(1) extra space.
The core challenge of this problem is to find the smallest k integers from an array of positive integers efficiently. This problem is significant in scenarios where we need to quickly identify the smallest elements, such as in data analysis, statistics, and competitive programming. A common pitfall is to use sorting, which would result in a time complexity of O(n log n), not meeting the required O(n log maxVal) complexity.
To solve this problem, we need to think beyond simple sorting. Here are a few approaches:
The naive solution would be to sort the array and then pick the first k elements. However, this approach has a time complexity of O(n log n), which is not optimal for this problem.
We can use a min-heap (priority queue) to keep track of the smallest k elements. This approach ensures that we only need to perform log(maxVal) operations for each element, resulting in the desired O(n log maxVal) time complexity.
Here is a step-by-step breakdown of the optimized algorithm:
#include <iostream>
#include <vector>
#include <queue>
#include <functional>
std::vector<int> smallestKIntegers(const std::vector<int>& nums, int k) {
// Min-heap to store the smallest k elements
std::priority_queue<int, std::vector<int>, std::greater<int>> minHeap;
// Iterate through each element in the array
for (int num : nums) {
minHeap.push(num); // Push the current element into the min-heap
if (minHeap.size() > k) {
minHeap.pop(); // Remove the largest element if size exceeds k
}
}
// Extract the smallest k elements from the min-heap
std::vector<int> result;
while (!minHeap.empty()) {
result.push_back(minHeap.top());
minHeap.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 << "Smallest " << k << " integers: ";
for (int num : result) {
std::cout << num << " ";
}
std::cout << std::endl;
return 0;
}
The time complexity of this approach is O(n log k) because each insertion and deletion operation in the min-heap takes O(log k) time, and we perform these operations n times. The space complexity is O(k) for storing the k smallest elements in the min-heap.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
When approaching such problems, consider the following tips:
In this blog post, we discussed how to find the smallest k integers from an array efficiently using a min-heap. 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.
For further reading and practice, consider the following resources: