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.
Your algorithm should run in O(n) time and use O(log n) extra space.
Given an array of positive integers nums, the task is to return the smallest k values from the array. The result can be in any order.
nums
.k
representing the number of smallest values to return.k
values from nums
.Input: nums = [5, 9, 3, 6, 2, 1, 3, 2, 7, 5], k = 4 Output: [1, 2, 2, 3]
The core challenge is to find the smallest k
values in an array efficiently. This problem is significant in scenarios where we need to quickly identify the smallest elements, such as in statistical analysis, data processing, and real-time systems.
Potential pitfalls include misunderstanding the requirement for the algorithm to run in O(n) time and using more space than allowed.
To solve this problem, we need to think about efficient ways to find the smallest elements without sorting the entire array, which would take O(n log n) time.
A naive solution would be to sort the array and then take the first k
elements. However, this approach is not optimal as it runs in O(n log n) time.
We can use a Min-Heap to efficiently find the smallest k
elements. A Min-Heap allows us to extract the minimum element in O(log n) time. By building a Min-Heap of the entire array, we can then extract the smallest k
elements in O(n + k log n) time, which simplifies to O(n) for large n
and small k
.
Here is a step-by-step breakdown of the optimized algorithm:
nums
.k
times.import heapq
def smallest_k_integers(nums, k):
# Step 1: Build a Min-Heap from the array
heapq.heapify(nums)
# Step 2: Extract the smallest element k times
result = []
for _ in range(k):
result.append(heapq.heappop(nums))
return result
# Example usage
nums = [5, 9, 3, 6, 2, 1, 3, 2, 7, 5]
k = 4
print(smallest_k_integers(nums, k)) # Output: [1, 2, 2, 3]
The time complexity of building the Min-Heap is O(n). Extracting the smallest element k
times takes O(k log n) time. Therefore, the overall time complexity is O(n + k log n), which simplifies to O(n) for large n
and small k
.
The space complexity is O(log n) due to the space required for the heap operations.
Potential edge cases include:
k
being larger than the length of the array.For these cases, the algorithm should handle them gracefully, either by returning an empty array or the entire array if k
is larger than the array length.
To test the solution comprehensively, consider the following test cases:
Using a testing framework like unittest
in Python can help automate and validate these test cases.
When approaching such problems, consider the following tips:
In this blog post, we discussed how to find the smallest k
integers in 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 coding skills.
For further reading and practice, consider the following resources: