8.3 Top-K Problem¶
Question
Given an unordered array nums of length \(n\), return the largest \(k\) elements in the array.
For this problem, we'll first introduce two solutions with relatively straightforward approaches, then introduce a more efficient heap-based solution.
8.3.1 Method 1: Iterative Selection¶
We can perform \(k\) rounds of traversal as shown in Figure 8-6, extracting the \(1^{st}\), \(2^{nd}\), \(\dots\), \(k^{th}\) largest elements in each round, with a time complexity of \(O(nk)\).
This method is only suitable when \(k \ll n\), because when \(k\) is close to \(n\), the time complexity approaches \(O(n^2)\), which is very time-consuming.
Figure 8-6 Traversing to find the largest k elements
Tip
When \(k = n\), we can obtain a complete sorted sequence, which is equivalent to the "selection sort" algorithm.
8.3.2 Method 2: Sorting¶
As shown in Figure 8-7, we can first sort the array nums, then return the rightmost \(k\) elements, with a time complexity of \(O(n \log n)\).
Clearly, this method "overachieves" the task, as we only need to find the largest \(k\) elements, without needing to sort the other elements.
Figure 8-7 Sorting to find the largest k elements
8.3.3 Method 3: Heap¶
We can solve the Top-k problem more efficiently using heaps, with the process shown in Figure 8-8.
- Initialize a min heap, where the heap top element is the smallest.
- First, insert the first \(k\) elements of the array into the heap in sequence.
- Starting from the \((k + 1)^{th}\) element, if the current element is greater than the heap top element, remove the heap top element and insert the current element into the heap.
- After traversal is complete, the heap contains the largest \(k\) elements.
Figure 8-8 Finding the largest k elements using a heap
Example code is as follows:
top_k.py
def top_k_heap(nums: list[int], k: int) -> list[int]:
"""Find the largest k elements in array based on heap"""
# Initialize min heap
heap = []
# Enter the first k elements of array into heap
for i in range(k):
heapq.heappush(heap, nums[i])
# Starting from the (k+1)th element, maintain heap length as k
for i in range(k, len(nums)):
# If current element is greater than top element, top element exits heap, current element enters heap
if nums[i] > heap[0]:
heapq.heappop(heap)
heapq.heappush(heap, nums[i])
return heap
top_k.cpp
/* Find the largest k elements in array based on heap */
priority_queue<int, vector<int>, greater<int>> topKHeap(vector<int> &nums, int k) {
// Python's heapq module implements min heap by default
priority_queue<int, vector<int>, greater<int>> heap;
// Enter the first k elements of array into heap
for (int i = 0; i < k; i++) {
heap.push(nums[i]);
}
// Starting from the (k+1)th element, maintain heap length as k
for (int i = k; i < nums.size(); i++) {
// If current element is greater than top element, top element exits heap, current element enters heap
if (nums[i] > heap.top()) {
heap.pop();
heap.push(nums[i]);
}
}
return heap;
}
top_k.java
/* Find the largest k elements in array based on heap */
Queue<Integer> topKHeap(int[] nums, int k) {
// Python's heapq module implements min heap by default
Queue<Integer> heap = new PriorityQueue<Integer>();
// Enter the first k elements of array into heap
for (int i = 0; i < k; i++) {
heap.offer(nums[i]);
}
// Starting from the (k+1)th element, maintain heap length as k
for (int i = k; i < nums.length; i++) {
// If current element is greater than top element, top element exits heap, current element enters heap
if (nums[i] > heap.peek()) {
heap.poll();
heap.offer(nums[i]);
}
}
return heap;
}
top_k.cs
/* Find the largest k elements in array based on heap */
PriorityQueue<int, int> TopKHeap(int[] nums, int k) {
// Python's heapq module implements min heap by default
PriorityQueue<int, int> heap = new();
// Enter the first k elements of array into heap
for (int i = 0; i < k; i++) {
heap.Enqueue(nums[i], nums[i]);
}
// Starting from the (k+1)th element, maintain heap length as k
for (int i = k; i < nums.Length; i++) {
// If current element is greater than top element, top element exits heap, current element enters heap
if (nums[i] > heap.Peek()) {
heap.Dequeue();
heap.Enqueue(nums[i], nums[i]);
}
}
return heap;
}
top_k.go
/* Find the largest k elements in array based on heap */
func topKHeap(nums []int, k int) *minHeap {
// Python's heapq module implements min heap by default
h := &minHeap{}
heap.Init(h)
// Enter the first k elements of array into heap
for i := 0; i < k; i++ {
heap.Push(h, nums[i])
}
// Starting from the (k+1)th element, maintain heap length as k
for i := k; i < len(nums); i++ {
// If current element is greater than top element, top element exits heap, current element enters heap
if nums[i] > h.Top().(int) {
heap.Pop(h)
heap.Push(h, nums[i])
}
}
return h
}
top_k.swift
/* Find the largest k elements in array based on heap */
func topKHeap(nums: [Int], k: Int) -> [Int] {
// Initialize min heap and build heap with first k elements
var heap = Heap(nums.prefix(k))
// Starting from the (k+1)th element, maintain heap length as k
for i in nums.indices.dropFirst(k) {
// If current element is greater than top element, top element exits heap, current element enters heap
if nums[i] > heap.min()! {
_ = heap.removeMin()
heap.insert(nums[i])
}
}
return heap.unordered
}
top_k.js
/* Element enters heap */
function pushMinHeap(maxHeap, val) {
// Negate element
maxHeap.push(-val);
}
/* Element exits heap */
function popMinHeap(maxHeap) {
// Negate element
return -maxHeap.pop();
}
/* Access top element */
function peekMinHeap(maxHeap) {
// Negate element
return -maxHeap.peek();
}
/* Extract elements from heap */
function getMinHeap(maxHeap) {
// Negate element
return maxHeap.getMaxHeap().map((num) => -num);
}
/* Find the largest k elements in array based on heap */
function topKHeap(nums, k) {
// Python's heapq module implements min heap by default
// Note: We negate all heap elements to simulate min heap using max heap
const maxHeap = new MaxHeap([]);
// Enter the first k elements of array into heap
for (let i = 0; i < k; i++) {
pushMinHeap(maxHeap, nums[i]);
}
// Starting from the (k+1)th element, maintain heap length as k
for (let i = k; i < nums.length; i++) {
// If current element is greater than top element, top element exits heap, current element enters heap
if (nums[i] > peekMinHeap(maxHeap)) {
popMinHeap(maxHeap);
pushMinHeap(maxHeap, nums[i]);
}
}
// Return elements in heap
return getMinHeap(maxHeap);
}
top_k.ts
/* Element enters heap */
function pushMinHeap(maxHeap: MaxHeap, val: number): void {
// Negate element
maxHeap.push(-val);
}
/* Element exits heap */
function popMinHeap(maxHeap: MaxHeap): number {
// Negate element
return -maxHeap.pop();
}
/* Access top element */
function peekMinHeap(maxHeap: MaxHeap): number {
// Negate element
return -maxHeap.peek();
}
/* Extract elements from heap */
function getMinHeap(maxHeap: MaxHeap): number[] {
// Negate element
return maxHeap.getMaxHeap().map((num: number) => -num);
}
/* Find the largest k elements in array based on heap */
function topKHeap(nums: number[], k: number): number[] {
// Python's heapq module implements min heap by default
// Note: We negate all heap elements to simulate min heap using max heap
const maxHeap = new MaxHeap([]);
// Enter the first k elements of array into heap
for (let i = 0; i < k; i++) {
pushMinHeap(maxHeap, nums[i]);
}
// Starting from the (k+1)th element, maintain heap length as k
for (let i = k; i < nums.length; i++) {
// If current element is greater than top element, top element exits heap, current element enters heap
if (nums[i] > peekMinHeap(maxHeap)) {
popMinHeap(maxHeap);
pushMinHeap(maxHeap, nums[i]);
}
}
// Return elements in heap
return getMinHeap(maxHeap);
}
top_k.dart
/* Find the largest k elements in array based on heap */
MinHeap topKHeap(List<int> nums, int k) {
// Initialize min heap, push first k elements of array to heap
MinHeap heap = MinHeap(nums.sublist(0, k));
// Starting from the (k+1)th element, maintain heap length as k
for (int i = k; i < nums.length; i++) {
// If current element is greater than top element, top element exits heap, current element enters heap
if (nums[i] > heap.peek()) {
heap.pop();
heap.push(nums[i]);
}
}
return heap;
}
top_k.rs
/* Find the largest k elements in array based on heap */
fn top_k_heap(nums: Vec<i32>, k: usize) -> BinaryHeap<Reverse<i32>> {
// BinaryHeap is a max heap, use Reverse to negate elements to implement min heap
let mut heap = BinaryHeap::<Reverse<i32>>::new();
// Enter the first k elements of array into heap
for &num in nums.iter().take(k) {
heap.push(Reverse(num));
}
// Starting from the (k+1)th element, maintain heap length as k
for &num in nums.iter().skip(k) {
// If current element is greater than top element, top element exits heap, current element enters heap
if num > heap.peek().unwrap().0 {
heap.pop();
heap.push(Reverse(num));
}
}
heap
}
top_k.c
/* Element enters heap */
void pushMinHeap(MaxHeap *maxHeap, int val) {
// Negate element
push(maxHeap, -val);
}
/* Element exits heap */
int popMinHeap(MaxHeap *maxHeap) {
// Negate element
return -pop(maxHeap);
}
/* Access top element */
int peekMinHeap(MaxHeap *maxHeap) {
// Negate element
return -peek(maxHeap);
}
/* Extract elements from heap */
int *getMinHeap(MaxHeap *maxHeap) {
// Negate all heap elements and store in res array
int *res = (int *)malloc(maxHeap->size * sizeof(int));
for (int i = 0; i < maxHeap->size; i++) {
res[i] = -maxHeap->data[i];
}
return res;
}
/* Extract elements from heap */
int *getMinHeap(MaxHeap *maxHeap) {
// Negate all heap elements and store in res array
int *res = (int *)malloc(maxHeap->size * sizeof(int));
for (int i = 0; i < maxHeap->size; i++) {
res[i] = -maxHeap->data[i];
}
return res;
}
// Function to find k largest elements in array using heap
int *topKHeap(int *nums, int sizeNums, int k) {
// Python's heapq module implements min heap by default
// Note: We negate all heap elements to simulate min heap using max heap
int *empty = (int *)malloc(0);
MaxHeap *maxHeap = newMaxHeap(empty, 0);
// Enter the first k elements of array into heap
for (int i = 0; i < k; i++) {
pushMinHeap(maxHeap, nums[i]);
}
// Starting from the (k+1)th element, maintain heap length as k
for (int i = k; i < sizeNums; i++) {
// If current element is greater than top element, top element exits heap, current element enters heap
if (nums[i] > peekMinHeap(maxHeap)) {
popMinHeap(maxHeap);
pushMinHeap(maxHeap, nums[i]);
}
}
int *res = getMinHeap(maxHeap);
// Free memory
delMaxHeap(maxHeap);
return res;
}
top_k.kt
/* Find the largest k elements in array based on heap */
fun topKHeap(nums: IntArray, k: Int): Queue<Int> {
// Python's heapq module implements min heap by default
val heap = PriorityQueue<Int>()
// Enter the first k elements of array into heap
for (i in 0..<k) {
heap.offer(nums[i])
}
// Starting from the (k+1)th element, maintain heap length as k
for (i in k..<nums.size) {
// If current element is greater than top element, top element exits heap, current element enters heap
if (nums[i] > heap.peek()) {
heap.poll()
heap.offer(nums[i])
}
}
return heap
}
top_k.rb
### Find largest k elements in array using heap ###
def top_k_heap(nums, k)
# Python's heapq module implements min heap by default
# Note: We negate all heap elements to simulate min heap using max heap
max_heap = MaxHeap.new([])
# Enter the first k elements of array into heap
for i in 0...k
push_min_heap(max_heap, nums[i])
end
# Starting from the (k+1)th element, maintain heap length as k
for i in k...nums.length
# If current element is greater than top element, top element exits heap, current element enters heap
if nums[i] > peek_min_heap(max_heap)
pop_min_heap(max_heap)
push_min_heap(max_heap, nums[i])
end
end
get_min_heap(max_heap)
end
A total of \(n\) rounds of heap insertions and removals are performed, with the heap's maximum length being \(k\), so the time complexity is \(O(n \log k)\). This method is very efficient; when \(k\) is small, the time complexity approaches \(O(n)\); when \(k\) is large, the time complexity does not exceed \(O(n \log n)\).
Additionally, this method is suitable for dynamic data stream scenarios. By continuously adding data, we can maintain the elements in the heap, thus achieving dynamic updates of the largest \(k\) elements.