Approach Summary
Optimal solution for Kth Largest Element in a Stream with O(nlogk) time complexity and O(k) space complexity.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n^2) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(nlogk) | O(k) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n^2)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class KthLargest {
private int k;
private PriorityQueue<Integer> minHeap;
public KthLargest(int k, int[] nums) {
this.k = k;
minHeap = new PriorityQueue<>();
for(int num: nums){
minHeap.add(num);
}
while(minHeap.size() > k){
minHeap.poll();
}
}
public int add(int val) {
minHeap.add(val);
if(minHeap.size() > k){
minHeap.poll();
}
return minHeap.peek();
}
}class KthLargest {
private:
int k;
priority_queue<int, vector<int>, greater<int>> minHeap;
public:
KthLargest(int k, vector<int>& nums) {
this->k = k;
for(int num: nums){
minHeap.push(num);
}
while(minHeap.size() > k){
minHeap.pop();
}
}
int add(int val) {
minHeap.push(val);
if(minHeap.size() > k){
minHeap.pop();
}
return minHeap.top();
}
};class Solution:
def kth_largest_element_in_a_stream(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Kth Largest Element in a Stream
*/
function kthLargestElementInAStream(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(nlogk)
Space: O(k)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class KthLargest {
private int k;
private PriorityQueue<Integer> minHeap;
public KthLargest(int k, int[] nums) {
this.k = k;
minHeap = new PriorityQueue<>();
for(int num: nums){
minHeap.add(num);
}
while(minHeap.size() > k){
minHeap.poll();
}
}
public int add(int val) {
minHeap.add(val);
if(minHeap.size() > k){
minHeap.poll();
}
return minHeap.peek();
}
}class KthLargest {
private:
int k;
priority_queue<int, vector<int>, greater<int>> minHeap;
public:
KthLargest(int k, vector<int>& nums) {
this->k = k;
for(int num: nums){
minHeap.push(num);
}
while(minHeap.size() > k){
minHeap.pop();
}
}
int add(int val) {
minHeap.push(val);
if(minHeap.size() > k){
minHeap.pop();
}
return minHeap.top();
}
};class Solution:
def kth_largest_element_in_a_stream(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Kth Largest Element in a Stream
*/
function kthLargestElementInAStream(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Heap / Priority Queue Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Heap / Priority Queue pattern.
Framework: Heap / Priority Queue • Interview Prep: Tested at FAANG •
Explore all Heap / Priority Queue problems & invariants →
How to Recognize This Pattern
- Common Heap interview pattern
- Optimal time: O(nlogk)
Complexity Analysis
Time Complexity
O(nlogk)
Space Complexity
O(k)
Tags
Heap Priority Queue
Frequently Asked Questions
What is the optimal time and space complexity for Kth Largest Element in a Stream?
The optimal solution for Kth Largest Element in a Stream achieves O(nlogk) runtime complexity and requires O(k) auxiliary space. Time Complexity is O(nlogk) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(k) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Kth Largest Element in a Stream requires the Heap / Priority Queue pattern?
In technical interviews, look for key problem signals: Common Heap interview pattern; Optimal time: O(nlogk). Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Heap / Priority Queue strategy.
What critical edge cases must be handled in Kth Largest Element in a Stream?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Kth Largest Element in a Stream?
Kth Largest Element in a Stream is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Google, Microsoft.