Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
arr = [2,3,4], the median is 3.arr = [2,3], the median is (2 + 3) / 2 = 2.5.Implement the MedianFinder class:
MedianFinder() initializes the MedianFinder object.void addNum(int num) adds the integer num from the data stream to the data structure.double findMedian() returns the median of all elements so far. Answers within 10-5 of the actual answer will be accepted.Example 1:
Input ["MedianFinder", "addNum", "addNum", "findMedian", "addNum", "findMedian"] [[], [1], [2], [], [3], []] Output [null, null, null, 1.5, null, 2.0] Explanation MedianFinder medianFinder = new MedianFinder(); medianFinder.addNum(1); // arr = [1] medianFinder.addNum(2); // arr = [1, 2] medianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2) medianFinder.addNum(3); // arr[1, 2, 3] medianFinder.findMedian(); // return 2.0
Constraints:
-105 <= num <= 105findMedian.5 * 104 calls will be made to addNum and findMedian.Follow up:
[0, 100], how would you optimize your solution?99% of all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?Problem summary: The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values. For example, for arr = [2,3,4], the median is 3. For example, for arr = [2,3], the median is (2 + 3) / 2 = 2.5. Implement the MedianFinder class: MedianFinder() initializes the MedianFinder object. void addNum(int num) adds the integer num from the data stream to the data structure. double findMedian() returns the median of all elements so far. Answers within 10-5 of the actual answer will be accepted.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers · Design
["MedianFinder","addNum","addNum","findMedian","addNum","findMedian"] [[],[1],[2],[],[3],[]]
sliding-window-median)finding-mk-average)sequentially-ordinal-rank-tracker)minimum-operations-to-make-median-of-array-equal-to-k)minimum-operations-to-make-subarray-elements-equal)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #295: Find Median from Data Stream
class MedianFinder {
private PriorityQueue<Integer> minQ = new PriorityQueue<>();
private PriorityQueue<Integer> maxQ = new PriorityQueue<>(Collections.reverseOrder());
public MedianFinder() {
}
public void addNum(int num) {
maxQ.offer(num);
minQ.offer(maxQ.poll());
if (minQ.size() - maxQ.size() > 1) {
maxQ.offer(minQ.poll());
}
}
public double findMedian() {
return minQ.size() == maxQ.size() ? (minQ.peek() + maxQ.peek()) / 2.0 : minQ.peek();
}
}
/**
* Your MedianFinder object will be instantiated and called as such:
* MedianFinder obj = new MedianFinder();
* obj.addNum(num);
* double param_2 = obj.findMedian();
*/
// Accepted solution for LeetCode #295: Find Median from Data Stream
type MedianFinder struct {
minq hp
maxq hp
}
func Constructor() MedianFinder {
return MedianFinder{hp{}, hp{}}
}
func (this *MedianFinder) AddNum(num int) {
minq, maxq := &this.minq, &this.maxq
heap.Push(maxq, -num)
heap.Push(minq, -heap.Pop(maxq).(int))
if minq.Len()-maxq.Len() > 1 {
heap.Push(maxq, -heap.Pop(minq).(int))
}
}
func (this *MedianFinder) FindMedian() float64 {
minq, maxq := this.minq, this.maxq
if minq.Len() == maxq.Len() {
return float64(minq.IntSlice[0]-maxq.IntSlice[0]) / 2
}
return float64(minq.IntSlice[0])
}
type hp struct{ sort.IntSlice }
func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}
/**
* Your MedianFinder object will be instantiated and called as such:
* obj := Constructor();
* obj.AddNum(num);
* param_2 := obj.FindMedian();
*/
# Accepted solution for LeetCode #295: Find Median from Data Stream
class MedianFinder:
def __init__(self):
self.minq = []
self.maxq = []
def addNum(self, num: int) -> None:
heappush(self.minq, -heappushpop(self.maxq, -num))
if len(self.minq) - len(self.maxq) > 1:
heappush(self.maxq, -heappop(self.minq))
def findMedian(self) -> float:
if len(self.minq) == len(self.maxq):
return (self.minq[0] - self.maxq[0]) / 2
return self.minq[0]
# Your MedianFinder object will be instantiated and called as such:
# obj = MedianFinder()
# obj.addNum(num)
# param_2 = obj.findMedian()
// Accepted solution for LeetCode #295: Find Median from Data Stream
use std::cmp::Reverse;
use std::collections::BinaryHeap;
struct MedianFinder {
minQ: BinaryHeap<Reverse<i32>>,
maxQ: BinaryHeap<i32>,
}
impl MedianFinder {
fn new() -> Self {
MedianFinder {
minQ: BinaryHeap::new(),
maxQ: BinaryHeap::new(),
}
}
fn add_num(&mut self, num: i32) {
self.maxQ.push(num);
self.minQ.push(Reverse(self.maxQ.pop().unwrap()));
if self.minQ.len() > self.maxQ.len() + 1 {
self.maxQ.push(self.minQ.pop().unwrap().0);
}
}
fn find_median(&self) -> f64 {
if self.minQ.len() == self.maxQ.len() {
let min_top = self.minQ.peek().unwrap().0;
let max_top = *self.maxQ.peek().unwrap();
(min_top + max_top) as f64 / 2.0
} else {
self.minQ.peek().unwrap().0 as f64
}
}
}
// Accepted solution for LeetCode #295: Find Median from Data Stream
class MedianFinder {
#minQ = new MinPriorityQueue<number>();
#maxQ = new MaxPriorityQueue<number>();
addNum(num: number): void {
const [minQ, maxQ] = [this.#minQ, this.#maxQ];
maxQ.enqueue(num);
minQ.enqueue(maxQ.dequeue());
if (minQ.size() - maxQ.size() > 1) {
maxQ.enqueue(minQ.dequeue());
}
}
findMedian(): number {
const [minQ, maxQ] = [this.#minQ, this.#maxQ];
if (minQ.size() === maxQ.size()) {
return (minQ.front() + maxQ.front()) / 2;
}
return minQ.front();
}
}
/**
* Your MedianFinder object will be instantiated and called as such:
* var obj = new MedianFinder()
* obj.addNum(num)
* var param_2 = obj.findMedian()
*/
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.