Solved leetcode problem 295- Find median from data stream.md (Hard)

dsa-solutions/lc-solutions/0200-0299/295- Find median from data stream.md

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+---
+id: median-finder
+title: Median Finder
+sidebar_label: Median Finder
+tags:
+- Heap
+- Data Structure
+- C++
+- Java
+- Python
+description: "This document provides a solution to the Median Finder problem, where you need to efficiently find the median of a stream of numbers."
+---
+
+## Problem
+The MedianFinder class is designed to efficiently find the median of a stream of numbers. You can add numbers to the stream using the `addNum` method and retrieve the median using the `findMedian` method.
+
+## Solution
+The approach uses two heaps:
+- A max heap to store the smaller half of the numbers
+- A min heap to store the larger half of the numbers
+
+The median can be found in constant time by looking at the tops of the heaps.
+
+### Step-by-Step Explanation
+
+1. **Initialize two heaps**:
+   - `maxHeap` for the lower half of the data (inverted to act like a max heap using negative values).
+   - `minHeap` for the upper half of the data.
+
+2. **Add number**:
+   - If the number is less than or equal to the maximum of `maxHeap`, push it to `maxHeap`.
+   - Otherwise, push it to `minHeap`.
+   - Balance the heaps if necessary to ensure `maxHeap` always has equal or one more element than `minHeap`.
+
+3. **Find median**:
+   - If the heaps have equal sizes, the median is the average of the roots of both heaps.
+   - Otherwise, the median is the root of `maxHeap`.
+
+## Code in Different Languages
+
+<Tabs>
+<TabItem value="cpp" label="C++">
+  <SolutionAuthor name="@User"/>
+
+## C++
+```cpp
+#include <queue>
+#include <vector>
+
+class MedianFinder {
+ public:
+  void addNum(int num) {
+    if (maxHeap.empty() || num <= maxHeap.top())
+      maxHeap.push(num);
+    else
+      minHeap.push(num);
+
+    // Balance the two heaps so that
+    // |maxHeap| >= |minHeap| and |maxHeap| - |minHeap| <= 1.
+    if (maxHeap.size() < minHeap.size())
+      maxHeap.push(minHeap.top()), minHeap.pop();
+    else if (maxHeap.size() - minHeap.size() > 1)
+      minHeap.push(maxHeap.top()), maxHeap.pop();
+  }
+
+  double findMedian() {
+    if (maxHeap.size() == minHeap.size())
+      return (maxHeap.top() + minHeap.top()) / 2.0;
+    return maxHeap.top();
+  }
+
+ private:
+  std::priority_queue<int> maxHeap;
+  std::priority_queue<int, std::vector<int>, std::greater<int>> minHeap;
+};
+
+int main() {
+  MedianFinder mf;
+  mf.addNum(1);
+  mf.addNum(2);
+  std::cout << mf.findMedian() << std::endl; // Output: 1.5
+  mf.addNum(3);
+  std::cout << mf.findMedian() << std::endl; // Output: 2
+}
+```
+</TabItem>
+<TabItem value="java" label="Java">
+  <SolutionAuthor name="@User"/>
+
+## Java
+```java
+import java.util.Collections;
+import java.util.PriorityQueue;
+import java.util.Queue;
+
+public class MedianFinder {
+  private Queue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
+  private Queue<Integer> minHeap = new PriorityQueue<>();
+
+  public void addNum(int num) {
+    if (maxHeap.isEmpty() || num <= maxHeap.peek())
+      maxHeap.offer(num);
+    else
+      minHeap.offer(num);
+
+    // Balance the two heaps so that
+    // |maxHeap| >= |minHeap| and |maxHeap| - |minHeap| <= 1.
+    if (maxHeap.size() < minHeap.size())
+      maxHeap.offer(minHeap.poll());
+    else if (maxHeap.size() - minHeap.size() > 1)
+      minHeap.offer(maxHeap.poll());
+  }
+
+  public double findMedian() {
+    if (maxHeap.size() == minHeap.size())
+      return (double) (maxHeap.peek() + minHeap.peek()) / 2.0;
+    return (double) maxHeap.peek();
+  }
+
+  public static void main(String[] args) {
+    MedianFinder mf = new MedianFinder();
+    mf.addNum(1);
+    mf.addNum(2);
+    System.out.println(mf.findMedian()); // Output: 1.5
+    mf.addNum(3);
+    System.out.println(mf.findMedian()); // Output: 2
+  }
+}
+```
+
+</TabItem>
+<TabItem value="python" label="Python">
+  <SolutionAuthor name="@User"/>
+
+## Python
+```python
+import heapq
+
+class MedianFinder:
+  def __init__(self):
+    self.maxHeap = []
+    self.minHeap = []
+
+  def addNum(self, num: int) -> None:
+    if not self.maxHeap or num <= -self.maxHeap[0]:
+      heapq.heappush(self.maxHeap, -num)
+    else:
+      heapq.heappush(self.minHeap, num)
+
+    # Balance the two heaps so that
+    # |maxHeap| >= |minHeap| and |maxHeap| - |minHeap| <= 1.
+    if len(self.maxHeap) < len(self.minHeap):
+      heapq.heappush(self.maxHeap, -heapq.heappop(self.minHeap))
+    elif len(self.maxHeap) - len(self.minHeap) > 1:
+      heapq.heappush(self.minHeap, -heapq.heappop(self.maxHeap))
+
+  def findMedian(self) -> float:
+    if len(self.maxHeap) == len(self.minHeap):
+      return (-self.maxHeap[0] + self.minHeap[0]) / 2.0
+    return -self.maxHeap[0]
+
+# Example usage
+mf = MedianFinder()
+mf.addNum(1)
+mf.addNum(2)
+print(mf.findMedian())  # Output: 1.5
+mf.addNum(3)
+print(mf.findMedian())  # Output: 2
+```
+</TabItem>
+</Tabs>
+
+# Complexity Analysis
+## Time Complexity: $O(log N)$ for addNum, $O(1)$ for findMedian
+### Reason:
+Adding a number involves heap insertion which takes $O(log N)$ time. Finding the median involves looking at the top elements of the heaps, which takes $O(1)$ time.
+
+## Space Complexity: $O(N)$
+### Reason:
+We are storing all the numbers in the two heaps.