algo to print continuous medians in a stream of integers

MAANG-DSA-Prep/medianOfStreamofIntegers.cpp

@@ -0,0 +1,56 @@
+#include <bits/stdc++.h>
+using namespace std;
+// TC:O(nLogn)
+// SC:O(n)
+
+void continuosMedian(int arr[], int n)
+{
+    float median;
+    // stores the left half of the list but in a max heap to get the greatest value from it
+    priority_queue<int> maxheap;
+    // stores the right half of the list but in a min heap to get the smallest value from it
+    priority_queue<int, vector<int>, greater<int>> minheap;
+    for (int i = 0; i < n; i++)
+    {
+        // Step 1:input elements in the maxHeap
+        int current = arr[i];
+        if (maxheap.empty() || maxheap.top() > current)
+            maxheap.push(current);
+        else
+            minheap.push(current);
+
+        // Step 2:maintain the difference <=1
+        if (minheap.size() > maxheap.size() + 1)
+        {
+            maxheap.push(minheap.top());
+            minheap.pop();
+        }
+        else
+        {
+            minheap.push(maxheap.top());
+            maxheap.pop();
+        }
+
+        // Step 3:decide for the way to find median based on following conditions
+        if (n % 2 == 1) // i.e elements are odd
+        {
+            if (minheap.size() > maxheap.size())
+                median = minheap.top();
+            else
+                median = maxheap.top();
+        }
+        else
+        {
+            median = (maxheap.top() + minheap.top()) / 2;
+        }
+        cout << median << " ";
+    }
+}
+int main()
+{
+    int arr[] = {5, 15, 1, 3, 2, 8, 7, 9, 10, 6, 11, 4};
+    int n = sizeof(arr) / sizeof(arr[0]);
+    cout << "Current Median at every stage:";
+    continuosMedian(arr, n);
+    return 0;
+}