add algorithm for distinct els in a window

Author: ShyrenMore

Hashing/distinct_el_in_window.cpp

@@ -0,0 +1,70 @@
+/*
+Given array and a number k, k<=n
+find count of distinct elements in every window of size k
+
+arr = [10, 20, 20, 10, 30, 40, 10]  k=4
+op: 2 3 4 3
+
+arr = [10, 10, 10, 10]  k=3
+op: 1   1
+
+arr = [10, 20, 30, 40]  k=3
+op: 3   3
+*/
+
+/*
+sol1: O(n-k * k * k) time
+Consider starting point of every window 
+There are total n-k+1 windows 
+so traverse all windows 
+for each window, count distinct els 
+We can see if an element is distinct or not 
+by having two loops
+the outer loop picks and element one by one and the 
+inner loop checks if same element has ocuured before
+*/
+
+#include <iostream>
+using namespace std;
+
+void printDistinct(int arr[], int n, int k)
+{
+    for (int i = 0; i <= n - k; i++)
+    {
+        int count = 0;
+        for (int j = 0; j < k; j++)
+        {
+            bool flag = false;
+            for (int p = 0; p < j; p++)
+                if(arr[i+j] == arr[i+p])
+                {
+                    flag = true;
+                    break;
+                }
+        }
+
+        cout << count << '\t';
+    }
+}
+
+/*
+sol2: O(n-k + k) = O(n) time, O(k) space
+Maintain frequency map of elements present in first window
+print siz of freq map
+Then traverse the windows,
+compute freq map of curr window using freq map of prev window
+
+for(int i = k; i < n; i++)
+{
+    - decrease frequency of arr[i-k]
+    - if freq of arr[i-k] becomes 0
+        remove it from map
+
+    - if arr[i] is absent in map
+        insert it
+      else 
+        increase it's frequency in map
+
+    - print size of map   
+}
+*/