From d8c35ba288fbbf6006cf281f10d00d6f67d1c954 Mon Sep 17 00:00:00 2001
From: KSM <mehrok.kabir.singh@gmail.com>
Date: Mon, 24 Oct 2022 10:26:04 +0530
Subject: [PATCH 1/2] Add least-number-of-unique-integers-after-k-removals

---
 ...r-of-unique-integers-after-k-removals.java | 38 +++++++++++++++++++
 1 file changed, 38 insertions(+)
 create mode 100644 Hacktoberfest2022/least-number-of-unique-integers-after-k-removals.java

diff --git a/Hacktoberfest2022/least-number-of-unique-integers-after-k-removals.java b/Hacktoberfest2022/least-number-of-unique-integers-after-k-removals.java
new file mode 100644
index 0000000..132dba8
--- /dev/null
+++ b/Hacktoberfest2022/least-number-of-unique-integers-after-k-removals.java
@@ -0,0 +1,38 @@
+/*
+Author: Kabir Singh Mehrok <https://github.com/KabirSinghMehrok>
+
+Title: Least Number of Unique Integers after K Removals
+Description: Given an array of integers arr and an integer k. Find the least number of unique integers after removing exactly k elements.
+*/
+
+class Solution {
+    public int findLeastNumOfUniqueInts(int[] arr, int k) {
+        int N = arr.length;
+        HashMap<Integer, Integer> h = new HashMap<Integer, Integer>();
+        for (int i=0; i<N; i++) h.put(arr[i], h.getOrDefault(arr[i], 0) + 1);
+        
+        // now make a set of values into arraylist
+        // sort it in ascending order
+        // sum values
+        // stop when >= k
+        
+        ArrayList<Integer> a = new ArrayList<Integer>(h.values());
+        Collections.sort(a);
+        
+        int sum = 0;
+        int i = 0;
+        for (i=0; i<a.size(); i++)
+        {
+            if (sum >= k) break;
+            sum += a.get(i);
+        }
+        
+        // now there can be two cases
+        // if sum > k, it implies that we have taken an extra element into account
+        // if sum == k, it implies we have exact amount of element
+        
+        if (sum > k) return a.size() - i + 1;
+        if (sum == k) return a.size() - i;
+        else return 0;
+    }
+}
\ No newline at end of file

From 6b62f86cf306a93efebff9f40aab195d22e7f186 Mon Sep 17 00:00:00 2001
From: KSM <mehrok.kabir.singh@gmail.com>
Date: Mon, 24 Oct 2022 10:29:41 +0530
Subject: [PATCH 2/2] Add count-good-nodes-in-binary-tree

---
 .../count-good-nodes-in-binary-tree.java      | 39 +++++++++++++++++++
 1 file changed, 39 insertions(+)
 create mode 100644 Hacktoberfest2022/count-good-nodes-in-binary-tree.java

diff --git a/Hacktoberfest2022/count-good-nodes-in-binary-tree.java b/Hacktoberfest2022/count-good-nodes-in-binary-tree.java
new file mode 100644
index 0000000..179c0d0
--- /dev/null
+++ b/Hacktoberfest2022/count-good-nodes-in-binary-tree.java
@@ -0,0 +1,39 @@
+/*
+Author: Kabir Singh Mehrok <https://github.com/KabirSinghMehrok>
+
+Title: Count Good Nodes in Binary Tree
+Description: Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.
+*/
+
+
+public class TreeNode {
+  int val;
+  TreeNode left;
+  TreeNode right;
+  TreeNode() {}
+  TreeNode(int val) { this.val = val; }
+  TreeNode(int val, TreeNode left, TreeNode right) {
+    this.val = val;
+    this.left = left;
+    this.right = right;
+  }
+}
+ 
+class Solution {
+    public int goodNodes(TreeNode root) {
+        // dfs traversal to every node
+        // keep track of the max val along the path
+        // if given val > maxval, +1 and recurse, with updated maxval
+        // else, traverse
+        return recurse(root, root.val);
+    }
+    
+    public int recurse(TreeNode root, int maxval)
+    {
+        if (root == null) return 0;
+        if (root.val >= maxval) {
+            return 1 + recurse(root.right, root.val) + recurse(root.left, root.val);
+        }
+        else return recurse(root.right, maxval) + recurse(root.left, maxval);
+    }
+}
\ No newline at end of file