Merge pull request #1619 from minji-go/week13

[minji-go] week 13 solutions

Author: minji-go

find-median-from-data-stream/minji-go.java

@@ -0,0 +1,45 @@
+/**
+ * <a href="https://leetcode.com/problems/find-median-from-data-stream/">week13-4. find-median-from-data-stream</a>
+ * <li>Description: Implement the MedianFinder class</li>
+ * <li>Topics: Two Pointers, Design, Sorting, Heap (Priority Queue), Data Stream</li>
+ * <li>Time Complexity: O(logN), Runtime 99ms   </li>
+ * <li>Space Complexity: O(N), Memory 63.68MB   </li>
+ */
+
+class MedianFinder {
+    PriorityQueue<Integer> head;
+    PriorityQueue<Integer> tail;
+
+    public MedianFinder() {
+        head = new PriorityQueue<>(Comparator.reverseOrder());
+        tail = new PriorityQueue<>();
+    }
+
+    public void addNum(int num) {
+        if(head.isEmpty() || num <= head.peek()) {
+            head.add(num);
+        } else {
+            tail.add(num);
+        }
+
+        if (head.size() > tail.size() + 1) {
+            tail.add(head.poll());
+        } else if (head.size() < tail.size()) {
+            head.add(tail.poll());
+        }
+    }
+
+    public double findMedian() {
+        if(head.size() == tail.size()){
+            return ((double)head.peek() + tail.peek()) / 2;
+        }
+        return head.peek();
+    }
+}
+
+/**
+ * Your MedianFinder object will be instantiated and called as such:
+ * MedianFinder obj = new MedianFinder();
+ * obj.addNum(num);
+ * double param_2 = obj.findMedian();
+ */

insert-interval/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/insert-interval/">week13-3. insert-interval</a>
+ * <li>Description: Return intervals after the insertion of 'new interval' given an array of non-overlapping intervals</li>
+ * <li>Topics: Array </li>
+ * <li>Time Complexity: O(N), Runtime 1ms       </li>
+ * <li>Space Complexity: O(N), Memory 45.02MB   </li>
+ */
+class Solution {
+    public int[][] insert(int[][] intervals, int[] newInterval) {
+        List<int[]> answer = new ArrayList<>();
+
+        int i = 0;
+        while (i < intervals.length && intervals[i][1] < newInterval[0]) {
+            answer.add(intervals[i++]);
+        }
+
+        while (i < intervals.length && newInterval[1] >= intervals[i][0]) {
+            newInterval[0] = Math.min(intervals[i][0], newInterval[0]);
+            newInterval[1] = Math.max(intervals[i][1], newInterval[1]);
+            i++;
+        }
+        answer.add(newInterval);
+
+        while (i < intervals.length) {
+            answer.add(intervals[i++]);
+        }
+
+        return answer.toArray(new int[answer.size()][]);
+    }
+}

kth-smallest-element-in-a-bst/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/kth-smallest-element-in-a-bst/">week13-3. kth-smallest-element-in-a-bst</a>
+ * <li>Description: Given the root of a binary search tree, and an integer k, return the kth smallest value</li>
+ * <li>Topics: Tree, Depth-First Search, Binary Search Tree, Binary Tree</li>
+ * <li>Time Complexity: O(N), Runtime 0ms   </li>
+ * <li>Space Complexity: O(H), Memory 44.5MB</li>
+ * <li>Note: If the BST is modified often (with frequent insertions or deletions), consider using an Augmented BST, Segment Tree, or TreeMap with additional metadata to efficiently support dynamic updates and k-th smallest queries in logarithmic time. </li>
+ */
+class Solution {
+    public int kthSmallest(TreeNode root, int k) {
+        return inorder(root, new AtomicInteger(k));
+    }
+
+    public Integer inorder(TreeNode node, AtomicInteger k) {
+        if (node == null) {
+            return null;
+        }
+
+        Integer value = inorder(node.left, k);
+        if (value != null) {
+            return value;
+        }
+
+        if (k.decrementAndGet() == 0) {
+            return node.val;
+        }
+
+        return inorder(node.right, k);
+    }
+
+}

lowest-common-ancestor-of-a-binary-search-tree/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/">week13-2. lowest-common-ancestor-of-a-binary-search-tree</a>
+ * <li>Description: find the lowest common ancestor (LCA) node of two given nodes in the BST. </li>
+ * <li>Topics: Tree, Depth-First Search, Binary Search Tree, Binary Tree </li>
+ * <li>Time Complexity: O(H), Runtime 5ms       </li>
+ * <li>Space Complexity: O(1), Memory 44.91MB   </li>
+ */
+
+class Solution {
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        TreeNode node = root;
+        while (node != null) {
+            if (p.val < node.val && q.val < node.val) {
+                node = node.left;
+            } else if (p.val > node.val && q.val > node.val) {
+                node = node.right;
+            } else {
+                return node;
+            }
+        }
+
+        throw new IllegalArgumentException();
+    }
+}