[Jeehay28] WEEK 12

non-overlapping-intervals/Jeehay28.js

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+// 🚀 Greedy Algorithm
+// ✅ Time Complexity: O(n log n), where n is the number of intervals
+// - Sorting the intervals: O(n log n)
+// - Iterating through intervals: O(n)
+
+// ✅ Space Complexity: O(1), No other data structures are used,
+
+/**
+ * @param {number[][]} intervals
+ * @return {number}
+ */
+var eraseOverlapIntervals = function (intervals) {
+  // ✅ Sorting by end time ensures that we keep intervals that finish the earliest, reducing the chances of overlap with the subsequent intervals.
+  // ❌ Sorting by start time would lead to a greedy choice too early, causing unnecessary removals.
+  intervals.sort((a, b) => a[1] - b[1]);
+
+  let removalCnt = 0;
+
+  let prevEnd = intervals[0][1];
+
+  for (let i = 1; i < intervals.length; i++) {
+    const [start, end] = intervals[i];
+
+    if (start < prevEnd) {
+      removalCnt += 1; // Increment removal count for an overlap
+    } else {
+      prevEnd = end;
+    }
+  }
+  return removalCnt;
+};
+

remove-nth-node-from-end-of-list/Jeehay28.js

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+// 🚀 Optimized Approach: Two-Pointer Method (One-Pass)
+// ✅ Time Complexity: O(N), where N is the number of nodes
+// ✅ Space Complexity: O(1)
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @param {number} n
+ * @return {ListNode}
+ */
+var removeNthFromEnd = function (head, n) {
+  let dummy = new ListNode(0, head);
+  let fast = dummy;
+  let slow = dummy;
+
+  for (let i = 0; i <= n; i++) {
+    fast = fast.next;
+  }
+
+  while (fast) {
+    fast = fast.next;
+    slow = slow.next;
+  }
+
+  slow.next = slow.next.next;
+
+  return dummy.next;
+};
+
+// ✅ Time Complexity: O(N), where N is the number of nodes
+// ✅ Space Complexity: O(1)
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @param {number} n
+ * @return {ListNode}
+ */
+// var removeNthFromEnd = function (head, n) {
+//   let length = 0;
+
+//   let node = head;
+
+//   // Compute the length of the linked list
+//   while (node) {
+//     length += 1;
+//     node = node.next;
+//   }
+
+//   // Create a dummy node pointing to head (helps handle edge cases)
+//   let dummy = new ListNode(0, head);
+//   node = dummy;
+
+//   // Move to the node just before the one to be removed
+//   for (let i = 0; i < length - n; i++) {
+//     node = node.next;
+//   }
+
+//   // Remove the nth node from the end by updating the next pointer
+//   node.next = node.next.next;
+
+//   // Return the modified linked list (excluding the dummy node)
+//   return dummy.next;
+// };

same-tree/Jeehay28.js

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+// 🚀 Iterative DFS (Stack)
+// ✅ Time Complexity: O(n), where n is the number of nodes in the tree
+// ✅ Space Complexity: O(n) (worst case), O(log n) (best case for balanced trees)
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {boolean}
+ */
+var isSameTree = function (p, q) {
+  let stack = [[p, q]];
+  while (stack.length > 0) {
+    const [p, q] = stack.pop();
+
+    if (p === null && q === null) continue;
+
+    if (p === null || q === null) return false;
+
+    if (p.val !== q.val) return false;
+
+    stack.push([p.left, q.left]);
+    stack.push([p.right, q.right]);
+  }
+
+  return true;
+};
+
+
+
+// 🚀 recursive approach
+// ✅ Time Complexity: O(n), where n is the number of nodes in the tree
+// ✅ Space Complexity: O(n) (worst case), O(log n) (best case for balanced trees)
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {boolean}
+ */
+// var isSameTree = function (p, q) {
+//   // Base case: If both trees are empty, they are the same
+//   if (p === null && q === null) return true;
+
+//   // If one of the trees is empty and the other is not, return false
+//   if (p === null || q === null) return false;
+
+//   // Compare the values of the current nodes
+//   if (p.val !== q.val) return false;
+
+//   // Recursively compare the left and right subtrees
+//   return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+// };
+
+

serialize-and-deserialize-binary-tree/Jeehay28.js

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+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+
+// ✅ Time Complexity: O(N), where N is the number of nodes
+// ✅ Space Complexity: O(N)
+
+/**
+ * Encodes a tree to a single string.
+ *
+ * @param {TreeNode} root
+ * @return {string}
+ */
+
+var serialize = function (root) {
+  // Serialization (Tree → String)
+  // Uses BFS (level-order traversal) to visit nodes level by level.
+  // Stores "null" for missing children to maintain structure.
+  // Output format: "1,2,3,null,null,4,5" (comma-separated values).
+
+  if (!root) return "";
+
+  let queue = [root];
+  let str = [];
+
+  while (queue.length > 0) {
+    let node = queue.shift();
+
+    if (node) {
+      str.push(node.val);
+      queue.push(node.left);
+      queue.push(node.right);
+    } else {
+      str.push("null");
+    }
+  }
+  return str.join(",");
+};
+
+// ✅ Time Complexity: O(N), where N is the number of nodes
+// ✅ Space Complexity: O(N)
+
+/**
+ * Decodes your encoded data to tree.
+ *
+ * @param {string} data
+ * @return {TreeNode}
+ */
+var deserialize = function (data) {
+  // Deserialization (String → Tree)
+  // Splits the serialized string into an array of values.
+  // Uses BFS (level-order traversal) to reconstruct the tree.
+
+  if (!data) return null;
+
+  let values = data.split(",");
+  let root = new TreeNode(parseInt(values[0])); // // Root node at index 0
+  let queue = [root];
+  let index = 1; // Start processing children from index 1
+
+  while (queue.length > 0) {
+    let node = queue.shift();
+
+    // Process left child (index points to left node value)
+    if (values[index] !== "null") {
+      node.left = new TreeNode(parseInt(values[index]));
+      queue.push(node.left);
+    }
+
+    index += 1; // Move to the next position
+
+    // Process right child (ensure index is within bounds)
+    if (index < values.length && values[index] !== "null") {
+      // Ensure we don't access an index out of bounds
+      node.right = new TreeNode(parseInt(values[index]));
+      queue.push(node.right);
+    }
+
+    index += 1; // Move to the next position
+  }
+
+  return root;
+};
+
+/**
+ * Your functions will be called as such:
+ * deserialize(serialize(root));
+ */