Merge pull request #1010 from Jeehay28/main

[Jeehay28] WEEK 10

Author: SamTheKorean

course-schedule/Jeehay28.js

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+/**
+ * @param {number} numCourses
+ * @param {number[][]} prerequisites
+ * @return {boolean}
+ */
+
+// ✅ Graph DFS (Depth-First Search) approach
+// Time Complexity: O(V + E), where V is the number of courses (numCourses) and E is the number of prerequisites (edges).
+// Space Complexity: O(V + E), where V is the number of courses and E is the number of prerequisites.
+
+var canFinish = function (numCourses, prerequisites) {
+  // prerequisites = [
+  // [1, 0], // Course 1 depends on Course 0
+  // [2, 1], // Course 2 depends on Course 1
+  // [3, 1], // Course 3 depends on Course 1
+  // [3, 2]  // Course 3 depends on Course 2
+  // ];
+
+  // graph = {
+  // 0: [1], // Course 0 is a prerequisite for Course 1
+  // 1: [2, 3], // Course 1 is a prerequisite for Courses 2 and 3
+  // 2: [3], // Course 2 is a prerequisite for Course 3
+  // 3: [] // Course 3 has no prerequisites
+  // };
+
+  // Build the graph
+  const graph = {};
+
+  // for(let i=0; i<numCourses; i++) {
+  //     graph[i] = []
+  // }
+
+  // Fill in the graph with prerequisites
+  for (const [course, prereq] of prerequisites) {
+    if (!graph[prereq]) {
+      graph[prereq] = [];
+    }
+    graph[prereq].push(course);
+  }
+
+  const visited = new Array(numCourses).fill(0);
+
+  const dfs = (course) => {
+    if (visited[course] === 1) return false; // cycle detected!
+    if (visited[course] === 2) return true; // already visited
+
+    visited[course] = 1;
+
+    // // Visit all the courses that depend on the current course
+    for (const nextCourse of graph[course] || []) {
+      if (!dfs(nextCourse)) {
+        return false; // cycle detected!
+      }
+    }
+
+    visited[course] = 2; // fully visited and return true
+    return true;
+  };
+
+  // Check for each course whether it's possible to finish it
+  for (let i = 0; i < numCourses; i++) {
+    if (!dfs(i)) {
+      return false; // cycle detected!
+    }
+  }
+
+  return true; // no cycles
+};
+

invert-binary-tree/Jeehay28.js

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+/**
+ * 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} root
+ * @return {TreeNode}
+ */
+
+// ✔️ Recursive Approach
+// Time Complexity: O(N), N = Total number of nodes (each node is processed once)
+// Space Complexity: O(H), H = Height of the tree (due to recursion stack depth)
+
+var invertTree = function (root) {
+  if (!root) return null;
+
+  [root.left, root.right] = [root.right, root.left];
+
+  invertTree(root.left);
+  invertTree(root.right);
+
+  return root;
+};
+
+/**
+ * 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} root
+ * @return {TreeNode}
+ */
+
+// ✔️ Stack → DFS approach
+// Time Complexity: O(N), N = Total number of nodes (each node is processed once)
+// Space Complexity: O(H), H = Height of the tree (due to recursion stack depth)
+
+// var invertTree = function (root) {
+//   let stack = [root];
+
+//   while (stack.length > 0) {
+//     const node = stack.pop();
+//     if (!node) continue;
+
+//     [node.left, node.right] = [node.right, node.left];
+//     stack.push(node.left);
+//     stack.push(node.right);
+//   }
+//   return root;
+// };
+
+
+/**
+ * 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} root
+ * @return {TreeNode}
+ */
+
+// ✔️ Queue → BFS
+// Time Complexity: O(N), N = Total number of nodes (each node is processed once)
+// Space Complexity: O(W), W = Maximum width of the tree
+// var invertTree = function (root) {
+//   let queue = [root];
+
+//   while (queue.length > 0) {
+//     const node = queue.shift();
+//     if (!node) continue;
+
+//     [node.left, node.right] = [node.right, node.left];
+//     queue.push(node.left);
+//     queue.push(node.right);
+//   }
+//   return root;
+// };
+
+

jump-game/Jeehay28.js

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+// 🚀 Greedy approach: much more efficient than the recursive approach
+// Time Complexity: O(n)
+// Space Complexity: O(1), No extra memory used
+
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var canJump = function (nums) {
+  // ➡️ The farthest position you can reach from any of the previous indices you have already visited.
+  let farthest = 0;
+
+  for (let i = 0; i < nums.length; i++) {
+    // You cannot reach this position even with your farthest jump value.
+    if (i > farthest) return false;
+
+    // Compare current maximum jump with the previous maximum.
+    farthest = Math.max(farthest, nums[i] + i);
+
+    // Check if you can reach the last index with the current farthest jump.
+    if (farthest >= nums.length - 1) return true;
+  }
+  return false;
+};
+
+
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+
+// Time Complexity: O(n)
+// Space Complexity: O(n)
+var canJump = function (nums) {
+  let lastIndex = nums.length - 1;
+
+  // Initialize memoization array to track visited indices
+  let memo = new Array(nums.length).fill(undefined);
+
+  const dfs = (i) => {
+    // Base case: if we've reached or surpassed the last index, return true
+    if (i >= lastIndex) return true;
+
+    // If the current index has already been visited, return the stored result
+    if (memo[i] !== undefined) return memo[i];
+
+    // Calculate the farthest position that can be reached from the current index
+    let maxJump = Math.min(nums[i] + i, lastIndex);
+
+    for (let j = i + 1; j <= maxJump; j++) {
+      if (dfs(j)) {
+        memo[i] = true;
+        return true;
+      }
+    }
+
+    memo[i] = false;
+    return false;
+  };
+
+  return dfs(0);
+};
+
+
+// 🌀 recursive approach
+// ⚠️ Time Complexity: O(2^n) - Exponential due to recursive branching without memoization
+// 🔵 Space Complexity: O(n) - Recursive call stack depth
+
+/**
+ * Check if you can jump to the last index from the first index.
+ * @param {number[]} nums - Array where nums[i] is the max jump length from position i.
+ * @return {boolean} True if you can reach the last index, otherwise false.
+ */
+
+// var canJump = function (nums) {
+//   const dfs = (start) => {
+//     // Base Case: Reached the last index
+//     if (start === nums.length - 1) {
+//       return true;
+//     }
+
+//     // Recursive Case: Try all possible jumps
+//     for (let i = 1; i <= nums[start]; i++) {
+//       // Example with nums = [2, 3, 1, 1, 4]:
+//       // start = 1, nums[1] = 3 (can jump 1, 2, or 3 steps)
+//       // Possible calls:
+//       //   dfs(1 + 1) -> check from index 2
+//       //   dfs(1 + 2) -> check from index 3
+//       //   dfs(1 + 3) -> reached the last index (success)
+
+//       if (dfs(start + i)) {
+//         return true;
+//       }
+//     }
+
+//     return false; // cannot reach the last index
+//   };
+
+//   return dfs(0);
+// };
+

merge-k-sorted-lists/Jeehay28.js

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+// Brute Force (Array Sorting) - Good for smaller cases
+// 🕒 Time Complexity: O(N log N), where N is the total number of nodes
+// 🗂️ Space Complexity: O(N)
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode[]} lists
+ * @return {ListNode}
+ */
+var mergeKLists = function (lists) {
+  const nums = [];
+
+  // ┌──────────────────────────────────────────────────────────────────┐
+  // │ Step 1: Extract values from all linked lists                     │
+  // │ We traverse each linked list and push node values into 'nums'.   │
+  // │ This flattens the K lists into a single array.                   │
+  // └──────────────────────────────────────────────────────────────────┘
+  for (let list of lists) {
+    while (list) {
+      nums.push(list.val);
+      list = list.next;
+    }
+  }
+
+  // ┌──────────────────────────────────────────────────────────────────┐
+  // │ Step 2: Sort the values                                          │
+  // │ JavaScript's default sort is lexicographical, so we use a custom │
+  // │ comparator to sort numbers correctly in ascending order.         │
+  // └──────────────────────────────────────────────────────────────────┘
+  nums.sort((a, b) => a - b);
+
+  // ┌──────────────────────────────────────────────────────────────────┐
+  // │ Step 3: Create a new sorted linked list                          │
+  // │ Initialize a dummy node, then iterate through the sorted array.  │
+  // │ For each value, create a new ListNode and append it to the list. │
+  // └──────────────────────────────────────────────────────────────────┘
+  let dummy = new ListNode(-1);
+  let current = dummy;
+
+  for (num of nums) {
+    current.next = new ListNode(num);
+    current = current.next;
+  }
+
+  // ┌──────────────────────────────────────────────────────────────────┐
+  // │ Step 4: Return the merged list                                   │
+  // │ We return dummy.next since dummy is a placeholder.               │
+  // └──────────────────────────────────────────────────────────────────┘
+  return dummy.next;
+};
+
+/**
+ * ┌──────────────────────────────────────────────────────────────────┐
+ * │ Time & Space Complexity                                          │
+ * ├──────────────────────────────────────────────────────────────────┤
+ * │ Operation             │ Complexity                               │
+ * ├──────────────────────────────────────────────────────────────────┤
+ * │ Extracting values     │ O(N) - N is the total number of nodes    │
+ * │ Sorting values        │ O(N log N) - JavaScript's sort uses Timsort │
+ * │ Building linked list  │ O(N) - Iterates through sorted array     │
+ * ├──────────────────────────────────────────────────────────────────┤
+ * │ Overall Time Complexity │ O(N log N)                             │
+ * ├──────────────────────────────────────────────────────────────────┤
+ * │ Space Complexity      │ O(N) - Storing all node values in array  │
+ * └──────────────────────────────────────────────────────────────────┘
+ */

search-in-rotated-sorted-array/Jeehay28.js

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+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+
+// ✅ Iterative Binary Search for Rotated Sorted Array
+// Time Complexity: O(log N)
+// Space Complexity: O(1)
+var search = function (nums, target) {
+  let left = 0;
+  let right = nums.length - 1;
+
+  while (left <= right) {
+    let pointer = Math.floor((left + right) / 2);
+
+    if (nums[pointer] === target) {
+      return pointer;
+    }
+
+    // Check if the left half is sorted
+    if (nums[left] <= nums[pointer]) {
+      // Target is in the sorted left half
+      if (nums[left] <= target && target < nums[pointer]) {
+        right = pointer - 1;
+      } else {
+        // Target is not in the left half, so search in the right half
+        left = pointer + 1;
+      }
+    } else {
+      // The right half must be sorted
+
+      // Target is in the sorted right half
+      if (nums[pointer] < target && target <= nums[right]) {
+        left = pointer + 1;
+      } else {
+        // Target is not in the right half, so search in the left half
+        right = pointer - 1;
+      }
+    }
+  }
+
+  return -1;
+};
+