LeetCode

This section collects Hot100 greedy tutorials, focusing on which local state is enough, why it stays valid while scanning, and how that local choice proves the global answer. Recommended Reading Order LeetCode 121: Best Time to Buy and Sell Stock, derive one-transaction greedy from the historical minimum price LeetCode 55: Jump Game, use the farthest reachable range to decide reachability LeetCode 45: Jump Game II, upgrade reachability into minimum jump layers LeetCode 435: Non-overlapping Intervals, turn minimum removals into maximum kept intervals LeetCode 452: Minimum Number of Arrows, derive interval greedy from shared intersections

This section collects Hot100 Union-Find templates and problems, focusing on set representatives, path compression, merge conditions, connectivity checks, and connected component counting.

This section collects Hot100 Trie tutorials, focusing on node fields, child traversal, end markers, and loop invariants.

Problem Requirement Input / Output Input: nums, with 1 <= nums.length <= 10 Value range: -10 <= nums[i] <= 10 nums may contain duplicates Output: return all unique subsets Ordering: the result order does not matter, but the same value sequence may appear only once Example Input: nums = [1,2,2] Output: [[],[1],[1,2],[1,2,2],[2],[2,2]] This tutorial starts with a correct but wasteful version, then derives sorting plus layer-level deduplication. Start From Duplicate Branches in [1,2,2] The smallest example that exposes the issue is: ...

Subtitle / Summary The easiest way to get lost in LeetCode 124 is to make one recursive return value carry too much meaning. The stable design is to separate two roles: the recursion returns only the best single-side gain to the parent, while the full path through the current node is used to update the global maximum. Reading time: 12-15 min Tags: Hot100, binary tree, tree DP, DFS, postorder SEO keywords: Binary Tree Maximum Path Sum, tree DP, single-side gain, postorder, DFS, LeetCode 124 Meta description: Learn LeetCode 124 from the single-side gain invariant, global path update, and negative-branch pruning, with runnable multi-language implementations. A — Algorithm Problem Restatement A path in a binary tree is a sequence of nodes such that: ...

Subtitle / Summary LeetCode 199 is not really about visual imagination. It is about translating a viewpoint problem into a level problem. Once you realize that the right side view is simply the last node of each level, the problem becomes a standard breadth-first traversal. Reading time: 10-13 min Tags: Hot100, binary tree, BFS, level order, queue SEO keywords: Binary Tree Right Side View, BFS, level order, queue, right-first DFS, LeetCode 199 Meta description: Learn LeetCode 199 from the core equivalence “right side view = last node of each level”, with step-by-step derivation, engineering mappings, and runnable multi-language implementations. A — Algorithm Problem Restatement Given the root root of a binary tree, imagine standing on its right side and return the values of the nodes you can see from top to bottom. ...

Subtitle / Summary The key to LeetCode 105 is not memorizing that preorder and inorder can reconstruct a tree. It is understanding what each traversal contributes: preorder tells you the root, inorder tells you the left/right boundary. Once those roles are clear, the whole problem becomes a clean indexed divide-and-conquer. Reading time: 12-15 min Tags: Hot100, binary tree, divide and conquer, hash map, preorder SEO keywords: Construct Binary Tree from Preorder and Inorder Traversal, preorder, inorder, divide and conquer, hash map, LeetCode 105 Meta description: Learn LeetCode 105 from traversal roles, indexed recursion, and hash-map root lookup, with runnable multi-language implementations. A — Algorithm Problem Restatement Given the preorder traversal preorder and inorder traversal inorder of a binary tree, reconstruct the tree and return its root. ...

Subtitle / Summary The key to LeetCode 108 is not recursion by itself. It is noticing that the problem wants two things at the same time: BST ordering and height balance. Once you read those two constraints together, “pick the middle element as the root” stops being a trick and becomes the natural construction rule. Reading time: 11-14 min Tags: Hot100, binary tree, BST, divide and conquer, recursion SEO keywords: Convert Sorted Array to Binary Search Tree, BST, balanced BST, divide and conquer, recursion, LeetCode 108 Meta description: Learn LeetCode 108 from the midpoint-construction idea, with step-by-step derivation, engineering mappings, and runnable multi-language implementations. A — Algorithm Problem Restatement Given an integer array nums sorted in strictly increasing order, convert it into a height-balanced binary search tree. ...

Subtitle / Summary The real difficulty of LeetCode 114 is not “flattening” a tree. It is rewiring pointers without destroying the structure you still need. Once you notice that the final linked list must follow preorder order, reverse preorder with a prev pointer becomes a very stable in-place solution. Reading time: 12-15 min Tags: Hot100, binary tree, preorder, in-place, recursion SEO keywords: Flatten Binary Tree to Linked List, preorder, in-place, reverse preorder, LeetCode 114 Meta description: Learn LeetCode 114 from preorder order and reverse-preorder rewiring, with step-by-step derivation, engineering mappings, and runnable multi-language implementations. A — Algorithm Problem Restatement Given the root root of a binary tree, flatten the tree into a linked list in place. ...

Subtitle / Summary LeetCode 230 is not really about tree traversal mechanics alone. It is about turning BST order into a useful query. Once you see that the k-th smallest value is simply the k-th node visited in inorder traversal, the problem becomes a very stable counting task. Reading time: 11-14 min Tags: Hot100, binary tree, BST, inorder traversal, stack SEO keywords: Kth Smallest Element in a BST, BST, inorder traversal, stack, LeetCode 230 Meta description: Learn LeetCode 230 from BST inorder ordering, explicit-stack counting, and early-stop traversal, with runnable multi-language implementations. A — Algorithm Problem Restatement Given the root root of a binary search tree and an integer k, return the k-th smallest value in the tree. ...

Subtitle / Summary LeetCode 2 is grade-school addition translated into a linked-list workflow. The only persistent state is the carry, and the only structural work is appending one new digit per round. Reading time: 12-15 min Tags: Hot100, linked list, carry, simulation SEO keywords: Add Two Numbers, linked-list carry, reverse order digits, LeetCode 2, Hot100 Meta description: A derivation-first ACERS guide to LeetCode 2 with carry propagation, dummy node construction, engineering mapping, and runnable multi-language code. A - Algorithm (Problem and Algorithm) Problem Restatement You are given two non-empty linked lists l1 and l2. Each list stores a non-negative integer in reverse order, and each node contains one digit. Add the two integers and return the sum as a linked list in the same reverse order. ...

Subtitle / Summary LeetCode 146 is not a memorize-the-template problem. It is the core cache-design exercise: how to support fast lookup and fast recency updates at the same time. Reading time: 14-18 min Tags: Hot100, LRU, hash table, doubly linked list SEO keywords: LRU Cache, hash table + doubly linked list, O(1) get put, LeetCode 146, Hot100 Meta description: A derivation-first ACERS guide to LRU Cache using a hash table plus doubly linked list, with engineering scenarios, pitfalls, and runnable multi-language code. A - Algorithm (Problem and Algorithm) Problem Restatement Design a data structure LRUCache with: ...

Subtitle / Summary The hard part of LeetCode 19 is not deleting a node. It is locating the predecessor of the n-th node from the end in a singly linked list without walking backward. Reading time: 12-15 min Tags: Hot100, linked list, two pointers, dummy node SEO keywords: Remove Nth Node From End of List, one-pass two pointers, dummy node, LeetCode 19, Hot100 Meta description: A derivation-first ACERS explanation of LeetCode 19 with fixed-gap two pointers, dummy node boundary handling, engineering mapping, and runnable multi-language implementations. A - Algorithm (Problem and Algorithm) Problem Restatement Given the head of a linked list, remove the n-th node from the end of the list and return the head of the modified list. ...