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‎hints/count-connected-components.md

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<br>
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<details class="hint-accordion">
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<summary>Recommended Time & Space Complexity</summary>
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<p>
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You should aim for a solution as good or better than <code>O(V + E)</code> time and <code>O(V + E)</code> space, where <code>V</code> is the number vertices and <code>E</code> is the number of edges in the graph.
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<br>
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<summary>Hint 1</summary>
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Assume there are no edges initially, so we have <code>n</code> components, as there are that many nodes given. Now, we should add the given edges between the nodes. Can you think of an algorithm to add the edges between the nodes? Also, after adding an edge, how do you determine the number of components left?
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<summary>Hint 2</summary>
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We can use the Union-Find (DSU) algorithm to add the given edges. For simplicity, we use Union-Find by size, where we merge the smaller component into the larger one. The Union-Find algorithm inserts the edge only between two nodes from different components. It does not add the edge if the nodes are from the same component. How do you find the number of components after adding the edges? For example, consider that nodes <code>0</code> and <code>1</code> are not connected, so there are initially two components. After adding an edge between these nodes, they become part of the same component, leaving us with one component.
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<summary>Hint 3</summary>
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We create an object of the DSU and initialize the result variable <code>res = n</code>, which indicates that there are <code>n</code> components initially. We then iterate through the given edges. For each edge, we attempt to connect the nodes using the union function of the DSU. If the union is successful, we decrement <code>res</code>; otherwise, we continue. Finally, we return <code>res</code> as the number of components.
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‎hints/course-schedule-ii.md

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<summary>Recommended Time & Space Complexity</summary>
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You should aim for a solution with <code>O(V + E)</code> time and <code>O(V + E)</code> space, where <code>V</code> is the number of courses (nodes) and <code>E</code> is the number of prerequisites (edges).
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<summary>Hint 1</summary>
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Consider the problem as a graph where courses represent the nodes, and <code>prerequisite[i] = [a, b]</code> represents a directed edge from <code>a</code> to <code>b</code>. We need to determine whether the graph contains a cycle. Why? Because if there is a cycle, it is impossible to complete the courses involved in the cycle. Can you think of an algorithm to detect a cycle in a graph and also find the valid ordering if a cycle doesn't exist?
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<summary>Hint 2</summary>
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We can use DFS to detect a cycle in a graph. However, we also need to find the valid ordering of the courses, which can also be achieved using DFS. Alternatively, we can use the Topological Sort algorithm to find the valid ordering in this directed graph, where the graph must be acyclic to complete all the courses, and the prerequisite of a course acts as the parent node of that course. How would you implement this?
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<summary>Hint 3</summary>
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We compute the indegrees of all the nodes. Then, we perform a BFS starting from the nodes that have no parents (<code>indegree[node] == 0</code>). At each level, we traverse these nodes, decrement the indegree of their child nodes, and append those child nodes to the queue if their indegree becomes <code>0</code>. We only append nodes whose indegree is <code>0</code> or becomes <code>0</code> during the BFS to our result array. If the length of the result array is not equal to the number of courses, we return an empty array.
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‎hints/course-schedule.md

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<summary>Recommended Time & Space Complexity</summary>
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<p>
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You should aim for a solution with <code>O(V + E)</code> time and <code>O(V + E)</code> space, where <code>V</code> is the number of courses (nodes) and <code>E</code> is the number of prerequisites (edges).
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You should aim for a solution with <code>O(V + E)</code> time and <code>O(V + E)</code> space, where <code>V</code> is the number of courses (nodes) and <code>E</code> is the number of prerequisites (edges).
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‎hints/min-cost-to-connect-points.md

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<summary>Recommended Time & Space Complexity</summary>
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You should aim for a solution as good or better than <code>O((n^2)logn)</code> time and <code>O(n^2)</code> space, where <code>n</code> is the number of points.
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<summary>Hint 1</summary>
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Think of this problem as a graph, where the given points represent nodes. We need to connect these nodes into a single component by creating edges. Can you think of an advanced graph algorithm that can be used to connect all points into one component?
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<summary>Hint 2</summary>
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We use Kruskal's algorithm along with Union-Find (DSU) to connect nodes into components. The final component forms the minimum spanning tree (MST), where the edges between nodes are weighted by the Manhattan distance, and the total weight of the tree is minimized. How would you implement this?
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<summary>Hint 3</summary>
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We create the possible edges by iterating through every pair of points and calculating the weights as the Manhattan distance between them. Next, we sort the edges in ascending order based on their weights, as we aim to minimize the cost. Then, we traverse through these edges, connecting the nodes and adding the weight of the edge to the total cost if the edge is successfully added. The final result will be the minimum cost.
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‎hints/network-delay-time.md

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<summary>Recommended Time & Space Complexity</summary>
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You should aim for a solution as good or better than <code>O(ElogV)</code> time and <code>O(V + E)</code> space, where <code>E</code> is the number of edges and <code>V</code> is the number of vertices (nodes).
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<summary>Hint 1</summary>
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As we are given the source node and we need to find the minimum time to reach all nodes, this represents the shortest path from the source node to all nodes. Can you think of a standard algorithm to find the shortest path from a source to a destination? Maybe a heap-based algorithm is helpful.
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<summary>Hint 2</summary>
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We can use Dijkstra's algorithm to find the shortest path from a source to destination. We end up finding the shortest paths from the source to the nodes that we encounter in finding the destination. So, to find shortest path for all nodes from the source, we need to perform Dijkstra's algorithm until the heap in this algorithm becomes empty. How would you implement this?
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<summary>Hint 3</summary>
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We use a Min-Heap as we need to find the minimum time. We create an adjacency list for the given times (weighted edges). We also initialize an array <code>dist[]</code> of size <code>n</code> (number of nodes) which represents the distance from the source to all nodes, initialized with infinity. We put <code>dist[source] = 0</code>. Then we continue the algorithm. After the heap becomes empty, if we don't visit any node, we return <code>-1</code>; otherwise, we return the time.
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‎hints/reconstruct-flight-path.md

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<summary>Recommended Time & Space Complexity</summary>
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You should aim for a solution with <code>O(ElogE)</code> time and <code>O(E)</code> space, where <code>E</code> is the number of tickets (edges).
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<summary>Hint 1</summary>
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Consider this problem as a graph where airports are the nodes and tickets are the edges. Since we need to utilize all the tickets exactly once, can you think of an algorithm that visits every edge exactly once? Additionally, how do you ensure the smallest lexical order is maintained?
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<summary>Hint 2</summary>
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We build an adjacency list from the given tickets, which represent directed edges. We perform a DFS to construct the result, but we first sort the neighbors' list of each node to ensure the smallest lexical order. Why? Sorting guarantees that during DFS, we visit the node with the smallest lexical order first.
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<summary>Hint 3</summary>
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DFS would be a naive solution, as it takes <code>O(E * V)</code> time, where <code>E</code> is the number of tickets (edges) and <code>V</code> is the number of airports (nodes). In this approach, we traverse from the given source airport <code>JFK</code>, perform DFS by removing the neighbor, traversing it, and then reinserting it back. Can you think of a better way? Perhaps an advanced algorithm that incorporates DFS might be helpful?
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<summary>Hint 4</summary>
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We can use Hierholzer's algorithm, a modified DFS approach. Instead of appending the node to the result list immediately, we first visit all its neighbors. This results in a post-order traversal. After completing all the DFS calls, we reverse the path to obtain the final path, which is also called Euler's path.
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‎hints/redundant-connection.md

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<summary>Recommended Time & Space Complexity</summary>
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You should aim for a solution as good or better than <code>O(V + E)</code> time and <code>O(V + E)</code> space, where <code>V</code> is the number vertices and <code>E</code> is the number of edges in the graph.
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<summary>Hint 1</summary>
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There will be only one edge that creates the cycle in the given problem. Why? Because the graph is initially acyclic, and a cycle is formed only after adding one extra edge that was not present in the graph initially. Can you think of an algorithm that helps determine whether the current connecting edge forms a cycle? Perhaps a component-oriented algorithm?
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<summary>Hint 2</summary>
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We can use the Union-Find (DSU) algorithm to create the graph from the given edges. While connecting the edges, if we fail to connect any edge, it means this is the redundant edge, and we return it. How would you implement this?
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We create an instance of the DSU object and traverse through the given edges. For each edge, we attempt to connect the nodes using the union function. If the union function returns <code>false</code>, indicating that the current edge forms a cycle, we immediately return that edge.
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‎hints/swim-in-rising-water.md

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You should aim for a solution with <code>O((n^2)logn)</code> time and <code>O(n^2)</code> space, where <code>n</code> is the number of rows or columns of the square matrix.
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<summary>Hint 1</summary>
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Think of this problem as a graph where each cell represents a node. You can move from one cell to its adjacent cell if the time is greater than or equal to the adjacent cell's elevation. Note that swimming does not cost time, but you may need to wait at a cell for the time to reach the required elevation. What do you notice about the path from <code>(0, 0)</code> to <code>(n - 1, n - 1)</code>? Perhaps a greedy approach would be useful here.
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<summary>Hint 2</summary>
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We can observe that the maximum elevation value along the path determines the time taken for that path. Therefore, we need to find the path where the maximum elevation is minimized. Can you think of an algorithm to find such a path from the source <code>(0, 0)</code> to the destination <code>(n - 1, n - 1)</code>? Perhaps a shortest path algorithm could be useful here.
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We can use Dijkstra's algorithm. We initialize a Min-heap and a matrix with infinity. We run the algorithm starting from the source <code>(0, 0)</code>, and we track the maximum elevation encountered along the paths. This maximum elevation is used as the key for comparison in Dijkstra's algorithm. If we encounter the destination <code>(n - 1, n - 1)</code>, we return the maximum elevation of the path that reached the destination.
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‎hints/valid-tree.md

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<summary>Recommended Time & Space Complexity</summary>
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You should aim for a solution as good or better than <code>O(V + E)</code> time and <code>O(V + E)</code> space, where <code>V</code> is the number vertices and <code>E</code> is the number of edges in the graph.
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<summary>Hint 1</summary>
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According to the definition of a tree, a tree is an undirected graph with no cycles, all the nodes are connected as one component, and any two nodes have exactly one path. Can you think of a recursive algorithm to detect a cycle in the given graph and ensure it is a tree?
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We can use the Depth First Search (DFS) algorithm to detect a cycle in the graph. Since a tree has only one component, we can start the DFS from any node, say node <code>0</code>. During the traversal, we recursively visit its neighbors (children). If we encounter any already visited node that is not the parent of the current node, we return false as it indicates a cycle. How would you implement this?
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We start DFS from node <code>0</code>, assuming <code>-1</code> as its parent. We initialize a hash set <code>visit</code> to track the visited nodes in the graph. During the DFS, we first check if the current node is already in <code>visit</code>. If it is, we return false, detecting a cycle. Otherwise, we mark the node as visited and perform DFS on its neighbors, skipping the parent node to avoid revisiting it. After all DFS calls, if we have visited all nodes, we return true, as the graph is connected. Otherwise, we return false because a tree must contain all nodes.
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‎hints/word-ladder.md

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<summary>Recommended Time & Space Complexity</summary>
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You should aim for a solution with <code>O((m ^ 2) * n)</code> time and <code>O((m ^ 2) * n)</code> space, where <code>n</code> is the number of words and <code>m</code> is the length of the word.
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<summary>Hint 1</summary>
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Consider the given problem in terms of a graph, treating strings as nodes. Think of a way to build edges where two strings have an edge if they differ by a single character. A naive approach would be to consider each pair of strings and check whether an edge can be formed. Can you think of an efficient way? For example, consider a string <code>hot</code> and think about the strings that can be formed from it by changing a single letter.
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To efficiently build edges, consider transforming each word into intermediate states by replacing one character with a wildcard, like <code>*</code>. For example, the word <code>hot</code> can be transformed into <code>*ot</code>, <code>h*t</code>, and <code>ho*</code>. These intermediate states act as "parents" that connect words differing by one character. For instance, <code>*ot</code> can connect to words like <code>cot</code>. For each word in the list, generate all possible patterns by replacing each character with <code>*</code> and store the word as a child of these patterns. We can run a <code>BFS</code> starting from the <code>beginWord</code>, visiting other words while avoiding revisiting by using a hash set.
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When visiting a node during BFS, if the word matches the <code>endWord</code>, we immediately return <code>true</code>. Otherwise, we generate the pattern words that can be formed from the current word and attempt to visit the words connected to these pattern words. We add only unvisited words to the queue. If we exhaust all possibilities without finding the <code>endWord</code>, we return <code>false</code>.
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