Sum of Distances in Tree

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from collections import defaultdict

class Solution:
    def sumOfDistancesInTree(self, n: int, edges: List[List[int]]) -> List[int]:
        graph = defaultdict(set)
        count = [1] * n  # Initialize count with 1s because a node is part of its own subtree
        answer = [0] * n  # Initialize answer with 0s

        # Build the graph
        for u, v in edges:
            graph[u].add(v)
            graph[v].add(u)

        # Postorder DFS to calculate count and initial answer
        def dfs(node = 0, parent = None):
            for child in graph[node]:
                if child != parent:
                    dfs(child, node)
                    count[node] += count[child]
                    answer[node] += answer[child] + count[child]

        # Preorder DFS to update answer
        def dfs2(node = 0, parent = None):
            for child in graph[node]:
                if child != parent:
                    answer[child] = answer[node] - count[child] + n - count[child]
                    dfs2(child, node)

        dfs()
        dfs2()
        return answer

Problem Classification

This problem falls under the domain of Graph Theory, which is a subfield of Computer Science and Mathematics. Specifically, it deals with undirected trees, which are a type of graph.

Here are the ‘What’ components of the problem:

  1. You are given an undirected, connected tree with ’n’ nodes labeled from 0 to n - 1 and n - 1 edges.
  2. The edges of the tree are provided in an array where each entry [ai, bi] represents an edge between nodes ai and bi.
  3. The task is to calculate the sum of the distances from each node to all other nodes in the tree and return this information in an array.

We can further classify the problem into the following categories:

  1. Graph Traversal: Since we’re dealing with a tree (a type of graph), one of the core parts of the problem is likely to involve traversing the tree. This could be done using methods such as depth-first search (DFS) or breadth-first search (BFS).

  2. Distance Calculation: The problem specifically asks for the sum of distances from each node to all others, so it involves some form of distance calculation within a tree.

  3. Data Structures: The problem makes use of arrays to represent the edges of the graph and the output. It may also be beneficial to use other data structures such as lists or dictionaries during the solution process.

  4. Optimization: Given the constraints of the problem (up to 3 * 10^4 nodes), an efficient solution is required. Simple brute force approaches may not work within the time limits, so this problem may also involve some form of optimization to solve effectively.

Constraints

Here are some characteristics and conditions from the problem statement that can be used to our advantage when solving the problem:

  1. Tree Structure: The problem defines the graph as a tree. In a tree, any two nodes are connected by exactly one path, meaning we don’t have to worry about multiple paths between nodes. This greatly simplifies the process of calculating distances.

  2. Undirected Edges: The edges in the tree are undirected. This means that the distance from node A to node B is the same as the distance from node B to node A, which might help us avoid some redundant calculations.

  3. Connected Nodes: The problem specifies that the tree is connected. This means there’s a path from any node to any other node, so we don’t have to check for disconnected nodes.

  4. Number of Edges: In a tree with n nodes, there are exactly n - 1 edges. This relationship can help with creating the tree from the given edges.

  5. Edge List Representation: The graph is represented as an edge list. Knowing how the graph is represented can help in understanding how to traverse it.

  6. Distance Calculation: We are interested in the sum of distances to all other nodes, not the individual distances. Therefore, once we have found the distance from a node to all other nodes, we can simply sum them.

  7. Constraints: The constraints of the problem (1 <= n <= 3 * 10^4) indicate that the solution needs to be more efficient than O(n^2), as a quadratic solution could potentially perform poorly for large inputs.

  8. Output: The output needs to be an array of length n, so a place to store the output is predefined and can be utilized right away.

Thought Process

The problem is asking for the sum of distances between each node and all other nodes in an undirected tree. This seems like a complex problem initially as it involves both the concepts of tree traversal and distance calculation. However, the problem can be broken down into simpler steps.

Looking at the problem, the first cue is that we have a tree data structure, which suggests a need for tree traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS).

Secondly, we are looking for distances between nodes, which hints at using the properties of trees, specifically parent-child relationships and subtree sizes.

Finally, the problem asks for the sum of distances for each node, implying that our solution needs to handle updates to all nodes.

With these insights, a reasonable approach would be to first represent the tree as a graph and perform a DFS to gather necessary information. This can then be followed by another DFS to update the answer for each node based on its parent’s answer.

Here’s a Python solution following this approach:

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from collections import defaultdict

class Solution:
    def sumOfDistancesInTree(self, n: int, edges: List[List[int]]) -> List[int]:
        graph = defaultdict(set)
        count = [1] * n  # Initialize count with 1s because a node is part of its own subtree
        answer = [0] * n  # Initialize answer with 0s

        # Build the graph
        for u, v in edges:
            graph[u].add(v)
            graph[v].add(u)

        # Postorder DFS to calculate count and initial answer
        def dfs(node = 0, parent = None):
            for child in graph[node]:
                if child != parent:
                    dfs(child, node)
                    count[node] += count[child]
                    answer[node] += answer[child] + count[child]

        # Preorder DFS to update answer
        def dfs2(node = 0, parent = None):
            for child in graph[node]:
                if child != parent:
                    answer[child] = answer[node] - count[child] + n - count[child]
                    dfs2(child, node)

        dfs()
        dfs2()
        return answer

This code first constructs a graph from the edges, then performs two DFS traversals. The first DFS calculates the sizes of the subtrees and the initial answer for each node. The second DFS adjusts the answer for each node based on its parent’s answer.

Solution Approach and Analysis

This problem involves finding the sum of distances between a node and all other nodes in a tree. Solving this problem efficiently involves a two-pass Depth-First Search (DFS).

Here’s a step-by-step approach:

  1. Build the Graph: First, we need to represent our tree as a graph using adjacency lists. We can create a list of sets, where each set contains the neighbors of a node.

  2. Calculate Subtree Sizes and Initial Answer: Perform a DFS from the root (node 0). Keep track of the number of nodes in the subtree of each node (including the node itself) and calculate the sum of distances from the root to all other nodes. This is the initial answer for the root.

  3. Calculate Answers for All Nodes: Perform a second DFS from the root. This time, use the sizes of subtrees and the previously calculated initial answer to find the answer for all nodes. The insight here is that when we move from a parent to a child, the child is closer by one to all nodes in its subtree and farther by one from the remaining nodes. We adjust the answer for the child node accordingly.

Consider the tree as a company hierarchy. The first DFS is like the CEO (root) getting a report of how many employees (nodes) each manager (parent node) has under them (subtree sizes) and how many layers of management (distance) are between the CEO and each employee. The second DFS is like each manager, starting from the CEO, then passing the information down to their subordinates (child nodes), adjusting the report based on their level in the hierarchy.

As the number of nodes increases, the time complexity of the solution remains linear since each node and edge is visited only once. The memory complexity is also linear because of storing the graph, subtree sizes, and answers.

Consider the example: n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]. The first DFS gives us subtree sizes: [6, 1, 4, 1, 1, 1] and the initial answer for root: 8. In the second DFS, we adjust this answer for each child node to get: [8, 12, 6, 10, 10, 10]. So, node 1 is at total distance 12 from all other nodes, node 2 at total distance 6, and so on.

Language Agnostic Coding Drills

  1. Concepts and drills identification:

    • Basic Python and programming understanding: This includes variables, arrays, and loops. Understanding these basics is necessary before proceeding to more complex concepts.

    • Graph theory and representation: In this problem, we represent the tree as a graph using adjacency lists. This is an essential concept in many computer science problems, particularly those dealing with networks or relationships.

    • Depth-First Search (DFS): This is a method for traversing trees and graphs. In this problem, we use DFS twice, once to find the subtree size and initial answer, and once to adjust the answer based on the parent node.

    • Recursion: DFS is implemented via recursion, so understanding this concept is crucial. In the context of this problem, recursion is used to backtrack and accumulate values up the tree.

    • Counting and updating tree nodes: The crux of this problem lies in these operations. We use the properties of trees to count subtree sizes and calculate the initial answer, and then adjust this answer based on the parent’s answer.

  2. Concepts ordered by increasing difficulty:

    • Basic Python and programming understanding
    • Graph theory and representation
    • Recursion
    • Depth-First Search (DFS)
    • Counting and updating tree nodes

    The first two concepts are fundamental in computer science, while recursion is a more advanced topic that requires some mathematical thinking. DFS is a well-known algorithm in graph theory and is a common interview topic. The last concept is problem-specific and can be quite tricky to grasp without a deep understanding of trees.

  3. Approach from problem statement to final solution:

    • Given the problem statement, identify that the problem is about trees and distances between nodes.
    • Represent the tree as a graph using adjacency lists.
    • Perform a postorder DFS traversal of the tree to calculate the subtree sizes and initial answer for each node.
    • Perform a preorder DFS traversal of the tree to adjust the answer for each node based on its parent’s answer.
    • Return the answer array as the final solution.

    Each coding drill plays an integral part in solving the problem. The graph representation allows us to traverse the tree, while the DFS and recursion concepts enable the postorder and preorder traversals. Counting and updating tree nodes is the strategy we use to solve the problem.

Targeted Drills in Python

  1. Python-based coding drills for identified concepts:

    a. Basic Python and programming understanding:

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    # Variable assignment
a = 5
# Basic loop
for i in range(a):
    print(i)
# Array manipulation
arr = [1, 2, 3, 4, 5]
arr.append(6)  # Appends 6 to the end of the array

    b. Graph theory and representation (Adjacency list):

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    edges = [[0,1],[1,2],[1,3]]  # Sample edges
graph = {}
for u, v in edges:
    if u in graph:
        graph[u].append(v)
    else:
        graph[u] = [v]
    if v in graph:
        graph[v].append(u)
    else:
        graph[v] = [u]

    c. Recursion:

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    def factorial(n):
    if n == 1 or n == 0:
        return 1
    else:
        return n * factorial(n - 1)
print(factorial(5))

    d. Depth-First Search (DFS):

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    graph = {0: [1], 1: [0, 2, 3], 2: [1], 3: [1]}  # Sample graph
visited = [False]*len(graph)
def dfs(node):
    visited[node] = True
    for neighbor in graph[node]:
        if not visited[neighbor]:
            dfs(neighbor)
dfs(0)

    e. Counting and updating tree nodes:

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    graph = {0: [1], 1: [0, 2, 3], 2: [1], 3: [1]}  # Sample graph
subtree_size = [0]*len(graph)
def dfs(node, parent):
    subtree_size[node] = 1
    for child in graph[node]:
        if child != parent:
            dfs(child, node)
            subtree_size[node] += subtree_size[child]
dfs(0, -1)
print(subtree_size)

  2. Problem-specific drills:

    In this problem, we need to calculate the initial answer for each node and then adjust this answer. For the initial answer, we can extend the last drill and accumulate the total sum of distances in a DFS. For the adjustment, we can perform another DFS, this time starting from the root node and using the parent’s answer to update the current node’s answer.

  3. Assembling the drills:

    Start by creating the graph representation from the given edges. Then, perform a DFS from any node to compute the initial answer for each node and the size of each subtree. Afterwards, perform another DFS from the root node to adjust the answer for each node based on its parent’s answer. Finally, return the answer array as the final solution.

    Each of these drills directly contributes to the final solution: the graph representation allows the tree to be traversed, the first DFS computes initial values for each node, and the second DFS adjusts these values based on the parent’s answer.

Q&A

My solution is in principle the same with the O(n) solutions posted by other people. But their explain is more or less based on directed tree while my intuition is based directly on undirected tree, as mentioned in the original problem.

In an undirected, every node is simply “connected” with each other, and their roles are equivalent. It is wired to call any of them as parent/children/root nodes. Although you can select an arbitrary node as root and create a directed tree, you don’t need to.

Explanation:

Disconnecting any edge, say [i, j], would break the tree into two subtrees, one with node i and another with node j Use count[i,j] to represent the number of nodes in subtree with node i if edge [i, j] was disconnected. Then count[j,i] = N - count[i,j] For any adjacent node pair i and j: sum[j] = sum[i] - count[j,i] + count[i,j] Calculate sum of distance for one arbitrary node, and use the fomula to calculate the others. Steps

Create node representation of tree from given edge list, O(n) time & space Calculate count[i,j], dp, O(n) time & space Calculate sum[0], bfs, O(n) time & space Calculate sum for all other nodes, bfs, O(n) time & space O(n) time & space in total.

In the given text, the author explains their approach to solving a problem related to undirected trees, which they believe is slightly different from others that are primarily based on directed trees.

An undirected tree is a type of graph where any two vertices are connected by exactly one path. Each edge is a two-way connection, unlike in a directed tree where edges have a direction (from a parent node to a child node).

Here’s a step-by-step elaboration of the approach mentioned:

  1. Disconnecting Edges: In this undirected tree, if we disconnect any edge, for instance [i, j], it will divide the tree into two subtrees: one containing node ‘i’ and the other containing node ‘j’.

  2. Counting Nodes: The author uses count[i,j] to denote the number of nodes in the subtree with node ‘i’ if edge [i, j] was disconnected. Similarly, count[j,i] would represent the total nodes minus count[i,j], as it’s the subtree remaining after node ‘i’ and its associated nodes are separated.

  3. Calculating Distances: The relationship between the sums of distances for any adjacent node pair ‘i’ and ‘j’ is defined as sum[j] = sum[i] - count[j,i] + count[i,j]. This equation represents that the sum of distances for node ‘j’ is the sum of distances for node ‘i’, subtracting the nodes associated with ‘j’, and then adding the nodes associated with ‘i’.

  4. Calculate distance for one node: Start by calculating the sum of distances for one node (let’s say node 0 for convenience).

  5. Calculate distance for all other nodes: Using the relationship found in step 3, calculate the sums of distances for all the other nodes in the tree.

The entire process is comprised of four primary steps:

  1. Create a node representation of the tree from the given edge list.
  2. Calculate the count of nodes for each edge disconnection scenario.
  3. Calculate the sum of distances for an arbitrary node.
  4. Use the relationship to calculate the sum of distances for all other nodes.

Each of these steps takes linear time (O(n)), as they each require traversing all nodes once. Hence, the total time complexity of the solution is linear, making it efficient for large inputs. Similarly, the space complexity is also linear (O(n)) because of the data structures used to hold the tree representation, the count of nodes, and the sum of distances.

The approach mentioned seems to be a sound and efficient way to solve the problem. However, as with any coding solution, it does have potential points of concern:

  1. Graph Representation: The success of this approach depends heavily on the correct creation of the graph (tree) representation from the given edges list. An incorrect graph can lead to wrong results.

  2. Breadth-First Search (BFS): The calculation of the sums of distances for all other nodes using BFS could potentially become complicated due to the undirected nature of the tree. However, it is manageable with the correct implementation and careful handling of the traversal.

  3. Memory Usage: The space complexity of this solution is O(n), which might pose a problem when dealing with very large trees.

  4. Count Calculation: The calculation of the count of nodes (count[i,j] and count[j,i]) after disconnecting each edge must be accurate, as these counts are directly used in the sum calculation formula.

These points are less about flaws in the approach and more about potential pitfalls to be aware of during implementation. As long as these points are carefully managed, the approach should provide the correct solution. It is always advisable to thoroughly test the implemented solution with various test cases to ensure its correctness and efficiency.

Shouldn’t it be presented as graph instead binary tree?

Yes, you are absolutely correct. In the given problem statement, the term “tree” is referring to the concept from graph theory, where a tree is an undirected graph in which any two vertices are connected by exactly one path. This means that any connected graph without simple cycles is a tree. The problem specifies that we are dealing with an undirected tree, not a binary tree.

In computer science, when people refer to a “tree”, they often mean a “binary tree”, which is a tree data structure in which each node has at most two children, referred to as the left child and the right child. However, in this context, the tree is not necessarily a binary tree. The “tree” in this problem is a more general concept. Each node could have an arbitrary number of children.

So, you’re correct that this problem should be approached as a graph problem, not a binary tree problem. You would want to use graph algorithms and concepts to solve it.

Since tree is a “subset” of graph, it’s always good to be more specific about the input structures!

Absolutely! Being precise about the data structures in question is critical when it comes to solving problems efficiently and effectively in computer science.

A “tree” is indeed a specific type of “graph”, but there are many different kinds of trees (binary trees, AVL trees, B-trees, etc.) and each type has its own specific properties and suitable use cases. The type of tree (or graph) can often hint at the type of algorithms or data structures that might be useful in a problem.

In the case of this problem, we are working with an undirected tree, which is a connected graph with no cycles. This information guides us to think about solutions that involve traversing the tree in some way, such as depth-first or breadth-first traversal, and rules out solutions that might apply to a different type of tree (like binary search for a binary search tree).

So, you’re absolutely right. When dealing with a problem, it’s important to be specific about the data structures in play, as this can lead you to the right approach and ultimately, the solution.

First of all, it ain’t a binary tree since, 2 has adjacent node 3,4 and 5 (total 3 nodes). Second, the idea of saying tree but not a graph might be indicating the input doesn’t contain any cycles as a tree doesn’t have cycles but the graph might have. Also, all trees are graphs but all graphs are not trees + saying a tree also gives us a hint to use tree traversals (inorder, preorder, postorder), if needed to solve the problem, because we rarely use the thought of preorder/postorder/inorder if question says graph! Or in some cases saying tree also puts a twist as well (Not here ig).

Your observations are correct!

  1. The problem statement indeed refers to an undirected tree, not a binary tree. In the given tree, each node can have an arbitrary number of adjacent nodes, not limited to two as in binary trees. This is an important distinction and informs the approach to solving the problem.

  2. Correct, the mention of “tree” and not “graph” in the problem statement implies the absence of cycles. In computer science, a tree is an acyclic, connected graph. Every tree is a graph, but not every graph is a tree because graphs can have cycles.

  3. Absolutely! Referring to the structure as a “tree” can guide us towards specific tree-related algorithms, like tree traversal algorithms (DFS, BFS, preorder, postorder, inorder), as you mentioned. However, the use of these methods would depend on the specifics of the problem.

  4. Yes, sometimes problem statements use specific terms like “tree” to introduce an additional layer of complexity or a twist. But in this case, it is a direct representation of the structure we’re working with.

The clues you picked up from the problem statement about the nature of the tree (not binary, undirected, connected, acyclic) are valuable insights that could be leveraged to approach the problem effectively. Good job!

Some of the node edges in the input data are “upside down”. It would be easier to model the input data as a graph data structure.

Indeed, the orientation of the edges in the input data doesn’t matter in this context since we’re dealing with an undirected tree (which is a type of undirected graph). Regardless of the order in which nodes are presented in each edge (i.e., whether an edge is “upside down” or not), the relationship is the same: the nodes are connected.

As such, you’re correct that modeling the input data as a graph data structure would be appropriate and likely make the problem easier to solve. You can create an adjacency list or an adjacency matrix to represent the graph. An adjacency list is a list where the index represents the node and the value is a list of all the nodes it’s connected to.

Here is a simple Python example of how you might transform the input data into an adjacency list:

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from collections import defaultdict

def create_adjacency_list(n, edges):
    adjacency_list = defaultdict(list)
    for node1, node2 in edges:
        adjacency_list[node1].append(node2)
        adjacency_list[node2].append(node1)
    return adjacency_list

n = 6
edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
adjacency_list = create_adjacency_list(n, edges)
print(adjacency_list)

This outputs:

defaultdict(<class 'list'>, {0: [1, 2], 1: [0], 2: [0, 3, 4, 5], 3: [2], 4: [2], 5: [2]})

In this dictionary, each key-value pair represents a node and all nodes it is directly connected to. Now, you can use this data structure to efficiently navigate and analyze the graph.

There is a technique known as “rerooting dp”. If you google you will find plenty of material. It is a common idea, at least in the competitive programming world. It is also pretty easy to spot, since such problems typically ask “find some value for each root”. Thanks to @nskybytskyi for pointing out in one of the comments. This really helped me understand these types of problems. This is another problem that can be solved using rerooting dp technique.

Yes, you are correct. “Rerooting” or “Rerootable” Dynamic Programming is a powerful technique that comes in handy when dealing with tree-like structures, especially when the problem statement asks for something “for each root”. This technique is widely used in competitive programming.

Here’s a simplified explanation of how rerooting works:

The rerooting technique involves performing a Depth-First Search (DFS) twice. In the first DFS, you start from an arbitrary root and calculate some kind of value (the specifics depend on the problem at hand) for each node as if the current root is the actual root. Then, you perform the second DFS. This time, you’re modifying the results of the first DFS as you move the root from the parent to the child, computing the final answer for each node along the way.

To illustrate, in the context of the problem you posted earlier, where you’re asked to find the sum of distances from each node to all other nodes:

  1. In the first DFS, you calculate the sum of distances from the root (which can be any arbitrary node as the starting point) to all other nodes in the tree, as well as the size of each subtree.

  2. In the second DFS (the rerooting step), you use the previously calculated sums and sizes to calculate the sum of distances for each node when it becomes the root. This is done by subtracting the sum of distances in the subtree of the child node from the total sum and then adding the sum of distances outside of this subtree.

This approach helps avoid recalculating the same information multiple times, which leads to a more efficient algorithm. However, it requires careful thought and a good understanding of tree structures and dynamic programming to implement correctly.

This one really is incredibly hard. I tried to solve it without using the appropriate data structure (some kind of DFS graph) and it is impossible. Margin of error is extremely high and keeping track of so many different variables/logic constraints makes this one incredibly difficult. Pays off to be really smart/outside-the-box-thinker or just really have a tight grasp on algorithm analysis. Otherwise, it’s pointless to try to solve this one without some actual Computer Science formal knowledge. Kudos to all those who’ve managed to solve it. I could really use a hint on this one myself.

Your observation is correct; this problem is indeed complex and requires a strong understanding of tree-based algorithms and dynamic programming. Don’t worry if you’re finding it challenging; it’s a sophisticated problem that even experienced programmers might find difficult. Here’s a hint to help you understand how you might approach it:

In this problem, we’re asked to find the sum of distances from each node to all other nodes in the tree. This involves understanding the concept of tree structures, specifically the concept of “rooting” a tree and the idea of parent and child nodes.

One way to approach this problem is to first arbitrarily select a root. For instance, we could select node 0 as the root. From this root, we calculate the sum of distances to all other nodes. This can be done using Depth-First Search (DFS) - a fundamental algorithm for tree and graph traversal.

After we calculate the sum of distances from our initial root to all other nodes, the real challenge comes: how do we find the sum of distances from all the other nodes without having to recompute everything from scratch? This is where the idea of “rerooting” comes in handy.

When we move our root from a parent node to one of its child nodes, the distances to all other nodes change. Specifically, the distances to the nodes in the subtree of the child node get one step closer, while the distances to the nodes outside this subtree get one step farther away.

Using this insight, we can calculate the sum of distances from the new root based on the sum of distances from the old root, without having to traverse the entire tree again. This significantly reduces the computational complexity of our algorithm and allows us to solve the problem efficiently.

Remember, this is a high-level approach and the actual implementation will require careful thought and coding. Take your time to understand each step and gradually build up your solution. Don’t hesitate to ask if you need further clarifications!

I am unable to understand why we are updating the dist of child nodes before making a dfs call of it?? ans[child] = ans[s] - sub[child] + (n - sub[child]); dfs_min(child, adj, n); like this?

The reason we update the distance (or answer) for the child nodes before making a recursive DFS call on the child nodes is because of the nature of the problem and how the “rerooting” technique works.

The ans[child] = ans[s] - sub[child] + (n - sub[child]); line is a formula that updates the sum of distances when the root is moved from parent node s to its child node child. This line reflects the intuition that when we move the root from parent to its child:

  • All nodes in the subtree of the child node get one step closer, so we subtract sub[child] (the total number of nodes in the subtree of child) from ans[s] (the sum of distances from the parent node).
  • All nodes outside of this subtree get one step farther away, so we add n - sub[child] (the total number of nodes outside of the subtree of child) to the updated sum of distances.

Once we update the distance for child node, we then recursively apply DFS on this child node to continue the process for its own children.

It’s essential to note that this DFS function is not just used for exploring the graph/tree but also for dynamically updating the sum of distances when rerooting the tree. This technique significantly reduces the time complexity of the algorithm because we do not need to recalculate the sum of distances from scratch for each node.

I hope this clarifies the idea for you. It’s indeed a tricky concept, so don’t hesitate to ask more if anything is unclear.

Think of it as kind of presum on tree:

eg. if each node contributes 1: 1 / 1 1 /
1 1 ‘presum on tree’ would look like that:

         4
    /          \
   1         3.    (1+1) from children and 1 was here already
              /  \ 
            1    1

The concept you’re referring to is typically called a “prefix sum” on a tree, and it’s indeed a powerful technique for tree-related problems.

In the problem we’ve been discussing, each node doesn’t “contribute” a static value (like 1) but varies based on its distance to other nodes. However, we can still leverage a similar “prefix sum” idea:

  • First, we calculate the “base” sum of distances for one node (for instance, the node 0). This is done by a Depth-First Search (DFS) where we count the number of nodes (the “size”) in each subtree, and sum the distances to the root node.

  • Then, we update the sum of distances for each node’s children, similar to calculating prefix sums based on the parent node’s sum. The update rule is, when we move the root from parent to its child, all nodes in the child’s subtree get one step closer, and all nodes outside of this subtree get one step farther.

This way, instead of calculating the sum of distances from scratch for each node, we “reuse” the calculation from its parent node and make an update based on the structure of the tree. This technique significantly optimizes the algorithm from O(n^2) to O(n).

Keep in mind, this concept might be a bit hard to grasp at first, especially if you’re new to tree or graph problems. However, once you get the hang of it, you’ll find it a powerful tool for a wide range of problems.

`vector sumOfDistancesInTree(int n, vector& edges) { vectormatrix(n,vector(n,INT_MAX)); for(int i=0;i<edges.size();i++) { int a=edges[i][0]; int b=edges[i][1]; matrix[a][b]=1; matrix[b][a]=1;

} for(int i=0;i<matrix.size();i++) { for(int j=0;j<matrix.size();j++) { if(i==j) { matrix[i][j]=0; } } }

int mod=10000000007;

for(int k=0;k<n;k++)
{
    for(int i=0;i<matrix.size();i++)
    {
        for(int j=0;j<matrix.size();j++)
        {
            matrix[i][j]=min(matrix[i][j],matrix[i][k]%mod+matrix[k][j]%mod);
        }
    }
}
vector<int>v;
for(int i=0;i<matrix.size();i++)
    {
        int c=0;
        for(int j=0;j<matrix.size();j++)
        {
           c+=matrix[i][j];
        }
      v.push_back(c);
    }
return v;

} `

I have tried floyd warshall coz that is the first approach i got in the head straight away, but showed tle after 64 testcases out of around 72

Yes, the Floyd-Warshall algorithm is an algorithm for finding the shortest paths between all pairs of nodes in a graph, which does fit the problem description here. However, its time complexity is O(n^3), where n is the number of nodes in the graph. Given that n can be up to 3 * 10^4 in this problem, it’s not surprising that your solution got Time Limit Exceeded (TLE) after 64 test cases.

The issue with Floyd-Warshall here is that it computes the shortest paths between every pair of nodes, which is more than what’s needed for this problem. This problem only asks for the sum of distances from each node to all other nodes. That’s why you were able to pass 64 test cases: Floyd-Warshall is doing too much work.

A more efficient solution would involve depth-first search (DFS) and dynamic programming. I’ve described the process in previous posts, and the key idea is to calculate the sum of distances for one node first, then use this calculation to find the sums for the other nodes. This approach has a time complexity of O(n), which is significantly faster than Floyd-Warshall’s O(n^3).

Can anyone Guess why in comprehension function is running twice only while I am passing 6 as range > check at last line in return

your inline code…your inline code…from collections import defaultdict class Solution: def sumOfDistancesInTree(self, n: int, edges: List[List[int]]) -> List[int]:

graph = defaultdict(list)

for  n , v  in edges :
    graph[n].append(v)
    graph[v].append(n)

def travel(i):

    visited = set([i])
    queue = deque([(i,0)])
    output = 0

    while queue:
        curr , dis = queue.popleft()

        for nx in graph[curr]:
            if nx not in visited:
                visited.add(nx)
                queue.append((nx,dis+1))

                output += (dis+1)

    return output

 return [travel(i) for i in range(n)]

I think there might be a confusion regarding the function invocation here. The function travel(i) is not running twice - it is actually running n times due to the list comprehension at the last line return [travel(i) for i in range(n)]. This statement will create a new list where each element is the result of travel(i) for each i in range(n).

In your function, travel(i) calculates the sum of distances from node i to all other nodes in the graph. So you need to run it for all nodes from 0 to n-1 to get the sum of distances for all nodes, hence it is running n times not twice.

If there’s any unexpected behavior or output, it might be due to the state of the data or a logical error in the code. If you can provide more specific details about what’s happening versus what you expect to happen, I’d be happy to help you debug it.

I’m Getting Runtime error(AddressSanitizer:DEADLYSIGNAL) error in my code ,Please help!!

class Solution { void fun(vector & arr,vector & ans,int i , int j ,int n){ if(ans[i][j]!=-1) { return ; } if(i==j){ ans[i][j]=0; return ; } if(i>=n||j>=n) return ; int anss = n+1; for(int k=0 ; k<n ;k++){ if(arr[i][k]==1){ if(ans[i][j]==-1)fun(arr,ans,k,j,n); anss=min(anss,1+ans[k][i]); } } ans[i][j]=anss; ans[j][i]=anss; return ; } public: vector sumOfDistancesInTree(int n, vector& edges) { vector v(n+1,0); vector arr (n+1,v); for(int i = 0 ; i<edges.size();i++){ arr[edges[i][0]][edges[i][1]]=1; arr[edges[i][1]][edges[i][0]]=1; } vector vv(n+1,-1); vector ans (n+1,vv); vector ret(n,0); for(int i = 0 ; i< n ; i++){ for(int j = 0 ; j<n ; j++){ if(ans[i][j]==-1)fun(arr,ans,i,j,n); } } for(int i = 0 ; i< n ; i++){ for(int j = 0 ; j<n ; j++){ ret[i]+=arr[i][j]; } } return ret; } };

The Runtime error(AddressSanitizer:DEADLYSIGNAL) usually indicates that you have an issue in your code related to memory management, such as a buffer overflow, an out-of-bounds access, use-after-free, or other types of bugs.

From the first look at your code, it seems like you may be accessing memory out of the bounds of your arrays in the fun function. You’ve specified the size of your arrays as n+1, but then you’re accessing up to index n in your loops. Remember that array indexing starts at 0, so you should only be accessing up to index n-1. Going up to n means you’re trying to access memory that is out of bounds of your array, which can cause a runtime error.

Moreover, in the fun function, you have anss=min(anss,1+ans[k][i]). However, when k is n-1, ans[k][i] is accessed without a check to confirm ans[k][i] is not -1, which could lead to incorrect results. A check should be added before using ans[k][i].

In addition, the ans matrix should be n x n not n+1 x n+1 unless there’s a specific reason for the extra row and column. The same applies to the arr matrix.

Please note that this code seems to be written in C++, not Python. As such, there could be other C++-specific issues that aren’t immediately evident from a Python perspective. Please consider consulting a C++ reference or community for further assistance if needed.

Correct these issues and see if it helps to resolve the runtime error. If the error persists, you may need to use a debugger or add additional logging to your code to help isolate where the issue is occurring.

Keep in mind that this problem requires an understanding of graph theory and tree data structures, as well as the efficient usage of dynamic programming to avoid redundant calculations and achieve acceptable performance for large inputs. The current approach seems to be inefficient and may lead to timeouts for large inputs. You might need to revisit the problem-solving strategy for this problem.