Maximum Score of a Node Sequence

Hard

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Graphs

There is an undirected graph with n nodes, numbered from 0 to n - 1.

You are given a 0-indexed integer array scores of length n where scores[i] denotes the score of node i. You are also given a 2D integer array edges where edges[i] = [a_i, b_i] denotes that there exists an undirected edge connecting nodes a_i and b_i.

A node sequence is valid if it meets the following conditions:

There is an edge connecting every pair of adjacent nodes in the sequence.
No node appears more than once in the sequence.

The score of a node sequence is defined as the sum of the scores of the nodes in the sequence.

Return the maximum score of a valid node sequence with a length of 4. If no such sequence exists, return -1.

Example 1:

Input: scores = [5,2,9,8,4], edges = [[0,1],[1,2],[2,3],[0,2],[1,3],[2,4]]
Output: 24
Explanation: The figure above shows the graph and the chosen node sequence [0,1,2,3].
The score of the node sequence is 5 + 2 + 9 + 8 = 24.
It can be shown that no other node sequence has a score of more than 24.
Note that the sequences [3,1,2,0] and [1,0,2,3] are also valid and have a score of 24.
The sequence [0,3,2,4] is not valid since no edge connects nodes 0 and 3.

Example 2:

Input: scores = [9,20,6,4,11,12], edges = [[0,3],[5,3],[2,4],[1,3]]
Output: -1
Explanation: The figure above shows the graph.
There are no valid node sequences of length 4, so we return -1.

Constraints:

n == scores.length
4 <= n <= 5 * 10⁴
1 <= scores[i] <= 10⁸
0 <= edges.length <= 5 * 10⁴
edges[i].length == 2
0 <= a_i, b_i <= n - 1
a_i != b_i
There are no duplicate edges.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the node values (e.g., can they be negative, zero, or floating-point numbers)?
What are the constraints on the number of nodes and edges in the graph? Is the graph guaranteed to be connected?
If there are multiple node sequences with the same maximum score, should I return any one of them, or is there a specific preference (e.g., the lexicographically smallest sequence)?
Is the graph directed or undirected?
Can a node appear more than once in a valid sequence, or must each node be unique within the sequence?

Brute Force Solution

Approach

To find the best possible sequence of nodes, the brute force method explores absolutely every path. We consider every possible combination, calculate its score, and then compare all scores to find the maximum.

Here's how the algorithm would work step-by-step:

Start by picking a first node to begin a potential sequence.
From that first node, try every possible second node connected to it to extend the sequence.
Continue extending the sequence, at each step trying every possible next node that's connected to the current last node in the sequence.
Keep going until the sequence has the required number of nodes.
Calculate the total score of this particular sequence by summing the scores of the nodes.
Remember the calculated total score for this particular sequence.
Repeat these steps starting from a different first node, and trying all different possible combinations from there.
After exploring every possible sequence, compare all the total scores that were remembered.
The highest total score among all the sequences is the maximum score.

Code Implementation

def maximum_score_of_a_node_sequence_brute_force(scores, graph, sequence_length):
    maximum_score = 0

    for start_node in range(len(scores)):
        maximum_score = find_sequences(
            start_node,
            [start_node],
            scores,
            graph,
            sequence_length,
            maximum_score
        )

    return maximum_score

def find_sequences(
    current_node,
    current_sequence,
    scores,
    graph,
    sequence_length,
    maximum_score
):
    # If sequence is complete, calculate and update max score.
    if len(current_sequence) == sequence_length:
        current_score = 0
        for node in current_sequence:
            current_score += scores[node]
        maximum_score = max(maximum_score, current_score)
        return maximum_score

    # Explore all possible next nodes to extend sequence
    for neighbor_node in graph[current_node]:
        new_sequence = current_sequence + [neighbor_node]

        maximum_score = find_sequences(
            neighbor_node,
            new_sequence,
            scores,
            graph,
            sequence_length,
            maximum_score
        )

    # After exhausting the neighbors, return max score
    return maximum_score

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach explores every possible sequence of k nodes in a graph. If we have n nodes, in the worst case, we're essentially considering every combination of k nodes, starting at each node and exploring all possible paths of length k-1. This means for each of the n nodes, we potentially explore all possible k-length sequences stemming from it. Therefore, the number of operations is proportional to n multiplied by a factor that grows exponentially with k, leading to O(n^k) where k is the number of nodes in the sequence.

Space Complexity

O(K^N) – The brute force solution explores all possible sequences of length N. In the worst-case, where each node is connected to every other node, we create a tree of possibilities. The depth of this recursion tree is N (the required sequence length). At each level of the tree, we are branching out to K possible nodes (where K is the number of nodes in the graph), potentially resulting in K^N sequences. To keep track of each potential sequence during the exploration, the recursion stack will grow to hold intermediate states for each node in the sequence, and the number of such nodes is K^N in the worst case. This results in an overall auxiliary space complexity of O(K^N).

Optimal Solution

Approach

The goal is to find the best path through a network to maximize the total score, moving from node to node. The clever approach involves looking at each node and choosing the best possible next three nodes to visit. This avoids exploring every single possible path, saving a lot of time.

Here's how the algorithm would work step-by-step:

For each node in the network, identify all possible sets of exactly three neighboring nodes that you could visit next.
For each set of three neighboring nodes, calculate the total score you would get by adding the scores of all four nodes (the starting node plus the three neighbors).
Choose the set of three neighboring nodes that gives you the highest total score.
Record this highest score, as well as the nodes you followed to get there (the 'path').
Repeat this process, starting from every node in the network. Keep track of the best score found so far, and its associated path.
After examining every node as a starting point, the path with the highest recorded score is the solution.

Code Implementation

def maximum_score_of_a_node_sequence(scores, edges):
    number_of_nodes = len(scores)
    maximum_score = -1
    best_path = []

    for start_node in range(number_of_nodes):
        neighbors = []
        for edge_start, edge_end in edges:
            if edge_start == start_node:
                neighbors.append(edge_end)
            elif edge_end == start_node:
                neighbors.append(edge_start)

        # Need at least 3 neighbors to form a sequence.
        if len(neighbors) < 3:
            continue

        neighbor_sets = []

        #Generate all possible sets of 3 neighbors
        for i in range(len(neighbors)):
            for j in range(i + 1, len(neighbors)):
                for k in range(j + 1, len(neighbors)):
                    neighbor_sets.append((neighbors[i], neighbors[j], neighbors[k]))

        for neighbor_set in neighbor_sets:
            neighbor1, neighbor2, neighbor3 = neighbor_set
            current_score = scores[start_node] + scores[neighbor1] + scores[neighbor2] + scores[neighbor3]

            #Find the neighbor set that yields the maximum score.
            if current_score > maximum_score:

                maximum_score = current_score
                best_path = [start_node, neighbor1, neighbor2, neighbor3]

    return maximum_score

Big(O) Analysis

Time Complexity

O(n*d^3) – The algorithm iterates through each of the n nodes in the network. For each node, it identifies all possible sets of three neighboring nodes. In the worst-case scenario, each node could have 'd' neighbors, where 'd' is the maximum degree of any node in the graph. Choosing three neighbors from 'd' neighbors involves combinations, which can be represented as d choose 3, or d! / (3! * (d-3)!). This simplifies to d * (d-1) * (d-2) / 6, which is O(d^3). Since this operation is performed for each of the n nodes, the overall time complexity is O(n * d^3).

Space Complexity

O(N) – The algorithm iterates through each node (N nodes), and for each node, it identifies all possible sets of three neighboring nodes. Although the number of neighboring node combinations for each node is limited to a constant number (at most degree of the graph choose 3), the algorithm needs to store the highest score found so far and its associated path which can have a maximum size of N (the number of nodes). This is because in the worst-case scenario, the highest scoring path could include all the nodes in the network. Therefore, we need to allocate memory for the path of length N, leading to O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty graph	Return 0 immediately, as there are no nodes to form a sequence.
Graph with only one node	Return the node's value, as it's the maximum possible score.
Graph with two nodes and one edge	Return the sum of the two node values if an edge exists, otherwise return the larger of the two.
Graph with no edges	Return the sum of the four highest node values in the graph, assuming at least four nodes exist.
Cycles in the graph	The algorithm must avoid infinite loops by using a visited set or by limiting the depth of the search for node sequences.
Nodes with very large values that could lead to integer overflow	Use a data type that supports larger numbers (e.g., long) or consider using modulo arithmetic to prevent overflow.
Graph with negative node values	The algorithm must correctly handle negative values when calculating the maximum score of a node sequence, potentially needing to prioritize sequences with fewer nodes if all have positive values.
Highly connected graph with a large number of nodes	The algorithm must be optimized to avoid excessive computation; consider prioritizing exploration of nodes with high degrees or values, and use pruning techniques to avoid exploring unpromising paths.

Null or empty graph

How to Handle:

Return 0 immediately, as there are no nodes to form a sequence.

Graph with only one node

How to Handle:

Return the node's value, as it's the maximum possible score.

Graph with two nodes and one edge

How to Handle:

Return the sum of the two node values if an edge exists, otherwise return the larger of the two.

Graph with no edges

How to Handle:

Return the sum of the four highest node values in the graph, assuming at least four nodes exist.

Cycles in the graph

How to Handle:

The algorithm must avoid infinite loops by using a visited set or by limiting the depth of the search for node sequences.

Nodes with very large values that could lead to integer overflow

How to Handle:

Use a data type that supports larger numbers (e.g., long) or consider using modulo arithmetic to prevent overflow.

Graph with negative node values

How to Handle:

The algorithm must correctly handle negative values when calculating the maximum score of a node sequence, potentially needing to prioritize sequences with fewer nodes if all have positive values.

Highly connected graph with a large number of nodes

How to Handle:

The algorithm must be optimized to avoid excessive computation; consider prioritizing exploration of nodes with high degrees or values, and use pruning techniques to avoid exploring unpromising paths.