Maximum Sum of Edge Values in a Graph

1. Generate all n! permutations of the values {1, 2, ..., n}.
2. For each permutation, treat it as an assignment of values to nodes 0, 1, ..., n-1.
3. For a given assignment (permutation), calculate the total score by summing the products of values for all connected nodes (edges).
4. Keep track of the maximum score found across all permutations.
5. Return the overall maximum score.

This method explores the entire solution space exhaustively. The core of the algorithm is a function that can generate all permutations of a set of numbers. A common way to implement this is using a recursive backtracking algorithm.

Here's a conceptual outline of the code:

1class Solution {2    long maxScore = 0;3 4    public long maximumScore(int n, int[][] edges) {5        List<Integer> values = new ArrayList<>();6        for (int i = 1; i <= n; i++) {7            values.add(i);8        }9 10        // Adjacency list for the graph11        List<List<Integer>> adj = new ArrayList<>();12        for (int i = 0; i < n; i++) {13            adj.add(new ArrayList<>());14        }15        for (int[] edge : edges) {16            adj.get(edge[0]).add(edge[1]);17            adj.get(edge[1]).add(edge[0]);18        }19 20        // Generate all permutations and calculate score for each21        permute(new int[n], new boolean[n + 1], 0, n, edges);22 23        return maxScore;24    }25 26    private void permute(int[] assignment, boolean[] used, int k, int n, int[][] edges) {27        if (k == n) {28            // A full assignment is created, calculate its score29            long currentScore = 0;30            for (int[] edge : edges) {31                currentScore += (long) assignment[edge[0]] * assignment[edge[1]];32            }33            maxScore = Math.max(maxScore, currentScore);34            return;35        }36 37        for (int val = 1; val <= n; val++) {38            if (!used[val]) {39                assignment[k] = val;40                used[val] = true;41                permute(assignment, used, k + 1, n, edges);42                used[val] = false; // backtrack43            }44        }45    }46}

This code generates permutations of values and assigns them to nodes 0 to n-1 in order. For each complete permutation, it calculates the score and updates the maximum.

1. Recognize that the graph structure (path or cycle) allows for a dynamic programming approach, similar to the Traveling Salesperson Problem (TSP).
2. Determine the sequence of nodes along the path or cycle, let's say p_0, p_1, ..., p_{n-1}.
3. Define a DP state dp[mask][i][v] representing the maximum score for assigning values from a subset S (represented by bitmask mask) to the first |S| nodes of the path, ending with node p_{i-1} being assigned value v.
4. The transitions would involve trying to extend the path by one node and one value, calculating the new score.
5. The state space would be n * n * 2^n, which is too large. A more standard TSP-like DP state would be dp[mask][i], the max score for a path of nodes in mask ending at node i, using values {1, ..., popcount(mask)}. This is still complex to formulate correctly and is exponential.

Let's try to formulate a DP. First, we find the node sequence of the path or cycle, p_0, p_1, ..., p_{n-1}. A possible DP state could be dp[i][mask], representing the maximum score for an assignment on the first i nodes of the path (p_0 to p_{i-1}), using the set of values represented by the bitmask mask.

To compute dp[i][mask], we would need to know which value was assigned to node p_{i-1} to calculate the edge product v(p_{i-2}) * v(p_{i-1}). This suggests the value assigned to the last node must be part of the state.

Let's define dp[i][v][mask] as the maximum score for the subproblem on the first i nodes (p_0, ..., p_{i-1}), where the values used are represented by mask, and node p_{i-1} is assigned value v.

dp[i][v][mask] = max_{u in mask, u != v} (dp[i-1][u][mask \setminus \{v\}] + u * v)

• i: index of the node in the path sequence (from 1 to n).
• v: value assigned to node p_{i-1} (from 1 to n).
• mask: a bitmask representing the set of values used so far.

The state space is O(n * n * 2^n), which is too large. Even with optimizations, this approach remains exponential and cannot solve the problem for the given constraints.

1. Build Graph and Find Degrees: Construct an adjacency list and an array to store the degree of each node. This takes O(n + m) time.
2. Identify Graph Structure: Check the degrees. If there are two nodes with degree 1, the graph is a path. If all nodes have degree 2, it's a cycle. This takes O(n).
3. Determine Node Sequence: Perform a traversal (like DFS) starting from an endpoint (for a path) or any node (for a cycle) to get the ordered sequence of nodes. This takes O(n).
4. Generate Optimal Value Sequence:
   • For a cycle: The values to assign are {1, 2, ..., n}. Generate an alternating sequence by taking all odd numbers in increasing order, followed by all even numbers in decreasing order (e.g., for n=6: 1, 3, 5, 6, 4, 2).
   • For a path: The n-2 inner nodes (degree 2) get values from {3, ..., n}, and the 2 endpoints (degree 1) get {1, 2}. First, generate the alternating sequence P for the inner values {3, ..., n}. Then, assign 1 and 2 to the endpoints based on the first and last values of P to maximize the terminal products.
5. Calculate Score: With the ordered node sequence and the optimal value sequence, map values to nodes and compute the total sum of edge products. This takes O(n).

This approach is linear in time and space, making it highly efficient.

Algorithm Steps:

1. Graph Representation: Build an adjacency list and a degree array from the edges input.
2. Structure Identification: Find a node with degree 1. If one exists, it's a path; start the traversal from there. Otherwise, it's a cycle; start from node 0.
3. Node Traversal: Create an ordered list of nodes (nodeSequence) by traversing the path/cycle.
4. Value Assignment Strategy:
   • If Cycle: Generate the alternating value sequence for {1, ..., n}. The total score is the sum of products of adjacent elements in this sequence, including the connection between the last and first elements.
   • If Path: a. Generate the alternating value sequence (innerValues) for the numbers {3, ..., n}. b. Let the first and last elements of innerValues be p_first and p_last. To maximize 1*p_x + 2*p_y, we should pair the larger of {1,2} with the larger of {p_first, p_last}. However, the problem is to decide which end of the path gets which value. The optimal choice is to pair the smaller endpoint value (1) with the end of the inner sequence that has the smaller value, and the larger endpoint value (2) with the end that has the larger value. Let's say p_first < p_last. The final value sequence will be 1, followed by innerValues, then 2. c. Calculate the sum of products for this final sequence.

Code Snippet:

1import java.util.*;2 3class Solution {4    public long maximumScore(int n, int[][] edges) {5        List<List<Integer>> adj = new ArrayList<>();6        for (int i = 0; i < n; i++) adj.add(new ArrayList<>());7        int[] degree = new int[n];8 9        for (int[] edge : edges) {10            adj.get(edge[0]).add(edge[1]);11            adj.get(edge[1]).add(edge[0]);12            degree[edge[0]]++;13            degree[edge[1]]++;14        }15 16        boolean isCycle = true;17        int startNode = 0;18        for (int i = 0; i < n; i++) {19            if (degree[i] == 1) {20                isCycle = false;21                startNode = i;22                break;23            }24        }25 26        List<Long> valueSequence;27        if (isCycle) {28            valueSequence = generateAlternatingSequence(1, n);29        } else {30            List<Long> innerValues = generateAlternatingSequence(3, n);31            long firstInner = innerValues.get(0);32            long lastInner = innerValues.get(innerValues.size() - 1);33            34            valueSequence = new ArrayList<>();35            if (firstInner < lastInner) {36                valueSequence.add(1L);37                valueSequence.addAll(innerValues);38                valueSequence.add(2L);39            } else {40                valueSequence.add(2L);41                valueSequence.addAll(innerValues);42                valueSequence.add(1L);43            }44        }45 46        long totalScore = 0;47        for (int i = 0; i < n - 1; i++) {48            totalScore += valueSequence.get(i) * valueSequence.get(i + 1);49        }50        if (isCycle) {51            totalScore += valueSequence.get(n - 1) * valueSequence.get(0);52        }53 54        return totalScore;55    }56 57    private List<Long> generateAlternatingSequence(long start, long end) {58        if (start > end) return new ArrayList<>();59        List<Long> smallHalf = new ArrayList<>();60        List<Long> largeHalf = new ArrayList<>();61        for (long i = start; i <= end; i++) {62            if ((i - start + 1) % 2 != 0) {63                smallHalf.add(i);64            } else {65                largeHalf.add(i);66            }67        }68        Collections.reverse(largeHalf);69        List<Long> result = new ArrayList<>(smallHalf);70        result.addAll(largeHalf);71        return result;72    }73}